In a game of throwing 3 coins,a player loses | vedclass

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

Question

(A) When $3$ coins are tossed,the sample space $S$ contains $2^3 = 8$ outcomes: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.Let $H$ be the number

Vedclass · Accepted Answer

$\frac{15}{2}$

Vedclass · Answer

(A) When $3$ coins are tossed,the sample space $S$ contains $2^3 = 8$ outcomes: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.Let $H$ be the number of heads and $T$ be the number of tails. The net gain $X$ is given by $X = 10T - 5H$. Since $H + T = 3$,we have $T = 3 - H$.Thus,$X = 10(3 - H) - 5H = 30 - 15H$.Calculating $X$ for each outcome:- $HHH (H=3): X = 30 - 15(3) = -15$- $HHT, HTH, THH (H=2): X = 30 - 15(2) = 0$- $HTT, THT, TTH (H=1): X = 30 - 15(1) = 15$- $TTT (H=0): X = 30 - 15(0) = 30$The probability distribution is:$x$$-15$$0$$15$$30$$P(x)$$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$The mean $E(X) = \Sigma x P(x) = (-15 	imes \frac{1}{8}) + (0 	imes \frac{3}{8}) + (15 	imes \frac{3}{8}) + (30 	imes \frac{1}{8})$$E(X) = \frac{-15 + 0 + 45 + 30}{8} = \frac{60}{8} = \frac{15}{2}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$0.1$	$0.5$	$0.2$	$-0.1$	$0.3$

In a game of throwing $3$ coins,a player loses $₹ 5$ for each head and gains $₹ 10$ for each tail. If a random variable $X: S \rightarrow R$ is defined as $X(a) = \text{net gain } (a \in S)$,then the mean of the random variable is (in rupees):

Similar Questions

If $P(X=x)=k\left(\frac{3}{8}\right)^{X}, x=1,2,3, \ldots$ is the probability distribution function of a discrete random variable $X$,then $k=$

$A$ coin is tossed three times. If $X$ denotes the absolute difference between the number of heads and the number of tails,then $P(X=1) = $

State which of the following is not a probability distribution of a random variable. Give reasons for your answer.

$X$ $0$ $1$ $2$ $3$ $4$

$P(X)$ $0.1$ $0.5$ $0.2$ $-0.1$ $0.3$

In a communication network,$98 \%$ of messages are transmitted with no error. If a random variable $X$ denotes the number of incorrectly transmitted messages,then the probability that at most one message is transmitted incorrectly out of $500$ messages sent,is

Let two cards be drawn at random from a pack of $52$ playing cards. Let $X$ be the number of aces obtained. Then the value of $E(X)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module