If E1 denotes the event of getting the sum 6 in throwing two dice and E2 the event of getting 2 in either of the throw, then P(E2/E1) is
Answer & explanation
Correct answer: option 4
Events favourable to E1 are:
(1, 5), (2, 4), (3, 3), (4, 2) and (1, 5)
Events E2 are (2, 4), (4, 2).
P(E2 / E1) = Probability of E2 when E1 has already occurred = $\frac{2}{5}$
Hence (D) is the correct answer.