An experiment succeeds twice as often as it f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The minimum number of times a fair coin needs to be tossed so that the probability of getting at least two heads is at least $0.96$ is:

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

If a fair coin is tossed $8$ times, then the probability of getting at most $2$ heads is...

Let $X$ be a random variable having a Binomial distribution $B(7, p)$. If $P(X=3) = 5 P(X=4)$,then the variance of $X$ is:

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Let $X$ denote the random variable of the number of jacks obtained in the two drawn cards. Then $P(X=1) + P(X=2)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module