The probability that a bulb produced by a fac | vedclass

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

Question

(B) Let $X$ be the number of bulbs that fuse after $150$ days. This follows a binomial distribution with $n = 5$ and $p = 0.05$.Here,$q = 1 - p = 0.95

Vedclass · Accepted Answer

$1 - (0.95)^4 	imes 1.25$

Vedclass · Answer

(B) Let $X$ be the number of bulbs that fuse after $150$ days. This follows a binomial distribution with $n = 5$ and $p = 0.05$.Here,$q = 1 - p = 0.95$.The probability mass function is $P(X = x) = ^nC_x q^{n-x} p^x = ^5C_x (0.95)^{5-x} (0.05)^x$.We need to find the probability that more than one bulb fuses,i.e.,$P(X > 1)$.$P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)]$.$P(X = 0) = ^5C_0 (0.95)^5 (0.05)^0 = (0.95)^5$.$P(X = 1) = ^5C_1 (0.95)^4 (0.05)^1 = 5 	imes (0.95)^4 	imes 0.05 = 0.25 	imes (0.95)^4$.Therefore,$P(X > 1) = 1 - [(0.95)^5 + 0.25(0.95)^4]$.Simplifying,$P(X > 1) = 1 - (0.95)^4 [0.95 + 0.25] = 1 - 1.2(0.95)^4$.

The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. Find the probability that out of $5$ such bulbs,more than one will fuse after $150$ days.

Similar Questions

For a Binomial variate $X$,the mean is $2$ and the variance is $1$. Then,the odds in favor of $X=0$ are:

The binomial distribution whose mean is $9$ and whose standard deviation is $\frac{3}{2}$ is equal to

$A$ box contains $15$ green and $10$ yellow balls. If $10$ balls are randomly drawn,one-by-one,with replacement,then the variance of the number of green balls drawn is:

The variance of the following probability distribution of $X$ is given by:
$X: 0, 1, 2, 3$
$P(X): q^3, 3pq^2, 3p^2q, p^3$
where $0 < p < 1$ and $q = 1 - p$.

The probability of getting either all heads or all tails for exactly the second time in the $3^{rd}$ trial,if in each trial three coins are tossed,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module