A die is thrown (2n+1) times. The probability | vedclass

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(D) Let $X$ be the number of times we get $1, 3,$ or $4$ in $(2n+1)$ throws. The probability of success $p$ in a single throw is $p = \frac{3}{6} = \f

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Let $X$ be the number of times we get $1, 3,$ or $4$ in $(2n+1)$ throws. The probability of success $p$ in a single throw is $p = \frac{3}{6} = \frac{1}{2}$.The probability of failure $q$ is $q = 1 - p = \frac{1}{2}$.Since the die is thrown $(2n+1)$ times,$X$ follows a binomial distribution $B(2n+1, \frac{1}{2})$.We want to find $P(X \le n) = \sum_{k=0}^{n} \binom{2n+1}{k} (\frac{1}{2})^{2n+1}$.Using the property of binomial coefficients,$\sum_{k=0}^{n} \binom{2n+1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = \frac{1}{2} 	imes 2^{2n+1} = 2^{2n}$.Thus,$P(X \le n) = 2^{2n} 	imes (\frac{1}{2})^{2n+1} = \frac{2^{2n}}{2^{2n+1}} = \frac{1}{2}$.

$A$ die is thrown $(2n+1)$ times. The probability of getting $1$ or $3$ or $4$ at most $n$ times on it is

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