In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses

#### Solution

The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equatl to 4 in the first throw and in the second throw he may get the number greater than 4 and quits the game.

In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.

Let X be the amount he wins/looses.

Then, X can take values -3, 3, 4, 5 such that

P (X = 5) = P(Getting number greater than 4 in first throw) = 1/3

P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) `= 4/6×2/6=2/9`

P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) `= 4/6×4/6×2/6=4/27`

P (X = -3) = P(Getting number less than equal to 4 in all three throws) `= 4/6×4/6×4/6=8/27`

X | 5 | 4 | 3 | -3 |

P(X) |
`1/3` |
`2/9` |
`4/27` |
`8/27` |

`E (X) = (5×1/3)+4 (2/9)+3(4/27)−3 (8/27)`

`=1/27(45+24+12−24)`

`=57/27`