According to a poll, \( 60 \% \) of a nation's women 18 years old or older stated that the minimum driving age should be 18 . (a) In a random sample of 15 women 18 years old or older, find the probability that exactly 10 believe that the minimum driving age should be 18 . The probability is \( \square \). (Round to four decimal places as needed.)

To solve this problem, we can use the binomial probability formula, which is defined as follows: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] where: - \( n \) = total number of trials (sample size), - \( k \) = number of successes (in this case, women who believe the driving age should be 18), - \( p \) = probability of success on an individual trial, - \( (1 - p) \) = probability of failure, and - \( \binom{n}{k} \) = number of combinations of n items taken k at a time, calculated as \( \frac{n!}{k!(n-k)!} \). Given: - \( n = 15 \) - \( k = 10 \) - \( p = 0.60 \) (60% supporting the age) We’ll calculate \( P(X = 10) \). 1. Calculate the binomial coefficient: \[ \binom{15}{10} = \frac{15!}{10!(15-10)!} = \frac{15!}{10!5!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \] 2. Now plug in the values into the probability formula: \[ P(X = 10) = 3003 \cdot (0.60)^{10} \cdot (0.40)^{5} \] 3. Calculate \( (0.60)^{10} \) and \( (0.40)^{5} \): \[ (0.60)^{10} \approx 0.0060466176 \] \[ (0.40)^{5} \approx 0.01024 \] 4. Now multiply everything together: \[ P(X = 10) = 3003 \cdot 0.0060466176 \cdot 0.01024 \approx 3003 \cdot 0.0000618664 \approx 0.185 \] Therefore, rounding to four decimal places, the answer is approximately: \[ P(X = 10) \approx 0.1851 \] So, the probability that exactly 10 out of 15 women believe that the minimum driving age should be 18 is \( \boxed{0.1851} \).