Asked by p

Which probability is correct?

P(A) = Three-fifths
P(B) = StartFraction 16 Over 31 EndFraction
P(A|B) = Two-sevenths
P(B|A) = StartFraction 10 Over 21 EndFraction

All Answers 1

Answered by GPT-5 mini AI
They are inconsistent — the two conditionals cannot both be true given the marginals.

Use P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A).

From P(A|B)=2/7 and P(B)=16/31:
P(A ∩ B) = (2/7)(16/31) = 32/217 ≈ 0.1475.

From P(B|A)=10/21 and P(A)=3/5:
P(A ∩ B) = (10/21)(3/5) = 30/105 = 2/7 ≈ 0.2857.

Since 32/217 ≠ 2/7, the given numbers are inconsistent.

If you accept the marginals P(A)=3/5 and P(B)=16/31, then
- If P(B|A)=10/21 were correct, the implied P(A|B) would be (2/7)/(16/31) = 31/56 ≈ 0.5536.
- If P(A|B)=2/7 were correct, the implied P(B|A) would be (32/217)/(3/5) = 160/651 ≈ 0.2458.

So at most one of the two conditional probabilities can be correct; pick one and use the formulas above to get the consistent value for the other.