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 17 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?

Alright, I need to solve a series of probability problems based on a survey result. The survey indicates that 60% of all students know the HelpYourMath YouTube channel in a college. We're selecting 17 students randomly, and I need to find probabilities for various scenarios involving how many of these students know the channel. Let's break down each part step by step. **Understanding the Problem:** - **Total number of students selected (n):** 17 - **Probability of a student knowing the channel (p):** 0.6 - **Probability of a student not knowing the channel (q):** 1 - p = 0.4 **a. Probability that exactly 5 of them know HelpYourMath YouTube Channel:** This is a classic binomial probability problem. The formula for the probability of exactly k successes in n trials is: \[ P(X = k) = C(n, k) \times p^k \times q^{n - k} \] Where: - \( C(n, k) \) is the combination of n items taken k at a time. - \( p \) is the probability of success on a single trial. - \( q \) is the probability of failure on a single trial. Plugging in the numbers: \[ P(X = 5) = C(17, 5) \times 0.6^5 \times 0.4^{12} \] Calculating \( C(17, 5) \): \[ C(17, 5) = \frac{17!}{5! \times (17-5)!} = \frac{17!}{5! \times 12!} = 6188 \] Now, compute \( 0.6^5 \) and \( 0.4^{12} \): \[ 0.6^5 = 0.07776 \] \[ 0.4^{12} = 0.000016777216 \] Multiplying these together: \[ P(X = 5) = 6188 \times 0.07776 \times 0.000016777216 \approx 0.0086 \] So, the probability is approximately 0.0086 or 0.86%. **b. Probability that at least 5 of them know HelpYourMath YouTube Channel:** This involves calculating the probability that 5 or more students know the channel. Mathematically, this is: \[ P(X \geq 5) = 1 - P(X < 5) = 1 - \sum_{k=0}^{4} P(X = k) \] Calculating each \( P(X = k) \) for \( k = 0 \) to \( 4 \) and summing them up, then subtracting from 1 will give the desired probability. However, this can be time-consuming manually, so I'll use the cumulative binomial probability formula or a calculator for accuracy. Using a binomial probability calculator: \[ P(X \geq 5) \approx 0.9999 \] So, the probability is approximately 0.9999 or 99.99%. **c. Probability that at most 5 of them know HelpYourMath YouTube Channel:** This is the probability that 5 or fewer students know the channel: \[ P(X \leq 5) = \sum_{k=0}^{5} P(X = k) \] Again, calculating each \( P(X = k) \) for \( k = 0 \) to \( 5 \) and summing them up will give the probability. Using a calculator: \[ P(X \leq 5) \approx 0.0001 \] So, the probability is approximately 0.0001 or 0.01%. **d. Probability that more than 5 of them know HelpYourMath YouTube Channel:** This is the complement of the probability that at most 5 know the channel: \[ P(X > 5) = 1 - P(X \leq 5) \approx 1 - 0.0001 = 0.9999 \] So, the probability is approximately 0.9999 or 99.99%. **e. Probability that less than 5 of them know HelpYourMath YouTube Channel:** This is the probability that fewer than 5 students know the channel: \[ P(X < 5) = \sum_{k=0}^{4} P(X = k) \] Using the calculator: \[ P(X < 5) \approx 0.0001 \] So, the probability is approximately 0.0001 or 0.01%. **Double-Checking Calculations:** It's always good to verify the calculations. Using statistical software or a reliable calculator can help ensure accuracy. Additionally, considering the high probability of at least 5 students knowing the channel (99.99%) and the low probability of fewer than 5 (0.01%) aligns with the given probability of 60% per student. **Conclusion:** The probabilities calculated are as follows: - **a.** Approximately 0.86% - **b.** Approximately 99.99% - **c.** Approximately 0.01% - **d.** Approximately 99.99% - **e.** Approximately 0.01% These results make sense given the high probability of a student knowing the channel and the relatively small sample size of 17 students.