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 $= (\frac{1}{3} 	imes 100) + (\frac{2}{9} 	imes 50) + (\frac{4}{27} 	imes 0) + (\frac{8}{27} 	imes -150) = \frac{100}{3} + \frac{100}{9} + 0 - \frac{1200}{27} = \frac{900 + 300 - 1200}{27} = 0$.

Outcome	Probability
$\omega_{1}$	$1/8$
$\omega_{2}$	$2/3$
$\omega_{3}$	$1/3$
$\omega_{4}$	$1/3$
$\omega_{5}$	$-1/4$
$\omega_{6}$	$-1/3$

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

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

Similar Questions

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome Probability

$\omega_{1}$ $1/8$

$\omega_{2}$ $2/3$

$\omega_{3}$ $1/3$

$\omega_{4}$ $1/3$

$\omega_{5}$ $-1/4$

$\omega_{6}$ $-1/3$

If the $c.d.f.$ (cumulative distribution function) is given by $F(x) = \frac{x-25}{10}$,then $P(27 \leq x \leq 33) = \_\_\_\_$

For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

$P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$

the standard deviation $(\sigma)$ is

$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$,then the mean of the random variable $X$ is

Let the mean and standard deviation of the probability distribution given by the table below be $\mu$ and $\sigma$ respectively. If $\sigma - \mu = 2$,then find the value of $\sigma$.
$X=x$ $-3$ $0$ $1$ $\alpha$
$P(X=x)$ $\frac{1}{4}$ $K$ $\frac{1}{4}$ $\frac{1}{3}$

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