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

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(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 numbe

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$\frac{15}{2}$

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(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}$.

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

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):

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