A random experiment is conducted five times. | vedclass

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(A) Given that the experiment is conducted $n = 5$ times. Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure. The mea

Vedclass · Accepted Answer

$\frac{64}{81}$

Vedclass · Answer

(A) Given that the experiment is conducted $n = 5$ times. Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure. The mean of a binomial distribution is $np$ and the variance is $npq$. Given $np - npq = \frac{5}{9}$. Substituting $n = 5$: $5p - 5pq = \frac{5}{9} \implies p - pq = \frac{1}{9}$. Since $1 - q = p$, we have $p(1 - q) = p^2 = \frac{1}{9}$, which gives $p = \frac{1}{3}$. Thus, $q = 1 - \frac{1}{3} = \frac{2}{3}$. The probability of getting at most two successes is $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$. Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$: $P(X = 0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = 1 	imes 1 	imes \frac{32}{243} = \frac{32}{243}$. $P(X = 1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 	imes \frac{1}{3} 	imes \frac{16}{81} = \frac{80}{243}$. $P(X = 2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 	imes \frac{1}{9} 	imes \frac{8}{27} = \frac{80}{243}$. Summing these probabilities: $P(X \leq 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243} = \frac{64}{81}$.

$A$ random experiment is conducted five times. If the number of successes of the experiment follows a binomial distribution such that the difference between the mean and variance of the successes is $\frac{5}{9}$, then the probability of getting at most two successes is

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