MAT 142 Summer Session 1
ANSWERS
Homework Day #15:
Name:______________________
Solving Probability Problems Using Combinations.
1. A bag contains four red balls and five green balls. You plan to select three balls at random. Determine
the probability of selecting three green balls.
C3 10 5


84 42
9 C3
5
 11.9%
2. Each of the digits 0 – 9 is written on a slip of paper and the slips of paper are placed in a hat. If three
slips of paper are selected at random, determine the probability that the three numbers selected are greater
than 4.
C3
10
1


C
120
12
10 3
5
 8.33%
3. You are dealt 5 cards from a standard deck of 52 cards. Determine the probability that you are dealt 5
red cards.
C5
65780
253


C
2598960
9996
52 5
26
 2.53%
Five men and six women are going to be assigned to a specific row of seats in a movie theatre. If the 11
tickets for the numbered seats are given out at random, determine the probability that…
4. five women are given the first five seats next to the center aisle.
C5
6
1


462 77
11 C5
6
 1.2987%
5. at least one women is in one of the first five seats.
P(at least one women infirst five seats) = 1 – P (no woman in first five seats)
= 1 – P(all men in first five seats)
C
1
461
= 1 5 5  1

 99.78%
C
462
462
11 5
6. exactly one woman is in one of the first five seats.
6
C1  5 C4 6  5 5


462 77
11 C5
 6.49%
Recall our work on Day 12 with Odds. Also recall the work we have been doing the past two days with
Poker hands. Use your incredibly detailed class notes to calculate the “Odds Against” being dealt the
following 5 card poker hands from a standard deck of cars. Use the decimal probabilities that we
calculated in your class notes…i.e the probability of getting one pair was .423….the probability of getting a
royal straight flush was .00000154. Express your final answer in the form “# to 1” where your # has been
rounded to three decimal places.
P(event NOT occurring ) 1  P(event occuring )
Remember odds against event 

P(event occuring )
P(event occuring )
7. One Pair
1  .423 .577

"1.364 to 1"
.423
.423
8. Two Pair
1  .048 .952

"19.833 to 1"
.048
.048
9. Three of a kind
1  .021 .979

"46.619 to 1"
.021
.021
10. Straight
1  .0039 .9961

"255.410 to 1"
.0039
.0039
11. Flush
1  .00197 .99803

"506.61 to 1"
.00197
.00197
12. Full House
1  .00144 .99856

"693.444 to 1"
.00144
.00144
13. Four of a Kind
1  .00024 .99976

"4165.667 to 1"
.00024
.00024
14. Straight Flush
1  .0000139 .9999861

"71,941.446 to 1"
.0000139
.0000139
15. Royal Straight Flush
1  .00000154 .99999846

"649,349.649 to 1"
.00000154
.00000154
Note: Because we used our “rounded” decimal probabilities from our notes, these odds are slightly off
BUT they give you a real good idea of how unlikely it is to be dealt some of these hands.