A player tosses 2 fair coins. He wins ₹ 5 if | vedclass

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

Question

(C) When a player tosses $2$ fair coins,the sample space is $S = \{HH, HT, TH, TT\}$.Let $X$ be a random variable that denotes the amount received by

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(C) When a player tosses $2$ fair coins,the sample space is $S = \{HH, HT, TH, TT\}$.Let $X$ be a random variable that denotes the amount received by the player.The possible values for $X$ are $5, 2,$ and $1$.The probabilities are:$P(X=5) = P(\{HH\}) = \frac{1}{4}$$P(X=2) = P(\{HT, TH\}) = \frac{2}{4} = \frac{1}{2}$$P(X=1) = P(\{TT\}) = \frac{1}{4}$The probability distribution is:$E(X) = \sum X P(X) = 5 	imes \frac{1}{4} + 2 	imes \frac{1}{2} + 1 	imes \frac{1}{4} = \frac{5}{4} + 1 + \frac{1}{4} = \frac{10}{4} = 2.5$$E(X^2) = \sum X^2 P(X) = 5^2 	imes \frac{1}{4} + 2^2 	imes \frac{1}{2} + 1^2 	imes \frac{1}{4} = \frac{25}{4} + 2 + \frac{1}{4} = \frac{34}{4} = 8.5$$	ext{Variance}(X) = E(X^2) - [E(X)]^2 = \frac{34}{4} - \left(\frac{10}{4}ight)^2 = \frac{34}{4} - \frac{100}{16} = \frac{136 - 100}{16} = \frac{36}{16} = \frac{9}{4}$.

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

$A$ player tosses $2$ fair coins. He wins ₹ $5$ if $2$ heads appear,₹ $2$ if one head appears,and ₹ $1$ if no head appears. Then the variance of his winning amount in ₹ is

Similar Questions

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$

The cumulative distribution function of a discrete random variable $X$ is given by the following table:
$X = x$ $-4$ $-2$ $0$ $2$ $4$ $6$ $8$ $10$
$F(X = x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.

$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ with probability $P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x$,where $k$ is a constant. Then $P(X=0)$ is

Bismuth has a half-life of $5$ days. If a sample originally has a mass of $800 \text{ mg}$, then the mass remaining after $30$ days will be: (in $\text{ mg}$)

The p.d.f. of a continuous random variable $X$ is $f(x)=\begin{cases} \frac{x^2}{18} & \text{if } -3 < x < 3 \\ 0 & \text{otherwise} \end{cases}$. Then $P[|X| < 2]=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module