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(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 3q$. Since $p + q = 1$,we have $3q + q = 1$,which implies

Vedclass · Accepted Answer

$\frac{81}{128}$

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(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 3q$. Since $p + q = 1$,we have $3q + q = 1$,which implies $4q = 1$,so $q = \frac{1}{4}$ and $p = \frac{3}{4}$.For a binomial distribution with $n = 5$ trials,the probability of $x$ successes is given by $P(X = x) = {}^nC_x p^x q^{n-x}$.We need to find the probability of at least $4$ successes,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.$P(X = 4) = {}^5C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 \cdot \frac{81}{256} \cdot \frac{1}{4} = \frac{405}{1024}$.$P(X = 5) = {}^5C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 \cdot \frac{243}{1024} \cdot 1 = \frac{243}{1024}$.Thus,$P(X \ge 4) = \frac{405}{1024} + \frac{243}{1024} = \frac{648}{1024} = \frac{81}{128}$.

The probability of securing a success in a trial is three times that of a failure. The probability of getting at least $4$ successes in $5$ trials is

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