A fair die is thrown twice. The probability that the first throw shows an even number and the second throw shows a multiple of 3 is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{6}\)
Let A be the event “the first throw shows an even number” and B the event “the second throw shows a multiple of 3”.
The even numbers are 2, 4, 6 and the multiples of 3 are 3, 6:
\( \displaystyle P(A) = \frac{3}{6} = \frac{1}{2} \)
\( \displaystyle P(B) = \frac{2}{6} = \frac{1}{3} \)
The two throws do not affect each other, so A and B are independent events:
\( \displaystyle P(A\cap B) = P(A)\times P(B) = \frac{1}{2}\times\frac{1}{3} = \frac{1}{6} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{3}\) — Incorrect. This is \(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\), the chance that at least one of the two things happens, not both.
Option (2) \(\displaystyle\frac{5}{6}\) — Incorrect. This adds \(\frac{1}{2}\) and \(\frac{1}{3}\); for both events together the probabilities are multiplied.
Option (3) \(\displaystyle\frac{1}{12}\) — Incorrect. The product \(\frac{1}{2}\times\frac{1}{3}\) is \(\frac{1}{6}\), not \(\frac{1}{12}\); of the 36 equally likely pairs, 3 × 2 = 6 are favourable.