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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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 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 trials (n):** 10 students
- **Probability of success (p):** 0.6 (60% of students know the channel)
- **Probability of failure (q):** 1 - p = 0.4

**a. Probability that exactly 3 of them know HelpYourMath YouTube Channel:**
This is a classic binomial probability problem. The formula for 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.

Plugging in the numbers:
$$ P(X = 3) = C(10, 3) \times 0.6^3 \times 0.4^7 $$
First, calculate $$ C(10, 3) $$:
$$ C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 $$
Now, compute $$ 0.6^3 $$:
$$ 0.6^3 = 0.216 $$
And $$ 0.4^7 $$:
$$ 0.4^7 = 0.0016384 $$
Multiply them all together:
$$ 120 \times 0.216 \times 0.0016384 $$
First, $$ 120 \times 0.216 = 25.92 $$
Then, $$ 25.92 \times 0.0016384 \approx 0.0423 $$
So, the probability is approximately 0.0423 or 4.23%.

**b. Probability that at least 3 of them know HelpYourMath YouTube Channel:**
"At least 3" means 3 or more. So, we need to calculate the probability of 3, 4, 5, ..., up to 10 students knowing the channel and sum them up.

Alternatively, it's easier to calculate the probability of fewer than 3 students knowing the channel and subtract that from 1.

First, calculate the probabilities for 0, 1, and 2 students knowing the channel.

- **For 0 students:**
$$ P(X = 0) = C(10, 0) \times 0.6^0 \times 0.4^{10} = 1 \times 1 \times 0.0001024 = 0.0001024 $$
- **For 1 student:**
$$ P(X = 1) = C(10, 1) \times 0.6^1 \times 0.4^9 = 10 \times 0.6 \times 0.000262144 = 0.001572864 $$
- **For 2 students:**
$$ P(X = 2) = C(10, 2) \times 0.6^2 \times 0.4^8 = 45 \times 0.36 \times 0.00065536 = 0.010616832 $$

Now, sum these probabilities:
$$ 0.0001024 + 0.001572864 + 0.010616832 = 0.012292096 $$

Subtract this from 1 to get the probability of at least 3 students knowing the channel:
$$ 1 - 0.012292096 = 0.987707904 $$
So, the probability is approximately 0.9877 or 98.77%.

**c. Probability that at most 3 of them know HelpYourMath YouTube Channel:**
"At most 3" means 0, 1, 2, or 3 students knowing the channel. We've already calculated these probabilities in part b:
$$ P(X \leq 3) = 0.0001024 + 0.001572864 + 0.010616832 + 0.0423 = 0.054592096 $$
So, the probability is approximately 0.0546 or 5.46%.

**d. Probability that more than 3 of them know HelpYourMath YouTube Channel:**
This is the complement of "at most 3" students knowing the channel.
$$ P(X > 3) = 1 - P(X \leq 3) = 1 - 0.054592096 = 0.945407904 $$
So, the probability is approximately 0.9454 or 94.54%.

**e. Probability that less than 3 of them know HelpYourMath YouTube Channel:**
This is the same as "at most 2" students knowing the channel, which we've already calculated in part b:
$$ P(X < 3) = 0.0001024 + 0.001572864 = 0.001675264 $$
So, the probability is approximately 0.0017 or 0.17%.

**Double-Checking Calculations:**
It's always good to verify the calculations to ensure accuracy. Recalculating a few key steps:
- For part a, the combination and probability calculations seem correct.
- For part b, the cumulative probability for 0, 1, and 2 students is accurate.
- The complement calculations in parts c and d also seem correct.

Everything adds up logically, and the probabilities make sense given the high success rate (60%) and the number of trials (10).

a. The probability that exactly 3 of them know HelpYourMath YouTube Channel is **0.0423** or **4.23%**. b. The probability that at least 3 of them know HelpYourMath YouTube Channel is **0.9877** or **98.77%**. c. The probability that at most 3 of them know HelpYourMath YouTube Channel is **0.0546** or **5.46%**. d. The probability that more than 3 of them know HelpYourMath YouTube Channel is **0.9454** or **94.54%**. e. The probability that less than 3 of them know HelpYourMath YouTube Channel is **0.0017** or **0.17%**.

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