Questions: It is known that 10 out of 15 members of the board of directors of a company are in favor of paying a bonus to its executives. Suppose three members are randomly selected by the media. What is the probability that all of them are in favor of a bonus?

Solution Steps

We need to find the probability that all three members selected by the media are in favor of paying a bonus to the executives. Given the total number of board members \( N = 15 \), the number of members in favor \( K = 10 \), and the number of members selected \( n = 3 \), we want to calculate \( P(X = 3) \).

The probability of drawing exactly \( k \) successes (members in favor) in \( n \) draws from a finite population is given by the hypergeometric distribution:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Substituting the values \( K = 10 \), \( k = 3 \), \( N = 15 \), \( n = 3 \):

\[ P(X = 3) = \frac{\binom{10}{3} \binom{5}{0}}{\binom{15}{3}} \]

Now we calculate the combinations:

• \( \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \)
• \( \binom{5}{0} = 1 \)
• \( \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \)

Substituting these values back into the probability formula:

\[ P(X = 3) = \frac{120 \times 1}{455} = \frac{120}{455} \]

Calculating the decimal value:

\[ P(X = 3) \approx 0.2637 \]

Final Answer