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? D.
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To solve the problems about the HelpYourMath YouTube channel and its recognition among students, we can use the binomial probability formula. The parameters for our case are \( n = 10 \) (the number of trials), \( k \) (the number of successes), and \( p = 0.6 \) (the probability of success). The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \] Where \( \binom{n}{k} \) is the binomial coefficient. Let's solve each of the parts: ### a. Probability that exactly 3 of them know HelpYourMath YouTube Channel: \[ P(X = 3) = \binom{10}{3} \cdot (0.6)^3 \cdot (0.4)^7 \] \[ = \frac{10!}{3! \cdot (10-3)!} \cdot (0.6)^3 \cdot (0.4)^7 \] Calculating the binomial coefficient: \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] Now calculate: \[ P(X = 3) = 120 \cdot (0.6)^3 \cdot (0.4)^7 \] \[ = 120 \cdot 0.216 \cdot 0.0016384 \approx 0.0394 \] ### b. Probability that at least 3 know HelpYourMath YouTube Channel: \[ P(X \geq 3) = 1 - P(X < 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) \] Calculating \( P(X = 0), P(X = 1), P(X = 2) \): 1. **For \( P(X = 0) \)**: \[ P(X = 0) = \binom{10}{0} \cdot (0.6)^0 \cdot (0.4)^{10} = 1 \cdot 1 \cdot 0.0001048576 \approx 0.0001 \] 2. **For \( P(X = 1) \)**: \[ P(X = 1) = \binom{10}{1} \cdot (0.6)^1 \cdot (0.4)^9 = 10 \cdot 0.6 \cdot 0.000262144 \approx 0.0016 \] 3. **For \( P(X = 2) \)**: \[ P(X = 2) = \binom{10}{2} \cdot (0.6)^2 \cdot (0.4)^8 = 45 \cdot 0.36 \cdot 0.00065536 \approx 0.0107 \] Now summing these probabilities: \[ P(X < 3) \approx 0.0001 + 0.0016 + 0.0107 \approx 0.0124 \] Thus, \[ P(X \geq 3) = 1 - 0.0124 \approx 0.9876 \] ### c. Probability that at most 3 know HelpYourMath YouTube Channel: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] \[ = 0.0001 + 0.0016 + 0.0107 + 0.0394 \approx 0.0518 \] ### d. Probability that more than 3 know HelpYourMath YouTube Channel: \[ P(X > 3) = 1 - P(X \leq 3) \approx 1 - 0.0518 = 0.9482 \] ### e. Probability that less than 3 know HelpYourMath YouTube Channel: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0124 \] ### Summary of Results - a. \( P(X = 3) \approx 0.0394 \) - b. \( P(X \geq 3) \approx 0.9876 \) - c. \(