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MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cards in the deck but we are βgiven thatβ
the card is not a face card.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cards in the deck but we are βgiven thatβ
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cards in the deck but we are βgiven thatβ
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cardsThere
in theare
deck
but we are βgiven thatβ
4 twos.
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cardsThere
in theare
deck
but we are βgiven thatβ
4 twos.
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
theβ¦and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cardsThere
in theare
deck
but we are βgiven thatβ
4 twos.
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
theβ¦and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
4
1
π π‘π€π πππππππ ππππ) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cardsThere
in theare
deck
but we are βgiven thatβ
4 twos.
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
theβ¦and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
4
1
π π‘π€π πππππππ ππππ) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cardsThere
in theare
deck
but we are βgiven thatβ
4 twos.
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
theβ¦and
12 face
cards
pictured
to the
right.space.
40 cards
in the
conditional
sample
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
4
1
π π‘π€π πππππππ ππππ) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
We are drawing a single card from a standard 52-card deck.
Find π π‘π€π πππππππ ππππ).
There are 52 cards in the deck but we are βgiven thatβ
the card is not a face card.
That means that the βgiven thatβ condition has
reduced the sample space from the entire deck
of cards to just the cards that are NOT one of
the 12 face cards pictured to the right.
So the conditional sample space consists of the
52 β 12= 40 cards that are NOT face cards.
4
1
π π‘π€π πππππππ ππππ) =
=
40 10
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
On the first draw, the
card could be a diamond
or not a diamond.
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
On the first draw, the
card could be a diamond
or not a diamond.
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume
2 cards are
There arethat
13 diamonds
& drawn from a standard 52-card deck.
If the cards are
drawn without replacement, find the
39 non-diamonds
in the
probability
of drawing
52 card deck
giving us a diamond followed by a non-diamond.
thesemeans
probabilities
That
that we need to compute this probability:
Diamond
1st
draw
NonDiamond
On the first draw, the
card could be a diamond
or not a diamond.
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st draw
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
Both of these
fractions can
be simplified.
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st draw
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
donβtfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
β1st is Diamondβ
and by
β2ndaisnon-diamond.
not Diamondβ
branch.
That means that we need to compute
this probability:
Diamond
1st draw
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
donβtfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
β1st is Diamondβ
and by
β2ndaisnon-diamond.
not Diamondβ
branch.
That means that we need to compute
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
donβtfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
probability of drawing a diamond
followed
β1st is Diamondβ
and by
β2ndaisnon-diamond.
not Diamondβ
branch.
That means that we need to compute
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
donβtfrom
needathe
entire tree
diagram.
theprobability
cards areofdrawn without
All replacement,
we really need isfind
the the
So,Ifthe
st is Diamondβ
ndaisnon-diamond.
probability
of
drawing
a
diamond
followed
by
β1
and
β2
not Diamondβ
drawing a nonbranch.
That
means
we
need
to
compute
this probability:
diamond
is 39 that
.
51
Diamond
1st
draw
NonDiamond
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional
Practice
Exercises
At thisProbability
point, we might
notice
that we
Assume that 2 cards are drawn
standard
52-card
deck.
donβtfrom
needathe
entire tree
diagram.
If the cards are drawn without
All replacement,
we really need isfind
the the
β¦whichof
reduces
probability
drawing a diamond
followed
β1st is Diamondβ
and by
β2ndaisnon-diamond.
not Diamondβ
13
to
17. that we need to compute
branch.
That means
this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
We are drawing without
replacement so after the
1st card was drawn
(which on this branch
was a diamond), there
are only 51 cards left (12
diamonds and 39 nondiamonds).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
πππππππ
πππππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
πππππππ
πππππππ
4 17
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
πππππππ
πππππππ
4 17
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
That means that we need to compute this probability:
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
without drawing the tree diagram at all.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn without replacement, find the
probability of drawing a diamond followed by a non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 14
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 14
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 β’ 13 17
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 β’ 13 17 =
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 β’ 13 17 = 13 68
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH
110
Secapplied
13.3 Conditional
Probability
Practice Exercises
We
could
have
the Product
Rule for Probabilities
directly
drawing
the from
tree diagram
at all. 52-card deck.
