Find the mean number of heads in three tosses | vedclass

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(B) Let $X$ denote the number of heads obtained in three tosses of a fair coin.The sample space $S$ is given by:$S = \{HHH, HHT, HTH, HTT, THH, THT, T

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$1.5$

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(B) Let $X$ denote the number of heads obtained in three tosses of a fair coin.The sample space $S$ is given by:$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$The random variable $X$ can take values $0, 1, 2, 3$.Calculating the probabilities for each value of $X$:$P(X=0) = P(TTT) = \frac{1}{8}$$P(X=1) = P(HHT) + P(HTH) + P(THH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$$P(X=2) = P(HHT) + P(HTH) + P(THH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$$P(X=3) = P(HHH) = \frac{1}{8}$The probability distribution is:$X$$P(X)$$0$$1/8$$1$$3/8$$2$$3/8$$3$$1/8$The mean $E(X)$ is calculated as:$E(X) = \sum X_i P(X_i)$$E(X) = (0 	imes \frac{1}{8}) + (1 	imes \frac{3}{8}) + (2 	imes \frac{3}{8}) + (3 	imes \frac{1}{8})$$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$$E(X) = \frac{12}{8} = 1.5$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$K$	$2K$	$3K$	$4K$	$5K$	$6K$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\lambda$	$2\lambda$	$3\lambda$	$4\lambda$

Find the mean number of heads in three tosses of a fair coin.

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$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$