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 - Mathematics

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

Concept: Conditional Probability
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