Question

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

Answer 1

p=$\frac{1}{100}$ and q=1-p=$1-\frac{1}{100}$=$\frac{99}{100}$
n=50
(a) P(at least one prize)=1-P(X=0)=$1-\bigg(\frac{99}{100}\bigg)^{50}$

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(b) P(exactly one prize)=P(X=1)
= $^{50}C_1\bigg(\frac{1}{100}\bigg)\bigg(\frac{99}{100}\bigg)^{49}=50*\frac{1}{100}\bigg(\frac{99}{100}\bigg)^{49}=\frac{1}{2}\bigg(\frac{99}{100}\bigg)^{49}$

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(c) P(at least twice)

=1-[P(X=0)+(X=1)]=1-$\bigg[\bigg(\frac{99}{100}\bigg)^{50}+\frac{1}{2}\bigg(\frac{99}{100}\bigg)^{49}\bigg]$
=1-$\bigg(\frac{99}{100}\bigg)^{49}\bigg(\frac{99}{100}+\frac{1}{2}\bigg)=1-\bigg(\frac{99}{100}\bigg)^{49}\bigg(\frac{149}{100}\bigg)$