The probability of an outcome A is 1/6 . The probability of the B outcome is 1/4 . If the probability of A or…

Analyze the question

We are given P(A) = 1/6, P(B) = 1/4, and P(A or B or both) = P(A ∪ B) = 1/12. We need P(A ∩ B).

Concept

The addition rule of probability:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

so

P(A ∩ B) = P(A) + P(B) − P(A ∪ B)

Solution

P(A ∩ B) = 1/6 + 1/4 − 1/12

Use the common denominator 12:

= 2/12 + 3/12 − 1/12 = 4/12 = 1/3

Check for a problem

However, the question as written is flawed, and you should know why:

• P(A ∩ B) can never be larger than P(A) or P(B). Here 1/3 is bigger than both 1/6 and 1/4, which is impossible.
• Also, P(A ∪ B) must be at least as large as P(A) or P(B). But 1/12 is smaller than 1/6 and 1/4, which is also impossible.

So the numbers in this question are inconsistent. The original JAMB question most likely had P(A or B or both) = 1/2 (or some other value), but the version in the question bank has a misprint. In the exam, use the formula and pick the option it produces, which is B.

(Check with 1/2 as the union: 1/6 + 1/4 − 1/2 would be negative, so that is not the original either. The original wording is uncertain, but the intended method is the addition rule.)

Tip

Always remember: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Rearrange it to find whichever part is missing. If the numbers are consistent, the intersection can never exceed the smaller of the two individual probabilities. Use that as a quick sanity check.