Probability - Hypergeometric

Question

ABC Toys Company employs $40$ people in the Assembly Department. Suppose that $16$ employees are female and $13$ employees just received merit raises. Only $4$ employees are female and received merit raises. Six employees are selected at random to form a committee to meet the management regarding shift starting times.

(i) What is the probability that at least four selected employees for the committee are female or received merit raises?

My answer:

$$P(\text{at least 4 are female or received merit raises}) = \frac{\binom{25}{4}\binom{15}{2}}{\binom{40}{6}} + \frac{\binom{25}{5}\binom{15}{1}}{\binom{40}{6}} + \frac{\binom{25}{6}}{\binom{40}{6}} = 0.5998$$

Is it right?

(ii) Given that at least two selected employees for the committee received merit raises, what is the probability that exactly four employees who received merit raises are finally selected in the committee?

$P(\text{Exactly 4 who received merit raises} \mid \text{At least two selected received merit})$

For (ii) I am not sure whether I am on the right track. Advice needed, thank you!

Answer 1

Note that we can write the following table for the population of $N = 40$ employees: $$\begin{array}{c|c|c|c} & \text{Raise} & \text{No Raise} & \\ \hline \text{Female} & 4 & 12 & 16 \\ \hline \text{Male} & 9 & 15 & 24 \\ \hline & 13 & 27 & 40 \end{array}$$

Thus, if $X$ is a random variable that counts the number of committee members who are female or received merit raises, then $$X \sim \operatorname{Hypergeometric}(N = 40, n = 6, m = 25), \quad \Pr[X = x] = \frac{\binom{m}{x} \binom{N-m}{n-x}}{\binom{N}{n}}$$ and we have $$\Pr[X \ge 4] = \sum_{x=4}^6 \frac{\binom{25}{x} \binom{15}{6-x}}{\binom{40}{6}}.$$ Thus your answer to the first part is correct.

For the second part, define $Y$ to be the random number of committee members who received merit raises, so that $$Y \sim \operatorname{Hypergeometric}(N = 40, n = 6, m = 13).$$ Then you want $$\Pr[Y = 4 \mid Y \ge 2] = \frac{\Pr[(Y = 4) \cap (Y \ge 2)]}{\Pr[Y \ge 2]} = \frac{\Pr[Y = 4]}{\Pr[Y \ge 2]} = \frac{\Pr[Y = 4]}{1 - \Pr[Y \le 1]}.$$ The numerator you should be able to work out easily. The denominator was written using the complementary probability $\Pr[Y \le 1]$ because it is easier to find $$\Pr[Y \le 1] = \Pr[Y = 0] + \Pr[Y = 1]$$ rather than $\Pr[Y \ge 2] = \Pr[Y = 2] + \Pr[Y = 3] + \cdots + \Pr[Y = 6]$.