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(D) Let $X$ denote the number of successes in $3$ trials. Since the die is tossed $3$ times,$n=3$.The probability of getting an even number (success)

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$\frac{1}{2}$

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(D) Let $X$ denote the number of successes in $3$ trials. Since the die is tossed $3$ times,$n=3$.The probability of getting an even number (success) in a single toss is $p = \frac{3}{6} = \frac{1}{2}$.The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.$X$ follows a binomial distribution $B(n, p)$,where $P(X=r) = { }^n C_r p^r q^{n-r}$.We need to find the probability of getting at least $2$ successes,which is $P(X \geq 2) = P(X=2) + P(X=3)$.$P(X=2) = { }^3 C_2 \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^{3-2} = 3 	imes \left(\frac{1}{2}ight)^3 = \frac{3}{8}$.$P(X=3) = { }^3 C_3 \left(\frac{1}{2}ight)^3 \left(\frac{1}{2}ight)^{3-3} = 1 	imes \left(\frac{1}{2}ight)^3 = \frac{1}{8}$.Therefore,$P(X \geq 2) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

$A$ die is tossed thrice. If the event of getting an even number is a success,then the probability of getting at least $2$ successes is

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