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 a dice is thrown $7$ times,then the probability of obtaining $5$ exactly $4$ times is

If $X$ is a binomial variate with the range $\{0, 1, 2, 3, 4, 5, 6\}$ and $P(X=2) = 4 P(X=4)$,then the parameter $p$ of $X$ is

If a dice is thrown $5$ times,then the probability of getting $6$ exactly three times is:

Each of the persons $A$ and $B$ independently tosses three fair coins. The probability that both of them get the same number of heads is:

How many times must a fair coin be tossed so that the probability of getting at least one head is at least $0.8$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module