Question

There are 20 candidates for a job position, of whom 14 are women and 6 are men. The hiring committee will randomly select 3 candidates and interview them one by one. Of course, a candidate can be interviewed only once.

(a) What is the probability that the second interview will be given to a woman, given that the first interview was given to another woman?

Answer: (Round to 3 decimal places.)

(b) What is the probability that all three interviews will be given to men? Answer: (Round to 3 decimal places.)

Answer 1

Answer 2

Given:

• Total candidates = 20 (14 women, 6 men)

• Committee selects 3 candidates without replacement.

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(a) Probability that the second interview is a woman, given the first was a woman:

• After selecting 1 woman, remaining candidates = 19 (13 women, 6 men).

• Probability = $\frac{13}{19}\approx \boxed{{\displaystyle 0.684}}$ (rounded to 3 decimal places).

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(b) Probability that all three interviews are men:

• Ways to choose 3 men out of 6: $\left(\genfrac{}{}{0px}{}{6}{3}\right)=20$.

• Total ways to choose any 3 candidates: $\left(\genfrac{}{}{0px}{}{20}{3}\right)=1140$.

• Probability = $\frac{20}{1140}=\frac{1}{57}\approx \boxed{{\displaystyle 0.018}}$ (rounded to 3 decimal places).

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Note:

• (a) Uses conditional probability (dependent events).

• (b) Uses combinations (hypergeometric distribution)