Questions: If a jury pool consists of 10 men and 12 women, what is the probability of selecting a 5 person jury consisting of all females? Round your answer to the nearest whole number Probability = % (Type an integer.)

Solution Steps

We need to calculate the probability of selecting a 5-person jury consisting entirely of females from a jury pool that consists of 10 men and 12 women.

The hypergeometric distribution is defined as follows:

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

Where:

• \(N\) = total number of items in the population = 22 (10 men + 12 women)
• \(K\) = total number of success items in the population = 12 (women)
• \(n\) = number of items drawn in the sample = 5 (jury members)
• \(k\) = number of success items in the sample = 5 (all females)

We need to compute the combinations:

1. \(\binom{K}{k} = \binom{12}{5}\)
2. \(\binom{N-K}{n-k} = \binom{10}{0}\)
3. \(\binom{N}{n} = \binom{22}{5}\)

Substituting the values into the formula gives:

\[ P(X = 5) = \frac{\binom{12}{5} \cdot \binom{10}{0}}{\binom{22}{5}} \]

Calculating the combinations:

• \(\binom{12}{5} = 792\)
• \(\binom{10}{0} = 1\)
• \(\binom{22}{5} = 26334\)

Thus, we have:

\[ P(X = 5) = \frac{792 \cdot 1}{26334} \approx 0.0301 \]

To express the probability as a percentage, we multiply by 100:

\[ P(X = 5) \times 100 \approx 3.01\% \]

Rounding to the nearest whole number gives:

\[ \text{Probability} = 3\% \]

Final Answer