A 1-rupee coin,a 2-rupee coin,a 5-rupee coin, | vedclass

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

Question

(D) Let $X_1, X_2, X_3, X_4$ be the values of the coins $1, 2, 5, 10$ respectively. Let $I_k$ be an indicator random variable such that $I_k = 1$ if t

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let $X_1, X_2, X_3, X_4$ be the values of the coins $1, 2, 5, 10$ respectively. Let $I_k$ be an indicator random variable such that $I_k = 1$ if the $k$-th coin shows heads and $I_k = 0$ if it shows tails. The probability of getting heads for each coin is $P(I_k = 1) = \frac{1}{2}$. The expected value of each indicator variable is $E[I_k] = 1 	imes \frac{1}{2} + 0 	imes \frac{1}{2} = \frac{1}{2}$. The total sum $S$ of the values of the coins showing heads is $S = 1 \cdot I_1 + 2 \cdot I_2 + 5 \cdot I_3 + 10 \cdot I_4$. By the linearity of expectation,$E[S] = 1 \cdot E[I_1] + 2 \cdot E[I_2] + 5 \cdot E[I_3] + 10 \cdot E[I_4]$. $E[S] = 1 	imes \frac{1}{2} + 2 	imes \frac{1}{2} + 5 	imes \frac{1}{2} + 10 	imes \frac{1}{2}$. $E[S] = \frac{1+2+5+10}{2} = \frac{18}{2} = 9$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x_i)$	$\alpha$	$\alpha$	$\alpha$	$\beta$	$\beta$	$0.3$

$A$ $1$-rupee coin,a $2$-rupee coin,a $5$-rupee coin,and a $10$-rupee coin are tossed simultaneously. The expected value of the sum of the values of the coins that show heads up is:

Similar Questions

The mean of the numbers obtained on throwing a die having written $1$ on three faces,$2$ on two faces,and $5$ on one face is

$A$ fair die is tossed twice in succession. If $X$ denotes the number of sixes in two tosses,then the probability distribution of $X$ is given by

The probability distribution of a random variable $X$ is given below:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x_i)$ $\alpha$ $\alpha$ $\alpha$ $\beta$ $\beta$ $0.3$

If $\mu$ and $\sigma^2$ represent the mean and variance of $X$ and $\mu=4.2$,then $\sigma^2+\mu^2=$

If $X$ is a Poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$,then the mean of $X$ is

If the probability mass function (p.m.f.) of a discrete random variable $X$ is given by $P(X=x) = \frac{c}{x^3}$ for $x = 1, 2, 3$ and $0$ otherwise,then $E(X)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module