A person throws two fair dice. He wins Rs., 1 | vedclass

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(D) The total number of outcomes when throwing two dice is $6 	imes 6 = 36$. $1$. For a doublet: The outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5),

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$\frac{1}{2}$ loss

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(D) The total number of outcomes when throwing two dice is $6 	imes 6 = 36$. $1$. For a doublet: The outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. There are $6$ such outcomes. Probability $P(	ext{doublet}) = \frac{6}{36}$. Gain $= +15$. $2$. For a sum of $9$: The outcomes are $(3,6), (4,5), (5,4), (6,3)$. There are $4$ such outcomes. Probability $P(	ext{sum } 9) = \frac{4}{36}$. Gain $= +12$. $3$. For any other outcome: The number of outcomes is $36 - (6 + 4) = 26$. Probability $P(	ext{other}) = \frac{26}{36}$. Loss $= -6$. Expected value $E(X) = \sum x_i p_i = (15 	imes \frac{6}{36}) + (12 	imes \frac{4}{36}) + (-6 	imes \frac{26}{36})$ $E(X) = \frac{90}{36} + \frac{48}{36} - \frac{156}{36} = \frac{90 + 48 - 156}{36} = \frac{-18}{36} = -\frac{1}{2}$. Since the result is negative,it represents a loss of $\frac{1}{2}$ $Rs$. Thus,the correct option is $(D)$.

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$A$ person throws two fair dice. He wins $Rs.\, 15$ for throwing a doublet (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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