Question

Dawn Corporation has 12 employees who hold managerial positions. Of them, 7 are female and 5 are male. The company is planning to send 3 of these 12 managers to a conference. If 3 managers are randomly selected out of 12, Find the probability that at most 2 of them are female.

Answer 1

Answer on Question #70838 – Math – Statistics and Probability

Question

Dawn Corporation has 12 employees who hold managerial positions. Of them, 7 are female and 5 are male. The company is planning to send 3 of these 12 managers to a conference. If 3 managers are randomly selected out of 12, Find the probability that at most 2 of them are female.

Solution

$A = \{$ not more than 2 women will go to the conference\}$

On the basis of the addition theorem and the multiplication theorem for probabilities, we get

$$P(A) = \frac{C_5^3}{C_{12}^3} + \frac{C_5^2 \cdot C_7^1}{C_{12}^3} + \frac{C_5^1 \cdot C_7^2}{C_{12}^3} = 0.841,$$

where $C_n^k = \frac{n!}{k! \times (n - k)!}$, $k! = 1 \times 2 \times 3 \times \ldots \times k$, $0! = 1$.

Answer: 0.841.

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