Assume that 2without
cards are
drawn
a standard
Assume that 2 cards are drawn from a standard 52-card deck.
IfIfππ the
are
without
find
the
1π π‘
π cards
1π π‘ ππ replacement,
πreplacement,
1π π‘ ππ π
2ππ
ππ drawn
πππ‘
2ππ ππ πππ‘
the
cards
are
drawn
without
find
π
πππ
=π
βπ
| the
probability
followed
aanon-diamond.
πππππππ of
πππππππ
πππππππ
πππππππaadiamond
πππππππ
probability
ofdrawing
drawing
diamond
followedby
by
non-diamond.
= 1 4 β’ 13 17 = 13 68
Diamond
1st
draw
NonDiamond
NonDiamond
1 13 13
1π π‘ ππ π
2ππ ππ πππ‘
π
πππ
= β
=
πππππππ
πππππππ
4 17 68
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
39
13
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
39
13
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced,39
there are39
still 52 cards 39
in the
52
1
39
13
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
39
52
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
1π π‘
π
2ππ ππ done
πππ‘ previously
themππ toπππ‘
what
we ππ
need
πLook at the calculations
πππ
=
π
β
π
|
39 probability
here.
We see that only
the
of3getting
a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
πππππππ
But
to
.
4
52 reduces
draw changes. Because the 1st card was replaced,39
there are39
still 52 cards 39
in the
52
1
39
13
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
39
52
Non-Diamond
Previous
calculations
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
1π π‘
π
2ππ ππ done
πππ‘ previously
themππ toπππ‘
what
we ππ
need
πLook at the calculations
πππ
=
π
β
π
|
39 probability
here.
We see that only
the
of3getting
a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
πππππππ
But
to
.
4
52 reduces
3
draw changes. Because the 1st card was replaced, there
are39
still 52 cards 39
in the
4
1
39
13
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
68
51
Diamond
3
4
Non-Diamond
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
aβ’ non-diamond
instead of .
= drawing
=
52
51
4
4
Diamond
Non-Diamond
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
calculate this new probability. We see that they are exactly the same EXCEPT,
because the 1st card is replaced before the 2nd draw, now the probability of
39
39
drawing a non-diamond next is instead of .
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
aβ’ non-diamond
3instead of 51.
= drawing
=
52
4
4
16
Diamond
Non-Diamond
110 the
Secsame
13.3 problem
Conditional
Probability
Practice
This MATH
is EXACTLY
except
we are drawing
withExercises
replacement.
Assume that 2 cards are drawn from a standard 52-card deck.
If the cards are drawn with replacement, find the probability
of drawing a diamond followed by a non-diamond.
Letβs examine the previous calculation and compare it to what we need to
So,they
the probability
of drawing
a
calculate this new probability. We see that
are exactly the
same EXCEPT,
because the 1st card is replaced before
the 2ndfollowed
draw, now
probability of
diamond
bythe
a39non-diamond
39
drawing a non-diamond nextwith
is replacement
instead of .is 3 16.
52
51
1π π‘ ππ π
1π π‘and
ππ πcompare 2ππ
ππ π
2ππ ππ done
πππ‘ previously
ππ πππ‘ 1π π‘
πLook at the calculations
πππ
=π
β π them to what
| we need
here.
We see that only
the probabilityπππππππ
of getting a non-diamond
on πππππππ
the second
πππππππ
πππππππ
πππππππ
draw changes. Because the 1st card was replaced, there are39
still 52 cards 39
in the
1
3
deck making the probability of
aβ’ non-diamond
3instead of 51.
= drawing
=
52
4
4
16
Diamond
Non-Diamond
3
16
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
Here we are βgiven thatβ the disk is pink.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
X X X
X X X X
Here we are βgiven thatβ the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
X X X
X X X X
Here we are βgiven thatβ the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
X X X
X X X X
Here we are βgiven thatβ the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
2
1
π π π‘ππ ππππ) =
=
10 5
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
You are to randomly pick one disk from a bag that contains
the disks shown below. Find π π π‘ππ ππππ).
