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This printable supports Common Core Mathematics Standard HSS-CP.B.9

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# Probability With Combinations and Permutations (Grades 11-12)

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## Probability With Combinations and Permutations

1.
Ms. Choi teaches English at a high school. She announces that the she will be randomly selecting 5 students out of the 36 students in the class to perform a play they are reading. There are 8 varsity athletes in the class. What is the probability that the group will consist completely of varsity athletes?
1. $1.2 xx 10^{-6}$
2. $1.5 xx 10^{-4}$
3. $1.4 xx 10^{-1}$
4. $2.2 xx 10^{-1}$
2.
Kevin has a box of 21 baseball cards. 3 of the cards are very rare. If Kevin randomly chooses 5 of his cards, what is the probability that he chooses all of his rare cards?
1. $4.9 xx 10^{-4}$
2. $7.5 xx 10^{-3}$
3. $1.0 xx 10^{-2}$
4. $1.4 xx 10^{-1}$
3.
The lock on my luggage has a 4-digit combination. Each digit can be 1-5. If you only have one chance, what is the probability that you guess the correct combination? (Round your answer to 2 decimal places.)
1. 0.16%
2. 0.83%
3. 5.67%
4. 2.35%
5. none of these are correct
4.
If Anna tosses a coin 10 times, what is the probability that she gets 4 heads?
1. 0.50
2. 0.07
3. 0.40
4. 0.21
5.
For a certain lottery, 4 numbers between 1 and 64 are chosen. Each number can only be used once. To win, the numbers chosen must match the winning numbers and their order. What is the probability of winning the lottery?
1. $6.3 xx 10^{-2}$
2. $1.6 xx 10^{-6}$
3. $6.6 xx 10^{-8}$
4. $1.2 xx 10^{-82}$
6.
Given the letters A, B, E, L, S, T, what is the probability that a random assortment of these letters will result in the words "stable" or "tables"?
1. $1.3 xx 10^{-1}$
2. $1.4 xx 10^{-3}$
3. $2.8 xx 10^{-3}$
4. $4.3 xx 10^{-5}$
7.
Given a standard deck of 52 cards, what is the probability that, if 5 cards are chosen, 3 are black (spades or clubs) and 2 are red (diamonds or hearts)?
1. $3.3 xx 10^{-1}$
2. $1.0 xx 10^{-1}$
3. $2.3 xx 10^{-3}$
4. $3.8 xx 10^{-6}$
8.
There are 26 slips of equally sized paper in a bag, each with a different letter of the alphabet. Margret will choose 8 of these, one at a time. What is the probability that she will spell the word "light" as she chooses pieces of paper? There can be no spaces between the letters of "light", and the letters before and/or after do not matter.
1. $6.7 xx 10^{-3}$
2. $5.1 xx 10^{-7}$
3. $6.4 xx 10^{-11}$
4. $6.1 xx 10^{-11}$
9.
A community organization, which has 25 members, needs to choose a new governing group. The governing group will have 4 positions- chairperson, treasurer, secretary, and marketing manager. Austin, Josh, Kaitlin, Tommy, Kora, and Stephanie are part of the organization. What is the probability that Austin or Josh are chosen as chairperson, and Kaitlin, Tommy, Kora, or Stephanie are chosen for the other 3 roles?
1. $2.4 xx 10^{-1}$
2. $3.8 xx 10^{-3}$
3. $1.6 xx 10^{-4}$
4. $9.5 xx 10^{-4}$
10.
At Karen's school, each locker comes with a lock that already has a combination. The locks use four numbers between 1 and 60 which aren't repeated. Karen is hoping that her locker combination has the numbers 4, 10, 22, and 50 which happen to have special significance for her. She doesn't care what order these numbers are in. She determines that there are $P(60,4)$ total possibilities for the locker combination, and $P(4,4)$ possibilities that include her numbers. Therefore, the probability that she gets her numbers is $2.1 xx 10^{-6}$. Is she correct, and if not, why?
1. No. Although justified in using permutations for the total number of possibilities, since order does matter, she should have used combinations to calculate the number of possibilities which include her numbers, since she doesn't care about the order for them. The probability should be $(C(4,4)) / (P(60,4)) = 8.5 xx 10^{-8}$.
2. No. Even though the end answer is correct, it is by chance. The total possibilities for locker combinations is $C(60,4)$ and the number of possibilities that include her numbers is $C(4,4)$. This just happens to also equal $2.1 xx 10^{-6}$.
3. No. The correct number of possibilities for the lock combination should be $60^4$. Therefore, the probability would be $(P(4,4)) / 60^4 = 1.9 xx 10^{-6}$.
4. Yes. Karen's method is correct.
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