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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $1, 2, 3, 	ext{ or } 4$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

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

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

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

For the random variable $X$ with the probability distribution given by the table:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $K$ $K + \frac{1}{7}$ $2K$ $\frac{2}{5}$

The mean of $X$ is:

If the probability density function (p.d.f.) of a continuous random variable $X$ is given by $f(x) = \begin{cases} k(9 + 8x - x^2), & \text{for } -1 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$, then the value of $k$ is:

The number of persons joining a cinema ticket counter in a minute follows a Poisson distribution with parameter $\lambda = 6$. The probability that at least one and at most five persons join the queue in a particular minute is:

Two dice are rolled. If a random variable $X$ is defined as the absolute difference of the two numbers that appear on them, then the mean of $X$ is

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{1}{2}$ if $0 < x < 2$ and $f(x) = 0$ otherwise. If $a = P(X < \frac{1}{2})$ and $b = P(X > \frac{3}{2})$,then the relation between $a$ and $b$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module