In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

When a die is thrown, then probability of getting a six =16then, probability of not getting a six =1−16=56If the man gets a six in the first throw, thenprobability of getting a six =16If he does not get a six in first throw, but gets a six in second throw, thenprobability of getting a six in the second throw =56×16=536If he does not get a six in the first two throws, but gets in the third throw, thenprobability of getting a six in the third throw =56×56×16=25216probability that he does not get a six in any of the three throws =56×56×56=125216In the first throw he gets a six, then he will receive Re 1.If he gets a six in the second throw, then he will receive Re (1 - 1) = 0If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1),that means he will lose Re 1 in this case.Expected value =16×1+(56×16)×0+(56×56×16)×(−1)=11216So, he will loose Rs11216. read less