An experiment succeeds twice as often as it f | vedclass

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(A) Let $p$ be the probability of success and $q$ be the probability of failure. Given that the experiment succeeds twice as often as it fails,we have

Vedclass · Accepted Answer

$\frac{496}{729}$

Vedclass · Answer

(A) Let $p$ be the probability of success and $q$ be the probability of failure. Given that the experiment succeeds twice as often as it fails,we have $p = 2q$. Since $p + q = 1$,we get $2q + q = 1$,which implies $3q = 1$,so $q = \frac{1}{3}$ and $p = \frac{2}{3}$.For $n = 6$ trials,the number of successes $X$ follows a binomial distribution $B(n, p) = B(6, \frac{2}{3})$.The probability of at least $4$ successes is $P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$.Using the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:$P(X=4) = \binom{6}{4} (\frac{2}{3})^4 (\frac{1}{3})^2 = 15 	imes \frac{16}{81} 	imes \frac{1}{9} = \frac{240}{729}$.$P(X=5) = \binom{6}{5} (\frac{2}{3})^5 (\frac{1}{3})^1 = 6 	imes \frac{32}{243} 	imes \frac{1}{3} = \frac{192}{729}$.$P(X=6) = \binom{6}{6} (\frac{2}{3})^6 (\frac{1}{3})^0 = 1 	imes \frac{64}{729} 	imes 1 = \frac{64}{729}$.Summing these probabilities: $P(X \geq 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}$.

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

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