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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/losss (in rupees) is

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$\frac{400}{3}$ gain

$\frac{400}{3}$ loss

$\frac{400}{9}$ loss

answer is B.

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Detailed Solution

Let A=event that 5 or 6 appear on die

B=event that any other number appear on die

$∴$ following events are possible

A, BA, BBA, BBB

Let X be the random variable denoting the amount won.

X	+100	+50	0	-150
P(x)	$\frac{1}{3}$	$\frac{2}{9}$	$\frac{4}{27}$	$\frac{8}{27}$

Mean $= \sum_{} xP (X)$

$= \frac{100}{3} + \frac{100}{9} + 0 - \frac{1200}{27} = 0$

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