A survey shows that about \( 60 \% \) (i.e., 0.6 ) of all students know the HelpYourMath YouTube channel in a college. Suppose that we randomly select 11 students from the college, please solve the following problems and round your answers to 4 decimal places. a. What is the probability that exactly 5 of them know HelpYourMath YouTube Channel? b. What is the probability that at least 5 of them know HelpYourMath YouTube Channel? c. What is the probability that at most 5 of them know HelpYourMath YouTube Channel? d. What is the probability that more than 5 of them know HelpYourMath YouTube Channel? e. What is the probability that less than 5 of them know HelpYourMath YouTube Channel?

To solve these problems, we can use the binomial probability formula, which is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (students selected), - \( k \) is the number of successes (students who know the channel), - \( p \) is the probability of success (0.6), - \( \binom{n}{k} \) is the binomial coefficient. In this case, \( n = 11 \) and \( p = 0.6 \). ### a. Probability that exactly 5 of them know the HelpYourMath YouTube Channel Calculating for \( k = 5 \): \[ P(X = 5) = \binom{11}{5} (0.6)^5 (0.4)^{11-5} \] First, calculate \( \binom{11}{5} \): \[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \] Then calculate the probability: \[ P(X = 5) = 462 \times (0.6)^5 \times (0.4)^6 \] Calculating powers: \((0.6)^5 = 0.07776\) \((0.4)^6 = 0.004096\) Now multiply: \[ P(X = 5) = 462 \times 0.07776 \times 0.004096 \approx 0.1495 \] ### b. Probability that at least 5 of them know HelpYourMath YouTube Channel This is calculated as: \[ P(X \geq 5) = 1 - P(X < 5) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4) \] Calculating \( P(X < 5) \) involves calculating probabilities for \( k = 0, 1, 2, 3, 4 \): After calculation, we get: \[ P(X = 0) \approx 0.0001, \quad P(X = 1) \approx 0.0028, \quad P(X = 2) \approx 0.0270, \quad P(X = 3) \approx 0.1384, \quad P(X = 4) \approx 0.2278 \] Summing these probabilities: \[ P(X < 5) \approx 0.0001 + 0.0028 + 0.0270 + 0.1384 + 0.2278 \approx 0.3961 \] Then: \[ P(X \geq 5) = 1 - 0.3961 \approx 0.6039 \] ### c. Probability that at most 5 of them know HelpYourMath YouTube Channel \[ P(X \leq 5) = P(X < 5) + P(X = 5) \] From above: \[ P(X \leq 5) = 0.3961 + 0.1495 \approx 0.5456 \] ### d. Probability that more than 5 of them know HelpYourMath YouTube Channel \[ P(X > 5) = 1 - P(X \leq 5) = 1 - 0.5456 \approx 0.4544 \] ### e. Probability that less than 5 of them know HelpYourMath YouTube Channel This is already been calculated as \( P(X < 5) \): \[ P(X < 5) \approx 0.3961 \] ### Summary a. \( P(X = 5) \approx 0.1495 \) b. \( P(X \geq 5) \approx 0.6039 \) c. \( P(X \leq 5) \approx 0.5456 \) d. \( P(X > 5) \approx 0.4544 \) e. \( P(X < 5) \approx 0.3961 \)