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InterviewSolution
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In a game, a man wins Rs 100 if he gets 5 or 6 on a throw of a fair die and loses Rs 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is: (a) `400/3` gain (b) `400/9` loss (c) 0 (d) `400/3` lossA. `(400)/(3)` lossB. `(400)/(9)` lossC. 0D. `(400)/(3)` gain |
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Answer» Correct Answer - C Let p and q represents the probability of success and failure in a trial respectively. Then, p=P(5 or 6)`=(2)/(6)=(1)/(3)` and `q=1-p=(4)/(6)=(2)/(3)`. Now, as the man decides to throw the die either till he gets a five or a six or to a maximum of three throws, so he can get the success in first, second and third throw or not get the success in any of the three throws. So, the expected gain/loss (in ₹) `=(pxx100)+qp(-50+100)+q^(2)p(-50-50+100)+q^(3)(-50-50-50)` `=((1)/(3)xx100)+((2)/(3)xx(1)/(3))(50)+((2)/(3))^(2)((1)/(3))(0)+((2)/(3))^(3)(-150)` `(100)/(3)+(100)/(9)+0-(1200)/(27)` `=(900+300-1200)/(27)=(1200-1200)/(27)=0` |
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