Question

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
(a) at least once
(b) exactly once
(c) at least twice ?

Answer 1

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 $p=\frac{1}{100}$
$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x{=}^{50}{C}_x{\left(\frac{99}{100}\right)}^{50-x}\cdot {\left(\frac{1}{100}\right)}^x$
(a) $P\left(winningatleastonce\right)=P(X\ge 1)$
$=1-P(X<1)$
$=1-P(X=0)$
$=1{-}^{50}{C}_x{\left(\frac{99}{100}\right)}^{50}$
$=1-1\cdot {\left(\frac{99}{100}\right)}^{50}$
$=1-{\left(\frac{99}{100}\right)}^{50}$
(b) $P\left(winningexactlyonce\right)=P(X=1)$
${=}^{50}{C}_1{\left(\frac{99}{100}\right)}^{49}\cdot {\left(\frac{1}{100}\right)}^1$
$=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}$
$=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$
(c) $P\left(atleasttwice\right)=P(X\ge 2)$
$=1-P(X<2)$
$=1-P(X\le 1)$
$=1-[P(X=0)+P(X=1)]$
$=[1-P(X=0)]-P(X=1)$
$=1-{\left(\frac{99}{100}\right)}^{50}-\frac{1}{2}\cdot {\left(\frac{99}{100}\right)}^{49}$
$=1-{\left(\frac{99}{100}\right)}^{49}\left[\frac{99}{100}+\frac{1}{2}\right]$
$=1-{\left(\frac{99}{100}\right)}^{49}\cdot \left(\frac{149}{100}\right)$
$=1-\left(\frac{149}{100}\right){\left(\frac{99}{100}\right)}^{49}$