If X sim B(35, p) such that 7 P(X=0)=P(X=1),t | vedclass

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(B) Given $X \sim B(n, p)$ with $n=35$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. Given $7 P(X=0) = P(X=1)

Vedclass · Accepted Answer

$3125$

Vedclass · Answer

(B) Given $X \sim B(n, p)$ with $n=35$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. Given $7 P(X=0) = P(X=1)$. Substituting the values: $7 \binom{35}{0} p^0 q^{35} = \binom{35}{1} p^1 q^{34}$. $7 q = 35 p \implies q = 5p$. Since $p+q=1$,we have $p+5p=1 \implies 6p=1 \implies p = \frac{1}{6}$ and $q = \frac{5}{6}$. Now,we need to find $\frac{P(X=15)}{P(X=20)} = \frac{\binom{35}{15} p^{15} q^{20}}{\binom{35}{20} p^{20} q^{15}} = \frac{\binom{35}{15}}{\binom{35}{20}} \cdot \frac{q^5}{p^5} = \frac{\binom{35}{15}}{\binom{35}{15}} \cdot (\frac{q}{p})^5$. Since $\binom{35}{15} = \binom{35}{20}$,the ratio is $(q/p)^5 = (5)^5 = 3125$.

If $X \sim B(35, p)$ such that $7 P(X=0)=P(X=1)$,then the value of $\frac{P(X=15)}{P(X=20)}$ is equal to

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