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(C) When a coin is tossed $3$ times,the total number of outcomes is $2^3 = 8$. Let $X$ be the number of heads obtained. $X$ follows a binomial distrib

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$\frac{3}{4}$

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(C) When a coin is tossed $3$ times,the total number of outcomes is $2^3 = 8$. Let $X$ be the number of heads obtained. $X$ follows a binomial distribution $B(n, p)$ where $n = 3$ and $p = \frac{1}{2}$. The probability of getting exactly $r$ heads is given by $P(X = r) = {}^nC_r p^r q^{n-r}$,where $q = 1 - p = \frac{1}{2}$. We need to find the probability of getting exactly $1$ head or $2$ heads,which is $P(X = 1) + P(X = 2)$. $P(X = 1) = {}^3C_1 \left( \frac{1}{2} ight)^1 \left( \frac{1}{2} ight)^2 = 3 	imes \frac{1}{8} = \frac{3}{8}$. $P(X = 2) = {}^3C_2 \left( \frac{1}{2} ight)^2 \left( \frac{1}{2} ight)^1 = 3 	imes \frac{1}{8} = \frac{3}{8}$. Therefore,the required probability is $P(X = 1) + P(X = 2) = \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4}$.

$A$ coin is tossed successively three times. The probability of getting exactly one head or $2$ heads is:

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