Transcript 16.4 Probability Problems Solved with Combinations

16.4 Probability Problems Solved
with
Combinations
16.4 Warm Up
Three cards are drawn from a well-shuffled standard deck of 52 cards,
one after the other and without replacement.
1. Find the probability of drawing
a) all clubs 13  12  11  11
b) no clubs 39  38  37  703
52 51 50
52 51 50
850
1700
c) exactly one club (hint: the club can occur on the 1st, 2nd or 3rd
drawing)  13 39 38  741
3 
 
52
51
50  1700

2. Evaluate:
a)
c)
13 C3
52 C3

11
850
13 C139 C2
52 C3

b)
39 C3
52 C3

703
1700
741
1700
3. Compare your answers to exercises 1 and 2. What do you notice?
They are the same!
4. Write and evaluate an expression using combinations to find the
probability of getting exactly 2 clubs. 13 C2 39 C1 117
52 C3

850
16.4: Solving Probability Problems w/ Combinations
Recall the definition of probability:
If E is an event from a sample space S of equally likely outcomes,
the probability of event E is:
P (E ) 
n(E ) # favorable

n(S )
# total
0  P(E)  1.
• Counting techniques can be used to determine the number of
favorable and total number.
• Combinations automatically handle situations where multiple cases
can occur (e.g., 3 cards with one ace: ANN, NAN & NNA).
Example: Three marbles are picked at random from a bag containing 4
red marbles and 5 white marbles. What is the probability that
1
4 C3 5 C0
a) all 3 are red
b) 2 red marbles 4 C2 5 C1  5

9 C3
c) 1 marble is red
9 C3
21
3+0=3
10
4 C15 C2

21
9 C3
d) no red marbles
4 C0 5 C3
9 C3
1+2=3
Notice that the number selected in the numerator is always 3.
14
2+1=3

5
42
0+3=3
Examples
1) Five cards are randomly chosen from a standard deck of 52 cards.
Find the probability that the following are chosen
a) all 4 aces 48 C14 C4
b) No aces
35,673
1
48 C5


52 C5
52 C5
54,145
c) exactly 4 diamonds
13 C4 39 C1
52 C5

54,145
d) four aces and one jack
4 C4 4 C1
143
13,328
52 C5

1
649,740
e) at least one ace
1 – P(no aces)  1
48 C5
52 C5

18,472
54,145
2) Three cards are dealt. Find the probability of getting either one ace or
two aces.
48
4 C148 C2
4 C2 48 C1
52 C3

52 C3

221
3) A carton contains 200 batteries, of which 5 are defective. If a random
sample of 5 batteries is chosen, what is the probability that at least
one is defective?
P(at least one defective) = 1 – P(no defective)  1 
195 C5
200 C5
 1  0.12  0.88
OR P(at least one defective) = P(1 def) + P(2 def) + P(3 def) + P(4 def) + P(5 def)
much more complicated…
4) Thirteen cards are dealt from a well shuffled standard card deck.
What is the probability of getting:
a) all cards from the same suit
b) 7 spades, 3 hearts, and 3 clubs
P(all any 1 suit) = P(all ♠) + P(all ♣) +
P(all ♥) + P(all ♦)
 13 C13

4
= 4·P(all ♠) 
 52 C13
13 C7 13 C3 13 C3
52 C13
 2.21X10 4

  6.30 X 10 12

c) all of the 12 face cards
12 C12 40 C1
52 C13
 6.20 X 1011
e) at least one diamond
P(at least one diamond) = 1 – P(no diamonds)  1 
39 C5
52 C5
 1  0.22  0.78
Should counting always be used?
6. A die is rolled twice
a) Find the probability that 2 sixes are rolled
2
 1
 1 
   
6
 36 
b) Find the probability that the face value is greater than 4 and that
the second is 2.
2 1 1
 
6 6 18
These are easier to do using probabilities and multiplication.