This is a binomial distribution problem. The binomial distribution model is appropriate here because it describes the number of successes in a fixed number of trials of a binary or yes/no type experiment.

Here, the probability of success (a student passing) is 0.10 (since 90% fail, 10% pass), the number of trials is 20 (the number of students), and we want to find the probability of at most 1 success (at most 1 student passing).

The probability mass function of a binomial distribution is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes

We want to find P(X

Question

This is a binomial distribution problem. The binomial distribution model is appropriate here because it describes the number of successes in a fixed number of trials of a binary or yes/no type experiment.

Here, the probability of success (a student passing) is 0.10 (since 90% fail, 10% pass), the number of trials is 20 (the number of students), and we want to find the probability of at most 1 success (at most 1 student passing).

The probability mass function of a binomial distribution is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes

We want to find P(X<=1), which is P(X=0) + P(X=1).

P(X=0) = C(20, 0) * (0.10^0) * ((1-0.10)^(20-0)) = 1 * 1 * (0.9^20) = 0.1216

P(X=1) = C(20, 1) * (0.10^1) * ((1-0.10)^(20-1)) = 20 * 0.10 * (0.9^19) = 0.2702

So, P(X<=1) = P(X=0) + P(X=1) = 0.1216 + 0.2702 = 0.3918

So, the closest answer is 0.392.

Knowee AI · Accepted Answer