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(B) An experiment succeeds twice as often as it fails. Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 2q$. Si

Vedclass · Accepted Answer

$\frac{496}{729}$

Vedclass · Answer

(B) An experiment succeeds twice as often as it fails. Let $p$ be the probability of success and $q$ be the probability of failure. Given $p = 2q$. Since $p + q = 1$,we have $2q + q = 1$,which implies $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$. We have $n = 6$ trials. Let $X$ be the number of successes,where $X \sim B(n, p)$. The required probability is $P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)$. Using the binomial distribution formula $P(X = k) = {^nC_k} p^k q^{n-k}$: $P(X = 4) = {^6C_4} (\frac{2}{3})^4 (\frac{1}{3})^2 = 15 	imes \frac{16}{81} 	imes \frac{1}{9} = \frac{240}{729}$. $P(X = 5) = {^6C_5} (\frac{2}{3})^5 (\frac{1}{3})^1 = 6 	imes \frac{32}{243} 	imes \frac{1}{3} = \frac{192}{729}$. $P(X = 6) = {^6C_6} (\frac{2}{3})^6 (\frac{1}{3})^0 = 1 	imes \frac{64}{729} 	imes 1 = \frac{64}{729}$. Adding these probabilities: $P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

An experiment succeeds twice as often as it fails. Then the probability,that in the next $6$ trials there will be at least $4$ successes,is

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