A dice is thrown (2n + 1) times. The probabil | vedclass

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(A) Let $X$ be the number of times $1, 3$ or $4$ occur on the die.Then $X$ follows a binomial distribution with parameters $N = 2n + 1$ and $p = \frac

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$\frac{1}{2}$

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(A) Let $X$ be the number of times $1, 3$ or $4$ occur on the die.Then $X$ follows a binomial distribution with parameters $N = 2n + 1$ and $p = \frac{3}{6} = \frac{1}{2}$.We have $q = 1 - p = \frac{1}{2}$.The probability of getting $1, 3$ or $4$ at most $n$ times is $P(X \le n) = \sum_{k=0}^{n} {}^{2n+1}C_k p^k q^{2n+1-k}$.Since $p = q = \frac{1}{2}$,this becomes $P(X \le n) = \sum_{k=0}^{n} {}^{2n+1}C_k (\frac{1}{2})^{2n+1}$.Let $S = \sum_{k=0}^{n} {}^{2n+1}C_k$. We know that $\sum_{k=0}^{2n+1} {}^{2n+1}C_k = 2^{2n+1}$.Since ${}^{2n+1}C_k = {}^{2n+1}C_{2n+1-k}$,we have $2S = \sum_{k=0}^{n} {}^{2n+1}C_k + \sum_{k=n+1}^{2n+1} {}^{2n+1}C_k = 2^{2n+1}$.Thus,$S = 2^{2n}$.Therefore,the required probability is $S 	imes (\frac{1}{2})^{2n+1} = 2^{2n} 	imes \frac{1}{2^{2n+1}} = \frac{1}{2}$.

$A$ dice is thrown $(2n + 1)$ times. The probability of getting $1, 3$ or $4$ at most $n$ times is:

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