SOLUTION: A group of 15 individuals is used for a biological case study. The group contains 3 people with blood type O, 5 with blood type A, 4 with blood type B, and 3 with blood type AB. Wh

Question 1195514: A group of 15 individuals is used for a biological case study. The group contains 3 people with blood type O, 5 with blood type A, 4 with blood type B, and 3 with blood type AB. What is the probability that a random sample of 7 will contain 2 persons with blood type O, 2 persons with blood type A, 2 persons with blood type B, and 1 person with blood type AB?

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

number of favorable outcomes P = --------------------------------- total number of possible outcomes

The favorable outcomes are choosing 2 of the 3 people with type O, AND 2 of the 5 with type A, AND 2 of the 4 with type B, AND 1 of the 3 with type AB:

The possible outcomes are choosing any 7 of the 15 people:

ANSWER:

Use a calculator....

Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
A group of 15 individuals is used for a biological case study.
The group contains 3 people with blood type O, 5 with blood type A,
4 with blood type B, and 3 with blood type AB.
What is the probability that a random sample of 7 will contain
2 persons with blood type O, 2 persons with blood type A,
2 persons with blood type B, and 1 person with blood type AB?
~~~~~~~~~~~~~~~

This problem is solved in two steps. First step is to compute the number of all possible different groups of 7 individuals that can be formed/selected from 15 individuals. This number is the number of combinations of 15 individuals taken 7 at a time = = = 6435. Second step is to compute the number of favorable groups of 7 individuals with given compositions. Two persons with blood type O can be chosen from 3 people with blood type O by = 3 different ways. Two persons with blood type A can be chosen from 5 people with blood type A by = 10 different ways. Two persons with blood type B can be chosen from 4 people with blood type B by = 6 different ways. One person with blood type AB can be chosen from 3 people with blood type AB by = 3 different ways. So, 3*10*6*3 = 540 different groups can be formed of the given composition. Then the probability under the problem's question is P = = = . ANSWER. The probability that a random sample of 7 will contain 2 persons with blood type O, 2 persons with blood type A, 2 persons with blood type B, and 1 person with blood type AB is = 0.0839 (rounded).

Solved.

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