Question

Half of a class took Form A of a test, and half took Form B. Of the students who took form B, 39% passed. What is the probability that a randomly chosen student took Form B and did not pass?

Answer 1

0.305 = 30.5% probability that a randomly chosen student took Form B and did not pass

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Took Form B.

Event B: Did not pass.

Half of a class took Form A of a test, and half took Form B.

This means that
`P(A) = 0.5`

Of the students who took form B, 39% passed.

So 100 - 39 = 61% did not pass, which means that
`P(B|A) = 0.61`.

What is the probability that a randomly chosen student took Form B and did not pass?

`P(B|A) = (P(A \cap B))/(P(A))`

`P(A \cap B) = P(B|A)*P(A) = 0.61*0.5 = 0.305`

0.305 = 30.5% probability that a randomly chosen student took Form B and did not pass