A person buys a lottery ticket in 50 lotterie | vedclass

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

Question

(A) Let $X$ represent the number of winning prizes in $50$ lotteries. The trials are Bernoulli trials. Clearly,$X$ has a binomial distribution with $n

Vedclass · Accepted Answer

$1 - \left(\frac{149}{100}\right)\left(\frac{99}{100}\right)^{49}$

Vedclass · Answer

(A) Let $X$ represent the number of winning prizes in $50$ lotteries. The trials are Bernoulli trials. Clearly,$X$ has a binomial distribution with $n = 50$ and $p = \frac{1}{100}$.$	herefore q = 1 - p = 1 - \frac{1}{100} = \frac{99}{100}$.$	herefore P(X = x) = ^{n}C_{x} q^{n-x} p^{x} = ^{50}C_{x} \left(\frac{99}{100}ight)^{50-x} \left(\frac{1}{100}ight)^{x}$.$P(	ext{at least twice}) = P(X \geq 2) = 1 - P(X $P(X = 0) = ^{50}C_{0} \left(\frac{99}{100}ight)^{50} \left(\frac{1}{100}ight)^{0} = \left(\frac{99}{100}ight)^{50}$.$P(X = 1) = ^{50}C_{1} \left(\frac{99}{100}ight)^{49} \left(\frac{1}{100}ight)^{1} = 50 \cdot \left(\frac{99}{100}ight)^{49} \cdot \frac{1}{100} = \frac{1}{2} \left(\frac{99}{100}ight)^{49}$.$P(X \geq 2) = 1 - \left[ \left(\frac{99}{100}ight)^{50} + \frac{1}{2} \left(\frac{99}{100}ight)^{49} ight]$.$= 1 - \left(\frac{99}{100}ight)^{49} \left[ \frac{99}{100} + \frac{1}{2} ight] = 1 - \left(\frac{99}{100}ight)^{49} \left[ \frac{99 + 50}{100} ight] = 1 - \left(\frac{149}{100}ight) \left(\frac{99}{100}ight)^{49}$.

$A$ person buys a lottery ticket in $50$ lotteries,in each of which his chance of winning a prize is $\frac{1}{100}$. What is the probability that he will win a prize at least twice?

Similar Questions

$A$ shooter can hit a given target with probability $\frac{1}{4}$. She keeps firing a bullet at the target until she hits it successfully three times and then she stops firing. The probability that she fires exactly six bullets lies in the interval

$A$ random variable $X$ follows a binomial distribution in which the difference between its mean and variance is $1$. If $2 P(X=2)=3 P(X=1)$,then $n^2 P(X>1)=$

The probability of a man hitting a target is $\frac{2}{3}$. What is the minimum number of times he must fire so that the probability of hitting the target at least once is more than $90 \%$?

An urn contains $25$ balls of which $10$ balls bear a mark $'X'$ and the remaining $15$ bear a mark $'Y'$. $A$ ball is drawn at random from the urn,its mark is noted down and it is replaced. If $6$ balls are drawn in this way,find the probability that not more than $2$ will bear $'Y'$ mark.

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of tens is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module