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(A) Let $S$ denote getting a six and $N$ denote not getting a six. The probabilities are $P(S) = 1/6$ and $P(N) = 5/6$.Case $1$: He gets a six in the

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(A) Let $S$ denote getting a six and $N$ denote not getting a six. The probabilities are $P(S) = 1/6$ and $P(N) = 5/6$.Case $1$: He gets a six in the first throw $(S)$.Probability $= 1/6$.Amount won $= +1$.Case $2$: He gets a six in the second throw $(NS)$.Probability $= (5/6) 	imes (1/6) = 5/36$.Amount won $= -1 + 1 = 0$.Case $3$: He gets a six in the third throw $(NNS)$.Probability $= (5/6) 	imes (5/6) 	imes (1/6) = 25/216$.Amount won $= -1 - 1 + 1 = -1$.Case $4$: He does not get a six in any of the three throws $(NNN)$.Probability $= (5/6)^3 = 125/216$.Amount won $= -1 - 1 - 1 = -3$.Expected Value $E = (1/6)(1) + (5/36)(0) + (25/216)(-1) + (125/216)(-3)$.$E = 1/6 - 25/216 - 375/216$.$E = (36 - 25 - 375) / 216 = -364 / 216 = -91 / 54$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$k-1$	$3k$	$k$	$3k$	$3k^2$	$k^2$	$k^2+k$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$X=x$	$1$	$2$	$3$
$P(X=x)$	$1/5$	$2/5$	$2/5$

In a game,a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decides to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins or loses. (in $/216$)

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In a game,a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decides to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins or loses. (in $/216$)

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