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A person throws two fair dice. He wins 15 for throwing a double (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/ loss (in Rs. of the person is

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$\frac{1}{4}$ loss

$\frac{1}{2}$ gain

$\frac{1}{2}$ loss

2 gain

answer is B.

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Detailed Solution

Number of cases for which the person

Wins 15 Rs = 6

Wins 12 Rs = 4

Loss 6 Rs = 26

X	15	12	-6
P(X=x)	$\frac{6}{36}$	$\frac{4}{36}$	$\frac{26}{36}$

$E (x) = \sum x . P (X = x)$

$= 15 \frac{(6)}{36} + 12 \times \frac{4}{36} - 6 \times \frac{26}{36} = \frac{90}{36} + \frac{48}{36} - \frac{156}{36} = \frac{90 + 48 - 156}{36} = \frac{- 18}{36} = \frac{- 1}{2}$

$∴$ The expected loss is $\frac{1}{2}$

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