If a random variable X follows the Binomial d | vedclass

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(A) Given $n = 33$,let the probability of success be $p$ and $q = 1 - p$.Given $3P(X=0) = P(X=1)$.Using the Binomial distribution formula $P(X=k) = {}

Vedclass · Accepted Answer

$1320$

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(A) Given $n = 33$,let the probability of success be $p$ and $q = 1 - p$.Given $3P(X=0) = P(X=1)$.Using the Binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:$3 \cdot {}^{33}C_{0} p^{0} q^{33} = {}^{33}C_{1} p^{1} q^{32}$$3q = 33p \implies q = 11p$.Since $p + q = 1$,we have $p + 11p = 1 \implies 12p = 1 \implies p = \frac{1}{12}$ and $q = \frac{11}{12}$.Thus,$\frac{q}{p} = 11$.Now,consider the expression $\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)}$.Using the formula $\frac{P(X=k)}{P(X=k+1)} = \frac{{}^{n}C_{k} p^{k} q^{n-k}}{{}^{n}C_{k+1} p^{k+1} q^{n-k-1}} = \frac{k+1}{n-k} \cdot \frac{q}{p}$,we evaluate the terms:$\frac{P(X=15)}{P(X=18)} = \frac{P(X=15)}{P(X=16)} \cdot \frac{P(X=16)}{P(X=17)} \cdot \frac{P(X=17)}{P(X=18)} = \left(\frac{16}{18} \cdot 11\right) \cdot \left(\frac{17}{17} \cdot 11\right) \cdot \left(\frac{18}{16} \cdot 11\right) = 11^3 = 1331$.Alternatively,$\frac{P(X=k)}{P(X=n-k)} = \frac{{}^{n}C_{k} p^{k} q^{n-k}}{{}^{n}C_{n-k} p^{n-k} q^{k}} = \left(\frac{q}{p}\right)^{n-2k}$.For the first term: $\frac{P(X=15)}{P(X=18)} = \frac{{}^{33}C_{15} p^{15} q^{18}}{{}^{33}C_{18} p^{18} q^{15}} = \left(\frac{q}{p}\right)^{18-15} = \left(\frac{q}{p}\right)^3 = 11^3 = 1331$.For the second term: $\frac{P(X=16)}{P(X=17)} = \frac{{}^{33}C_{16} p^{16} q^{17}}{{}^{33}C_{17} p^{17} q^{16}} = \left(\frac{q}{p}\right)^{17-16} = \left(\frac{q}{p}\right)^1 = 11$.Therefore,the value is $1331 - 11 = 1320$.

If a random variable $X$ follows the Binomial distribution $B(33, p)$ such that $3P(X=0) = P(X=1)$,then the value of $\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)}$ is equal to

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