0 0 votes Two players, $A$ and $B$, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player $A$ starts the game, the probability that $A$ wins the game is $5/11$ $1/2$ $7/13$ $6/11$ Probability & Statistics gate2015-ee-1 probability-and-statistics conditional-probability + – Misbah Ghaya 738 views answer comment Share Follow 0 reply Please log in or register to add a comment.
1 1 vote since the player A only wins when he gets ‘6’ and player B must not get 6. so the probability(P(a)) of player A getting 6 is 1/6. the probability(P(b)) of player B not getting 6 is 5/6. The probability(1-P(a)) of player A not getting 6 is 5/6. the final Probability(P) is : P=P(a)+(1-P(a)).P(b).P(a)+……… (because for every alternative Player A must-win such that P(a) is taken) by solving we get P=6/11. dheeraj2310 answered Aug 12, 2020 dheeraj2310 comment Share Follow 0 reply Please log in or register to add a comment.
