If in 6 trials, X is a binomial random variab | vedclass

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(D) For a binomial distribution, $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $n = 6$, $p$ is the probability of success, and $q = 1 - p$ is the proba

Vedclass · Accepted Answer

$0.75$

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(D) For a binomial distribution, $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $n = 6$, $p$ is the probability of success, and $q = 1 - p$ is the probability of failure.Given $9P(X = 4) = P(X = 2)$, we have $9 \binom{6}{4} p^4 q^2 = \binom{6}{2} p^2 q^4$.Since $\binom{6}{4} = \binom{6}{2} = 15$, the equation simplifies to $9 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$.Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$), we get $9p^2 = q^2$.Taking the square root of both sides, $3p = q$.Since $p + q = 1$, substitute $q = 3p$ into the equation: $p + 3p = 1 \implies 4p = 1 \implies p = 0.25$.The probability of failure $q = 1 - p = 1 - 0.25 = 0.75$.

If in $6$ trials, $X$ is a binomial random variable which follows the relation $9P(X = 4) = P(X = 2)$, then the probability of failure is...

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