X X X
X X X X
Here we are βgiven thatβ the disk is pink.
That means that we can eliminate all non-pink disks from the
conditional sample space.
2
1
π π π‘ππ ππππ) =
=
10 5
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
The hint reminds us that we can get a red and a green either by getting the
green first and the red second or by getting the red first and the green second.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
thedraw
first draw,
the probability that youOnwill
a
we ball?
can get either
red ball and a green
green, blue or red.
G
B
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
8
3
22
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
This reduces.
8
3
22
G
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
This reduces.
4
3
11
G
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110There
Sec 13.3
Conditional
are 22
balls in all.Probability Practice Exercises
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110
Sec
13.3areConditional
Practice Exercises
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
G
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
On
the second4
the probability that you will draw
a
draw, we can still11 B
red ball and a green ball? get either green,
blue or red. 3
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls
red
What
Noticeand
that3we
onlyballs.
need the
2 is
branches containing
bothwill
R and
G. a
the probability
that you
draw
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls
red
What
Noticeand
that3we
onlyballs.
need the
2 is
branches containing
bothwill
R and
G. a
the probability
that you
draw
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice
Exercises
Now
there
only 21 ballsProbability
left. Find the
probability
of drawing
G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
red after drawing green.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice
Exercises
Now
there
only 21 ballsProbability
left. Find the
probability
of drawing
G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
red after drawing green.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
3
21
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
This reduces.
4
11
B
3
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
3
21
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
This reduces.
4
11
B
3
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
1
7 R
G
4
11
Find the probability
of drawing
B
a green after drawing red.B
(Hint: There are 2 ways this can happen.)
3
R
22
G
Letβs use a tree diagram to
calculate this probability.
R
B
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
G
B
1
7 R
G
4
11
Find the probability
of drawing
B
a green after drawing red.B
(Hint: There are 2 ways this can happen.)
3
R
22
G
Letβs use a tree diagram to
calculate this probability.
R
B
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
7
B
R
G
B
R
G
B
R
MATH 110
Sec
13.3areConditional
Practice Exercises G
Now
there
only 21 ballsProbability
left.
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1 11
3 11
2
β
π π
πππ πΊ =
+
β
=
2 14
7
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
3 11
2
π π
πππ πΊ =
+
β
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
3 11
2
π π
πππ πΊ =
+
β
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
3 11
2
π π
πππ πΊ =
+
β
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
3 11
2
π π
πππ πΊ =
+
β
=
14
22 21
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
1
2
π π
πππ πΊ =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
1
2
π π
πππ πΊ =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
1
2
π π
πππ πΊ =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises G
Assume that you are drawing two
balls without replacement from an
urn containing 11 green balls, 8
blue balls and 3 red balls. What is
the probability that you will draw a
red ball and a green ball?
(Hint: There are 2 ways this can happen.)
Letβs use a tree diagram to
calculate this probability.
