The given formula represents a binomial probability distribution function:

$P(x) = \binom{n}{x} \cdot (p)^x \cdot (1 - p)^{n - x}$

Where:

• $\binom{n}{x}$ is the binomial coefficient, representing the number of ways to choose $x$ successes out of $n$ trials.
• $p$ is the probability of success on a single trial.
• $1 - p$ is the probability of failure on a single trial.
• $n$ is the total number of trials.
• $x$ is the number of successes.

In your case, the formula is: $P(x) = 10Cx \cdot (0.63)^x \cdot (0.37)^{10 - x}$ This implies:

• $n = 10$ (representing the total number of students)
• $p = 0.63$ (probability of success for each student in Math 247)
• $1 - p = 0.37$ (probability of failure)

Now, the problem asks for the probability that all 10 students succeed in Math 247. This implies $x = 10$.

To find the probability for $x = 10$, substitute into the formula:

$P(10) = \binom{10}{10} \cdot (0.63)^{10} \cdot (0.37)^{0}$

Since $\binom{10}{10} = 1$ and $(0.37)^{0} = 1$, the expression simplifies to:

$P(10) = 1 \cdot (0.63)^{10} \cdot 1 = (0.63)^{10}$

Now, calculate $(0.63)^{10}$:

$P(10) \approx (0.63)^{10} \approx 0.01698$

Thus, the probability that all 10 students succeed in Math 247 is approximately 0.01698.

Do you have any further questions or would you like more details?

Here are five follow-up questions:

1. How would the probability change if the success rate were 0.7 instead of 0.63?
2. What is the probability that exactly 8 out of 10 students succeed?
3. How does the binomial coefficient $\binom{n}{x}$ influence the overall probability?
4. How would you calculate the probability that fewer than 5 students succeed?
5. What happens to the distribution when the number of trials $n$ increases to 20?

Tip: Binomial probability distributions are useful when you have a fixed number of independent trials with two possible outcomes (success or failure), and the probability of success remains constant across all trials.