wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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.

Open in App
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) = 13
​P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) = 46×26=29
​P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) = 46×46×26=427
​P (X = -3) = P(Getting number less than equal to 4 in all three throws) = 46×46×46=827
X 5 4 3 -3
P(X) 13 29 427 827

E (X) = 5×13+4 29+3427-3 827=12745+24+12-24=5727
Expected value of the amount he wins/loses is 5727.

flag
Suggest Corrections
thumbs-up
16
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Probability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon