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? a 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: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where: - \( n \) = total number of trials (students selected) - \( k \) = number of successes (students that know the channel) - \( p \) = probability of success (students knowing the channel) - \( \binom{n}{k} \) = binomial coefficient In this case, \( n = 11 \) and \( p = 0.6 \). ### a. Probability that exactly 5 know HelpYourMath YouTube Channel \[ P(X = 5) = \binom{11}{5} (0.6)^5 (0.4)^{6} \] Calculating the binomial coefficient and substituting the values: \[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = 462 \] \[ P(X = 5) = 462 \times (0.6)^5 \times (0.4)^6 \approx 462 \times 0.07776 \times 0.004096 \approx 0.1470 \] So, the probability that exactly 5 students know the channel is approximately **0.1470**. ### b. Probability that at least 5 know HelpYourMath YouTube Channel To find this, we calculate \( P(X \geq 5) \): \[ P(X \geq 5) = 1 - P(X < 5) = 1 - P(X \leq 4) \] Now we need to find \( P(X \leq 4) \) which is the sum of probabilities from \( X = 0 \) to \( X = 4 \). Calculating all of these probabilities and summing them will give the result. \[ P(X = 0) \approx 0.0005 \\ P(X = 1) \approx 0.0131 \\ P(X = 2) \approx 0.1152 \\ P(X = 3) \approx 0.2613 \\ P(X = 4) \approx 0.2272 \] Calculating: \[ P(X \leq 4) \approx 0.0005 + 0.0131 + 0.1152 + 0.2613 + 0.2272 = 0.6173 \] So, \[ P(X \geq 5) = 1 - P(X \leq 4) \approx 1 - 0.6173 \approx 0.3827 \] Thus, the probability that at least 5 students know the channel is approximately **0.3827**. ### c. Probability that at most 5 know HelpYourMath YouTube Channel This is straightforward: \[ P(X \leq 5) = P(X < 5) + P(X = 5) \] From the previous result: \[ P(X < 5) = P(X \leq 4) \approx 0.6173 \] Thus, \[ P(X \leq 5) \approx 0.6173 + 0.1470 \approx 0.7643 \] The probability that at most 5 students know the channel is approximately **0.7643**. ### d. Probability that more than 5 know HelpYourMath YouTube Channel Using: \[ P(X > 5) = 1 - P(X \leq 5) \] \[ P(X > 5) = 1 - 0.7643 \approx 0.2357 \] So, the probability that more than 5 students know the channel is approximately **0.2357**. ### e. Probability that less than 5 know HelpYourMath YouTube Channel Using: \[ P(X < 5) = P(X \leq 4) \] So, this is already calculated: \[ P(X < 5) \approx 0.6173 \] In conclusion: - a. \( P(X = 5) \approx 0.1470 \) - b. \( P(X \geq 5) \approx 0.3827 \) - c. \( P(X \leq 5) \approx 0.7643 \) - d. \( P(X > 5) \approx 0.2357 \) - e. \( P(X < 5) \approx