Out of 100 people selected at random,10 have | vedclass

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

Question

(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.Then,the probability of a person not having a common c

Vedclass · Accepted Answer

$0.9185$

Vedclass · Answer

(B) Let the probability of a person having a common cold be $p = \frac{10}{100} = \frac{1}{10}$.Then,the probability of a person not having a common cold is $q = 1 - p = \frac{9}{10}$.We are selecting $n = 5$ persons. Let $X$ be the number of persons having a common cold. $X$ follows a binomial distribution $B(n, p) = B(5, 0.1)$.We need to find the probability that at most one person has a common cold,which is $P(X \le 1) = P(X = 0) + P(X = 1)$.Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}ight)^{0} \left(\frac{9}{10}ight)^{5} = 1 	imes 1 	imes \frac{59049}{100000} = 0.59049$.$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}ight)^{1} \left(\frac{9}{10}ight)^{4} = 5 	imes \frac{1}{10} 	imes \frac{6561}{10000} = \frac{32805}{100000} = 0.32805$.Therefore,$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$.Rounding to four decimal places,the probability is $0.9185$.

Out of $100$ people selected at random,$10$ have a common cold. If five persons are selected at random from the group,then the probability that at most one person will have a common cold is

Similar Questions

$A$ contest consists of predicting the results (win,draw,or defeat) of $7$ football matches. $A$ sent his entry by predicting at random. The probability that his entry will contain exactly $4$ correct predictions is:

If a random variable $X$ follows the Binomial distribution $B(33, p)$ such that $3P(X=0) = P(X=1)$,then the value of $\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)}$ is equal to

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

In a binomial distribution,the probability of success is $\frac{1}{4}$ and the standard deviation is $3$. Then,its mean is

$A$ die is thrown $(2n+1)$ times. The probability of getting $1$ or $3$ or $4$ at most $n$ times on it is

Vedclass Test Series

Exam Paper Generator

Online Exam Module