In a game,a man wins ₹ 40 if he gets 5 or 6 o | vedclass

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(C) Let $S$ be the event of getting $5$ or $6$ (success) and $F$ be the event of getting $1, 2, 3,$ or $4$ (failure).Probability of success $P(S) = \f

Vedclass · Accepted Answer

$0$

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(C) Let $S$ be the event of getting $5$ or $6$ (success) and $F$ be the event of getting $1, 2, 3,$ or $4$ (failure).Probability of success $P(S) = \frac{2}{6} = \frac{1}{3}$.Probability of failure $P(F) = 1 - \frac{1}{3} = \frac{2}{3}$.The game stops if he gets $S$ or after $3$ throws.Possible outcomes:$1$. Success on 1st throw: $S$. Probability $P_1 = \frac{1}{3}$. Gain = $₹ 40$.$2$. Success on 2nd throw: $FS$. Probability $P_2 = \frac{2}{3} 	imes \frac{1}{3} = \frac{2}{9}$. Gain = $40 - 20 = ₹ 20$.$3$. Success on 3rd throw: $FFS$. Probability $P_3 = \frac{2}{3} 	imes \frac{2}{3} 	imes \frac{1}{3} = \frac{4}{27}$. Gain = $40 - 20 - 20 = ₹ 0$.$4$. Failure on all $3$ throws: $FFF$. Probability $P_4 = \frac{2}{3} 	imes \frac{2}{3} 	imes \frac{2}{3} = \frac{8}{27}$. Gain = $-20 - 20 - 20 = -₹ 60$.Expected gain $E = (40 	imes \frac{1}{3}) + (20 	imes \frac{2}{9}) + (0 	imes \frac{4}{27}) + (-60 	imes \frac{8}{27})$.$E = \frac{40}{3} + \frac{40}{9} + 0 - \frac{480}{27} = \frac{360 + 120 - 480}{27} = \frac{0}{27} = 0$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

In a game,a man wins $₹ 40$ if he gets $5$ or $6$ on a throw of a fair die and loses $₹ 20$ for getting any other number on the die. If he decides to throw the die either until he gets a $5$ or $6$ or to a maximum of $3$ throws,then his expected gain/loss (in rupees) is:

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