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.
Let p and q respectively be the probabilities of getting a six and not getting a six.
In a throw of a die the probability of getting a six is given by,
p = 1 6
The probability of not getting a six is given by,
q = 5 6
Three cases will occur.
Case 1:
If he gets a six in first throw.
The probability of getting a six is given by,
p = 1 6
Amount received is Rs 1.
Case-2:
If he does not get a six in the first throw and gets in the second throw.
The probability of getting a six is given by,
= 1 6 × 5 6 = 5 36
Amount received is given by,
Rs 1 − Rs 1 = 0 Rs
Case-3:
If he does not get a six in the first two throws and gets in the third throw.
The probability of getting a six is given by,
= 1 6 × 5 6 × 5 6 = 25 216
Amount received is given by,
Rs 1 − Rs 1 − Rs 1 = − 1 Rs
The expected value he can win,
E = 1 6 ( 1 ) + ( 5 6 × 1 6 ) ( 0 ) + ( ( 5 6 ) 2 × 1 6 ) ( − 1 ) = 1 6 − 25 216 = 36 − 25 216 = 11 216
Thus, the expected value he can win is 11 216 .
Let
In a throw of a die the probability of getting a six is given by,
The probability of not getting a six is given by,
Three cases will occur.
Case 1:
If he gets a six in first throw.
The probability of getting a six is given by,
Amount received is Rs 1.
Case-2:
If he does not get a six in the first throw and gets in the second throw.
The probability of getting a six is given by,
Amount received is given by,
Case-3:
If he does not get a six in the first two throws and gets in the third throw.
The probability of getting a six is given by,
Amount received is given by,
The expected value he can win,
Thus, the expected value he can win is
