Jury selection problem

Question

A jury pool has 22 men and 25 women, from which 12 jurors will be selected. Assuming that each person is equally likely to be chosen and that the jury is selected at random, find the probability that the jury consists of

(a) all men

  

(b) all women

  

(c) 8 men and 4 women

  

(d) 6 men and 6 women

  

Answer 1

a). There are 47 people to choose from. 22 of them are men. So the probability of the first juror chosen is a man is 22/47.

Now there are 21 men out of 46 people. So the probability of the second juror chosen is a man is 21/46.

Now there are 20 men out of 45 people. So the probability of the third juror chosen is a man is 20/45.

and so on.

So the probability that all 12 jurors are men is

22/47 X 21/46 X 20/45 X 19/44 X 18/43 X 17/42 X 16/41 X 15/40 X 14/ 39 X 13/38 X 12/37 X 11/36

This is much easier to calculate if it is written as one fraction and then reduce all common factors.

22 X 21 X 20 X 19 X 18 X 17 X 16 X 15 X 14 X 13 X 12 X 11

47 X 46 X 45 X 44 X 43 X 42 X 41 X 40 X 39 X 38 X 37 X 36

The final answer is 251328 / 20308236768. OR 0.000012375668202.

Using the combination formula _nC_r = n! / (n-r)!

The solution can be expressed as a fraction using this formula.

₂₂C_12.X ₂₅C₀ = (22! / (22-12)! ) X (25! / (25-0)! ) / (47! / (47-12)!) = 0.000012375668202

₄₇C₁₂

b. All women means 25 women, choose 12 or ₂₅C_12. and choose no men ₂₂C₀ , divided by the total of 47 people, choose 12 or ₄₇C₁₂

₂₅C₁₂ X ₂₂C_{0 =} 25! / (25 - 12)! X 25! / (25-25)! / 47! / (47-12)! = 0000099524604418

₄₇C₁₂

c. 8 men and 4 women will be

₂₂C_{8 X 25}C₄

₄₇C₁₂

d. 6 men and 6 women will be

₂₂C₆ X ₂₅C₆

₄₇C₁₂