The correct option (a) (5/11)
Explanation:
probability of getting 1 in each trial = (1/6)
probability of not getting 1 = (5/6)
P[getting 1 in even chances] = P[getting 1 in 2nd, 4th or 6th toss]
= (5/6)(1/6) + (5/6)3(1/6) + (5/6)5(1/6) + ........
= (1/6)[(5/6) + (5/6)3 + ........ ∞]
= (1/6)[sum of infinite G.P.].... Sn = [a/(1 – r)]
= (1/6)[(5/6) / {1 – (25/36)}]
= (1/6)[(5/6) × (36/11)]
= (5/11).