G
4
3
11
1
7
B
22
R
π π
πππ πΊ = π 1π π‘ πΊ πππ 2ππ π
+ π(1π π‘ π
πππ 2ππ πΊ)
1
1
2
π π
πππ πΊ =
+
=
14
14
14
=
B
R
G
B
R
G
B
R
1
7
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
π π π’π ππ 7 ππ 1π π‘ ππππ =
1
6
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
and π π π’π ππ 7 ππ 2ππ ππππ =
16
26
36
46
56
66
1
6
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
π π π’π ππ 7 ππ 1π π‘ ππππ = and π π π’π ππ 7 ππ 2ππ ππππ =
6
6
1 1
1
π π’π ππ 7 ππ
π π’π ππ 7
π π’π ππ 7
π
=π
βπ
= β =
6 6
36
πππ‘β πππππ
ππ 1π π‘ ππππ
ππ 2ππ ππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
π π π’π ππ 7 ππ 1π π‘ ππππ = and π π π’π ππ 7 ππ 2ππ ππππ =
6
6
1 1
1
π π’π ππ 7 ππ
π π’π ππ 7
π π’π ππ 7
π
=π
βπ
= β =
6 6
36
πππ‘β πππππ
ππ 1π π‘ ππππ
ππ 2ππ ππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
π π π’π ππ 7 ππ 1π π‘ ππππ = and π π π’π ππ 7 ππ 2ππ ππππ =
6
6
1 1
1
π π’π ππ 7 ππ
π π’π ππ 7
π π’π ππ 7
π
=π
βπ
= β =
6 6
36
πππ‘β πππππ
ππ 1π π‘ ππππ
ππ 2ππ ππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
π π π’π ππ 7 ππ 1π π‘ ππππ = and π π π’π ππ 7 ππ 2ππ ππππ =
6
6
1 1
1
π π’π ππ 7 ππ
π π’π ππ 7
π π’π ππ 7
π
=π
βπ
= β =
6 6
36
πππ‘β πππππ
ππ 1π π‘ ππππ
ππ 2ππ ππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
If we roll a pair of dice two times. What is the probability that
a total of 7 is rolled each time?
Letβs examine the 36 element, equally-likely sample space, S.
So, for each roll of the two dice:
6
1
π π π’π ππ 7 = =
36
6
The probability of getting a sum of 7 on the
2nd roll is not affected by what happened on
the 1st rollβ¦so the rolls are independent.
1
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
1
π π π’π ππ 7 ππ 1π π‘ ππππ = and π π π’π ππ 7 ππ 2ππ ππππ =
6
6
1 1
1
π π’π ππ 7 ππ
π π’π ππ 7
π π’π ππ 7
π
=π
βπ
= β =
6 6
36
πππ‘β πππππ
ππ 1π π‘ ππππ
ππ 2ππ ππππ
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q β© R),
associated with the tree diagram.
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q β© R),
associated with the tree diagram.
π π β© π
= π(π πππ π
)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q β© R),
associated with the tree diagram.
π π β© π
= π(π πππ π
)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q β© R),
associated with the tree diagram.
π π β© π
= π(π πππ π
)
π π β© π
= (0.4)(0.8)
R
M
0.4 N
S
R
S
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
Find the probability, P(Q β© R),
associated with the tree diagram.
π π β© π
= π(π πππ π
)
π π β© π
= (0.4)(0.8)
R
M
0.4 N
S
R
S
π π β© π
= 0.32
R
Q
S
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
It is clear that the outcome of the penny flip has no effect on the
probability of any outcome of the nickel flip.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the nickel
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
It is clear that the outcome of the penny flip has no effect on the
probability of any outcome of the nickel flip.
Events A and B are independent.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
Be careful! You are only flipping each coin once and both events
mention only the penny. If the penny comes up heads (and you are just
tossing it once), then the probability the it comes up tails is 0.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
A penny and a nickel are flipped.
Event A: Heads on the penny.
Event B: Tails on the penny
Are A and B dependent or are they independent events?
Events for which the occurrence of one event effects the probability of the
occurrence of the other event are said to be DEPENDENT EVENTS.
Events for which the occurrence of one event has no effect the probability of
the occurrence of the other event are said to be INDEPENDENT EVENTS.
Be careful! You are only flipping each coin once and both events
mention only the penny. If the penny comes up heads (and you are just
tossing it once), then the probability the it comes up tails is 0.
Events A and B are dependent.
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
π π πππ£πππ ππππ ππππππ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
π π πππ£πππ ππππ ππππππ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
π π πππ£πππ ππππ ππππππ) =
59
MATH 110 Sec 13.3 Conditional Probability Practice Exercises
The table shows the results of a restaurant survey. Find the
probability that the service was good, given that the meal was dinner.
MEALS
SERVICE SERVICE
GOOD
POOR TOTAL
Lunch
Dinner
28
42
22
17
50
59
TOTAL
70
39
109
42
π π πππ£πππ ππππ ππππππ) =
59