This problem is an example of a binomial probability problem, where we are dealing with the probability that a certain number of students will withdraw from a course, given that 25% (or 0.25) of students typically withdraw.

Key Information:

• The probability of a student withdrawing, $p = 0.25$.
• The number of students, $n = 8$.
• We are asked to find the probability that fewer than two students withdraw, i.e., $P(X < 2)$, where $X$ is the number of students who withdraw.

This can be expressed as: $P(X < 2) = P(X = 0) + P(X = 1)$

The probability of exactly $k$ students withdrawing in a binomial distribution is given by the formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where:

• $\binom{n}{k}$ is the binomial coefficient (combinations),
• $p$ is the probability of success (in this case, a student withdrawing),
• $1 - p$ is the probability of failure (a student not withdrawing).

We will calculate $P(X = 0)$ and $P(X = 1)$.

1. Probability of 0 Students Withdrawing:

$P(X = 0) = \binom{8}{0} (0.25)^0 (0.75)^8 = 1 \cdot 1 \cdot (0.75)^8 \approx 0.1001$

2. Probability of 1 Student Withdrawing:

$P(X = 1) = \binom{8}{1} (0.25)^1 (0.75)^7 = 8 \cdot 0.25 \cdot (0.75)^7 \approx 0.2670$

Total Probability:

$P(X < 2) = P(X = 0) + P(X = 1) \approx 0.1001 + 0.2670 = 0.3671$

Thus, the probability that fewer than two students withdraw is approximately 0.3671 or 36.71%.

Would you like more details on the calculations or any clarifications?

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Here are 5 related questions to expand on this topic:

1. What is the probability that exactly two students withdraw?
2. What is the probability that at least one student withdraws?
3. How would the probability change if the withdrawal rate increased to 30%?
4. How would you calculate the expected number of students to withdraw?
5. How does the variance of the binomial distribution help in understanding this problem?

Tip: In binomial problems, recognizing "fewer than" or "at least" conditions can help in quickly setting up the correct sum of probabilities.