In throwing of a die, let A be the event 'an odd humber turns up' B be the event 'a number divisible by 3 turns up' and C be the event 'a number ≤ 4 turns up' Then find the probability that exactly two of A, B and C occur.

Engineering

Mathematics

Conditional Probability

Question

$\frac{2}{3}$

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{5}{6}$

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Event A = {1, 3, 5} event B = {3, 6} and event C = {1, 2, 3, 4}
∴ P(exactly two of A, B and C occur) = P(A∩B) + P(B∩C) + P(C∩A) − 3P(A∩B∩C)

$= \frac{1}{6} + \frac{1}{6} + \frac{2}{6} - 3 \times \frac{1}{6} = \frac{1}{6}$

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