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.
pls xplain the steps
When a die is thrown, probability of getting a six P= 1/6
Probability of not getting six Q=1-1/6 =5/6
(1) If he gets a six in first throw then probability of getting a six = 1/6
(2) If he does not get a six in first throw but he gets a six in second throw, its probability=5/6* 1/6 =5/36
(3) Probability that he does not get a six in first two throws and he get a six in third throw =5/6 *5/6 *1/6 = 25/216
Probability that he does not get a six in any of the three throws = (5/6)3 =125/216
In first throw if he gets a six, he will receive Rs 1
If he gets a six in second throw, he will receive Rs (1-1) =0
If he gets a six in third throw, he will receive = Rs (-1-1+1) = Rs (-1) = he will lose Rs 1.
Expected value is, he will loose Rs 11/216.