A and B are independent events such that \(P(A\cap B) = \displaystyle\frac{1}{8}\) and \(P(A'\cap B') = \displaystyle\frac{3}{8}\). If P(A) > P(B), then P(A) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Let P(A) = a and P(B) = b. For independent events:
\( \displaystyle ab = \frac{1}{8} \)
The complements of independent events are also independent, so
\( \displaystyle (1-a)(1-b) = \frac{3}{8} \)
\( \displaystyle 1-(a+b)+ab = \frac{3}{8} \)
\( \displaystyle a+b = 1+\frac{1}{8}-\frac{3}{8} = \frac{3}{4} \)
So a and b are the roots of the quadratic with sum \(\frac{3}{4}\) and product \(\frac{1}{8}\):
\( \displaystyle x^{2}-\frac{3}{4}x+\frac{1}{8} = 0 \)
\(8x^{2}-6x+1 = 0\)
\((2x-1)(4x-1) = 0\)
\( \displaystyle x = \frac{1}{2} \text{ or } x = \frac{1}{4} \)
Since P(A) > P(B), \(P(A) = \frac{1}{2}\) and \(P(B) = \frac{1}{4}\).
Check: \(\frac{1}{2}\times\frac{1}{4} = \frac{1}{8}\) and \(\frac{1}{2}\times\frac{3}{4} = \frac{3}{8}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{4}\) — Incorrect. This is the smaller root, P(B).
Option (3) \(\displaystyle\frac{3}{4}\) — Incorrect. This is P(A) + P(B), the sum of the two roots.
Option (4) \(\displaystyle\frac{3}{8}\) — Incorrect. This is the given value of \(P(A'\cap B')\), not P(A).