It is observed that 25% of the cases related | vedclass

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

Question

(D) Let $p$ be the probability of a case being solved,so $p = 25\% = \frac{1}{4}$.Let $q$ be the probability of a case not being solved,so $q = 1 - p

Vedclass · Accepted Answer

$\frac{19}{4096}$

Vedclass · Answer

(D) Let $p$ be the probability of a case being solved,so $p = 25\% = \frac{1}{4}$.Let $q$ be the probability of a case not being solved,so $q = 1 - p = \frac{3}{4}$.Given $n = 6$ trials,we use the binomial distribution formula $P(X=r) = {^nC_r} p^r q^{n-r}$.We need the probability that at least $5$ cases are solved,which is $P(X \ge 5) = P(X=5) + P(X=6)$.$P(X=5) = {^6C_5} \left(\frac{1}{4}ight)^5 \left(\frac{3}{4}ight)^1 = 6 	imes \frac{1}{1024} 	imes \frac{3}{4} = \frac{18}{4096}$.$P(X=6) = {^6C_6} \left(\frac{1}{4}ight)^6 \left(\frac{3}{4}ight)^0 = 1 	imes \frac{1}{4096} 	imes 1 = \frac{1}{4096}$.Therefore,$P(X \ge 5) = \frac{18}{4096} + \frac{1}{4096} = \frac{19}{4096}$.

It is observed that $25\%$ of the cases related to child labour reported to the police station are solved. If $6$ new cases are reported,then the probability that at least $5$ of them will be solved is . . . . . .

Similar Questions

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Then,the probability of getting exactly six successes in this distribution is:

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively,then $P(X < 3)$ equals to

The mean and variance of a binomial distribution are $5$ and $4$ respectively. Then what is $P(X=1)$?

Let $X \sim B(6, 1/2)$,then $P[|X-4| \leq 2]$ is

If $X$ is a binomial variate with mean $\frac{16}{5}$ and variance $\frac{48}{25}$,then $P(X \leq 2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module