The probability that a certain kind of compon | vedclass

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

Question

(C) Let $X$ be the number of components that survive the test. Here,$n = 4$ and $p = \frac{2}{3}$.Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.$X$

Vedclass · Accepted Answer

$\frac{33}{3^4}$

Vedclass · Answer

(C) Let $X$ be the number of components that survive the test. Here,$n = 4$ and $p = \frac{2}{3}$.Then $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.$X$ follows a binomial distribution $B(n, p)$,so $P(X = k) = \binom{n}{k} p^k q^{n-k}$.We need to find the probability that at most $2$ components survive,which is $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$.$P(X = 0) = \binom{4}{0} (\frac{2}{3})^0 (\frac{1}{3})^4 = 1 	imes 1 	imes \frac{1}{81} = \frac{1}{81}$.$P(X = 1) = \binom{4}{1} (\frac{2}{3})^1 (\frac{1}{3})^3 = 4 	imes \frac{2}{3} 	imes \frac{1}{27} = \frac{8}{81}$.$P(X = 2) = \binom{4}{2} (\frac{2}{3})^2 (\frac{1}{3})^2 = 6 	imes \frac{4}{9} 	imes \frac{1}{9} = \frac{24}{81}$.Summing these probabilities: $P(X \le 2) = \frac{1}{81} + \frac{8}{81} + \frac{24}{81} = \frac{33}{81} = \frac{33}{3^4}$.

The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most $2$ components out of $4$ tested will survive is

Similar Questions

The probability that a person wins a prize on a lottery ticket is $\frac{1}{4}$. If he purchases $5$ lottery tickets at random,then the probability that he wins at least one prize is

Find the probability distribution of the number of doublets in three throws of a pair of dice.

Bismuth has a half-life period of $5 \text{ days}$. $A$ sample originally has a mass of $1000 \text{ mg}$. What is the mass of Bismuth remaining after $30 \text{ days}$ (in $.625$)?

If a random variable $X \sim B\left(n=5, p=\frac{1}{3}\right)$,then $P(2 < X < 4) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module