Question 1195514
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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?
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<pre>
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

    {{{C[15]^7}}} = {{{(15!)/(7!*8!)}}} = {{{(15*14*13*12*11*10*9)/(1*2*3*4*5*6*7)}}} = 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 {{{C[3]^2}}} = 3 different ways.

Two persons with blood type  A can be chosen from 5 people with blood type  A by {{{C[5]^2}}} = 10 different ways.

Two persons with blood type  B can be chosen from 4 people with blood type  B by {{{C[4]^2}}} = 6 different ways.

One person  with blood type AB can be chosen from 3 people with blood type AB by {{{C[3]^1}}} = 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 = {{{favorable_groups_of_7/total_groups_of_7}}} = {{{540/6435}}} = {{{12/143}}}.


<U>ANSWER</U>.  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  {{{12/143}}} = 0.0839 (rounded).
</pre>

Solved.