In one city, the probability that a person will pass his or her driving test on the first attempt is 0.60 . Ten people are selected at random from among those taking their driving test for the first time. What is the probability that among these 10 people, the number passing the test is more than 7 ? You may use the binomial table or calculate the probabilities using the formula. 0.3822 0.1672 0.7031 0.8328 0.5332

Let \( X \) be the number of people who pass the test out of 10. Since each test is independent and the probability of passing is 0.60, \( X \) follows a binomial distribution: \[ X \sim \text{Binomial}(n=10, p=0.6) \] We are asked to find the probability that more than 7 people pass the test, i.e., \[ P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) \] The probability mass function for a binomial random variable is \[ P(X = k) = \binom{10}{k} (0.6)^k (0.4)^{10-k} \] **Step 1: Compute \(P(X = 8)\)** \[ P(X = 8) = \binom{10}{8} (0.6)^8 (0.4)^2 \] \[ \binom{10}{8} = \frac{10!}{8!2!} = 45 \] \[ P(X = 8) = 45 \times (0.6)^8 \times (0.4)^2 \] **Step 2: Compute \(P(X = 9)\)** \[ P(X = 9) = \binom{10}{9} (0.6)^9 (0.4)^1 \] \[ \binom{10}{9} = \frac{10!}{9!1!} = 10 \] \[ P(X = 9) = 10 \times (0.6)^9 \times (0.4) \] **Step 3: Compute \(P(X = 10)\)** \[ P(X = 10) = \binom{10}{10} (0.6)^{10} (0.4)^0 \] \[ \binom{10}{10} = 1 \] \[ P(X = 10) = 1 \times (0.6)^{10} \] **Step 4: Evaluate the individual probabilities** Using a calculator for the powers (approximate values): - \( (0.6)^8 \approx 0.01678 \) - \( (0.4)^2 = 0.16 \) - \( (0.6)^9 \approx 0.01007 \) - \( (0.6)^{10} \approx 0.00604 \) Now substitute these into the expressions: \[ P(X = 8) \approx 45 \times 0.01678 \times 0.16 \approx 45 \times 0.0026848 \approx 0.1208 \] \[ P(X = 9) \approx 10 \times 0.01007 \times 0.4 \approx 10 \times 0.004028 \approx 0.04028 \] \[ P(X = 10) \approx 0.00604 \] **Step 5: Sum the probabilities** \[ P(X > 7) \approx 0.1208 + 0.04028 + 0.00604 \approx 0.16712 \] Thus, the probability that more than 7 people pass the driving test is approximately \[ \boxed{0.1672} \]