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 10 students from the college, please solve the following problems and round your answers to 4 decimal places. a. What is the probability that exactly 3 of them know HelpYourMath YouTube Channel? b. What is the probability that at least 3 of them know HelpYourMath YouTube Channel? c. What is the probability that at most 3 of them know HelpYourMath YouTube Channel? d. What is the probability that more than 3 of them know HelpYourMath YouTube Channel? e. What is the probability that less than 3 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 10 students randomly, and we 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:** We have a binomial distribution scenario here. The probability of success (a student knowing the channel) is \( p = 0.6 \), and the number of trials (students selected) is \( n = 10 \). We need to find probabilities for different numbers of successes. **Part a: Probability that exactly 3 of them know HelpYourMath YouTube Channel** This is a straightforward binomial probability calculation. The formula for the probability of exactly \( k \) successes in \( n \) trials is: \[ P(X = k) = C(n, k) \times p^k \times (1-p)^{n-k} \] Where \( C(n, k) \) is the combination of \( n \) items taken \( k \) at a time. Plugging in the numbers: \[ P(X = 3) = C(10, 3) \times 0.6^3 \times 0.4^7 \] Calculating \( C(10, 3) \): \[ C(10, 3) = \frac{10!}{3! \times (10-3)!} = 120 \] So, \[ P(X = 3) = 120 \times 0.216 \times 0.0016384 \approx 0.0414 \] Rounded to four decimal places: 0.0414 **Part b: Probability that at least 3 of them know HelpYourMath YouTube Channel** "At least 3" means 3 or more students know the channel. This is the sum of probabilities from 3 to 10 successes: \[ P(X \geq 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] \] Calculating each: \[ P(X = 0) = C(10, 0) \times 0.6^0 \times 0.4^{10} = 1 \times 1 \times 0.0001024 \approx 0.0001 \] \[ P(X = 1) = C(10, 1) \times 0.6^1 \times 0.4^9 = 10 \times 0.6 \times 0.000262144 \approx 0.0015 \] \[ P(X = 2) = C(10, 2) \times 0.6^2 \times 0.4^8 = 45 \times 0.36 \times 0.0016777216 \approx 0.0262 \] Adding them up: \[ P(X < 3) \approx 0.0001 + 0.0015 + 0.0262 = 0.0278 \] Therefore, \[ P(X \geq 3) = 1 - 0.0278 = 0.9722 \] Rounded to four decimal places: 0.9722 **Part c: Probability that at most 3 of them know HelpYourMath YouTube Channel** "At most 3" means 0 to 3 students know the channel. This is the sum of probabilities from 0 to 3 successes: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] We've already calculated \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \). Calculating \( P(X = 3) \) as before: \[ P(X = 3) \approx 0.0414 \] Adding them up: \[ P(X \leq 3) \approx 0.0001 + 0.0015 + 0.0262 + 0.0414 = 0.0702 \] Rounded to four decimal places: 0.0702 **Part d: Probability that more than 3 of them know HelpYourMath YouTube Channel** "More than 3" means 4 or more students know the channel. This is the sum of probabilities from 4 to 10 successes: \[ P(X > 3) = 1 - P(X \leq 3) \] From part c, \( P(X \leq 3) \approx 0.0702 \), so: \[ P(X > 3) = 1 - 0.0702 = 0.9298 \] Rounded to four decimal places: 0.9298 **Part e: Probability that less than 3 of them know HelpYourMath YouTube Channel** "Less than 3" means 0 to 2 students know the channel. This is the sum of probabilities from 0 to 2 successes: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \] We've already calculated these: \[ P(X < 3) \approx 0.0001 + 0.0015 + 0.0262 = 0.0278 \] Rounded to four decimal places: 0.0278 **Summary of Results:** a. Probability that exactly 3 know: 0.0414 b. Probability that at least 3 know: 0.9722 c. Probability that at most 3 know: 0.0702 d. Probability that more than 3 know: 0.9298 e. Probability that less than 3 know: 0.0278