In a game,a man wins Rs. 100 if he gets 5 or | vedclass

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(B) Let $w$ be the probability of getting $5$ or $6$,so $w = \frac{2}{6} = \frac{1}{3}$.Let $L$ be the probability of getting $1, 2, 3, 	ext{ or } 4$

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(B) Let $w$ be the probability of getting $5$ or $6$,so $w = \frac{2}{6} = \frac{1}{3}$.Let $L$ be the probability of getting $1, 2, 3, 	ext{ or } 4$,so $L = \frac{4}{6} = \frac{2}{3}$.The game stops if he gets $5$ or $6$ or after $3$ throws.Case $1$: Wins on $1^{st}$ throw: Probability $= w = \frac{1}{3}$,Gain $= 100$.Case $2$: Wins on $2^{nd}$ throw: Probability $= L 	imes w = \frac{2}{3} 	imes \frac{1}{3} = \frac{2}{9}$,Gain $= -50 + 100 = 50$.Case $3$: Wins on $3^{rd}$ throw: Probability $= L^2 	imes w = (\frac{2}{3})^2 	imes \frac{1}{3} = \frac{4}{27}$,Gain $= -50 - 50 + 100 = 0$.Case $4$: Loses on all $3$ throws: Probability $= L^3 = (\frac{2}{3})^3 = \frac{8}{27}$,Gain $= -50 - 50 - 50 = -150$.Expected Value $= (\frac{1}{3} 	imes 100) + (\frac{2}{9} 	imes 50) + (\frac{4}{27} 	imes 0) + (\frac{8}{27} 	imes -150) = \frac{100}{3} + \frac{100}{9} + 0 - \frac{1200}{27} = \frac{900 + 300 - 1200}{27} = 0$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

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 until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is

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