Ten bulbs are drawn successively,with replace | vedclass

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

Question

(C) Let $X$ be the number of defective bulbs drawn in $n = 10$ trials. This follows a binomial distribution $B(n, p)$.Here,$n = 10$ and the probabilit

Vedclass · Accepted Answer

$1-\left(\frac{9}{10}\right)^{10}$

Vedclass · Answer

(C) Let $X$ be the number of defective bulbs drawn in $n = 10$ trials. This follows a binomial distribution $B(n, p)$.Here,$n = 10$ and the probability of a bulb being defective is $p = 10 \% = \frac{1}{10}$.The probability of a bulb not being defective is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.We want to find the probability that at least one bulb is defective,which is $P(X \ge 1)$.Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.The probability of getting $0$ defective bulbs in $10$ trials is given by the binomial formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.For $k = 0$,$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{10}ight)^{0} \left(\frac{9}{10}ight)^{10-0} = 1 	imes 1 	imes \left(\frac{9}{10}ight)^{10}$.Therefore,$P(X \ge 1) = 1 - \left(\frac{9}{10}ight)^{10}$.

Ten bulbs are drawn successively,with replacement,from a lot containing $10 \%$ defective bulbs. The probability that there is at least one defective bulb is:

Similar Questions

If getting a head on a coin when it is tossed is considered as success,then the probability of having more number of failures when ten fair coins are tossed simultaneously is

Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. What is the probability that none is a spade?

$A$ man takes a step forward with probability $0.4$ and one step backward with probability $0.6$. Then the probability that at the end of eleven steps he is one step away from the starting point is

$A$ fair die is thrown up to $20$ times. The probability that on the $10^{th}$ throw,the fourth six appears is:

For a binomial distribution with mean $6$ and variance $2$,find $P(X \geq 2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module