The probability of getting a success in a tri | vedclass

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(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 5q$. Since $p + q = 1$, we have $5q + q = 1$, which implie

Vedclass · Accepted Answer

$\frac{26}{6^5}$

Vedclass · Answer

(B) Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 5q$. Since $p + q = 1$, we have $5q + q = 1$, which implies $6q = 1$, so $q = \frac{1}{6}$ and $p = \frac{5}{6}$.For $n = 5$ trials, the probability of getting at most one success is given by $P(X \le 1) = P(X = 0) + P(X = 1)$.Using the binomial distribution formula $P(X = k) = ^nC_k p^k q^{n-k}$:$P(X = 0) = ^5C_0 \left(\frac{5}{6}ight)^0 \left(\frac{1}{6}ight)^5 = 1 	imes 1 	imes \frac{1}{6^5} = \frac{1}{6^5}$.$P(X = 1) = ^5C_1 \left(\frac{5}{6}ight)^1 \left(\frac{1}{6}ight)^4 = 5 	imes \frac{5}{6} 	imes \frac{1}{6^4} = \frac{25}{6^5}$.Therefore, $P(X \le 1) = \frac{1}{6^5} + \frac{25}{6^5} = \frac{26}{6^5}$.Thus, option $(b)$ is correct.

The probability of getting a success in a trial is five times that of a failure. The probability of getting at most one success in $5$ trials is:

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