 # If you buy a lottery ticket in 50 lotteries, in Kelly Nelson 2021-12-19 Answered
If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is $\frac{1}{100}$, what is the (approximate) probability that you will win a prize (a) at least once, (b) exactly once, and (c) at least twice?
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Step 1
$P\left\{X=i\right\}\approx {e}^{-\lambda }\frac{{\lambda }^{i}}{i!}$
Given: $n=50$
$p=\frac{1}{100}=0.01$
So, $\lambda =np=50×0.01=0.5$
a. Probability that you will win a prize at least once
$P\left\{X\ge 1\right\}\approx 1-P\left\{X=0\right\}=1-{e}^{-0.5}\frac{{0.5}^{0}}{0!}=0.3935$
b. Probability that you will win a prize exactly once
$P\left\{X=0\right\}\approx {e}^{-0.5}\frac{{0.5}^{1}}{1!}=0.3033$
c. Probability that you will win a prize at least twice
$P\left\{X\ge 2\right\}\approx 1-P\left\{X=0\right\}-P\left\{X=1\right\}=1-{e}^{-0.5}\frac{{0.5}^{0}}{0!}-{e}^{-0.5}\frac{{0.5}^{1}}{1!}=0.0902$
###### Did you like this example? Jonathan Burroughs
Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.
Clearly, X has a binomial distribution with
$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}$
$\therefore \left(X=x\right){=}^{n}{C}_{x}{q}^{n-x}{p}^{x}{=}^{50}{C}_{x}{\left(\frac{99}{100}\right)}^{50-x}\cdot {\left(\frac{1}{100}\right)}^{x}$
a) P (winning at least once) $=P\left(X\ge 1\right)$
$=1-P\left(X<1\right)$
$=1-P\left(X=0\right)$
$=1{-}^{50}{C}_{x}{\left(\frac{99}{100}\right)}^{50}$
$=1-1\cdot {\left(\frac{99}{100}\right)}^{50}$
$=1-{\left(\frac{99}{100}\right)}^{50}$
b) P (winning exactly once) $=P\left(X=1\right)$
${=}^{50}{C}_{1}{\left(\frac{99}{100}\right)}^{49}\cdot {\left(\frac{1}{100}\right)}^{1}$
$=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}$
$=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$
c) P (at least twice) $=P\left(X\ge 2\right)$
$=1-P\left(X<2\right)$
$=1-P\left(X\le 1\right)$
$=1-\left[P\left(X=0\right)+P\left(X=1\right)\right]$
$=\left[1-P\left(X=0\right)\right]-P\left(X=1\right)$
$=1-{\left(\frac{99}{100}\right)}^{50}-\frac{1}{2}\cdot {\left(\frac{99}{100}\right)}^{49}$
$=1-{\left(\frac{99}{100}\right)}^{49}\left[\frac{99}{100}+\frac{1}{2}\right]$
$=1-{\left(\frac{99}{100}\right)}^{49}\cdot \left(\frac{149}{100}\right)$

Let X:Number of times he wins a prize
Winning a prize on lottery is a Bernoulli trial
So, X has a binomial distribution
$P\left(X=x\right){=}^{n}{C}_{x}{q}^{n-x}{p}^{x}$
Here, $n=\text{number of lotteries}=50$
$p=\text{Probability of winning a prize}=\frac{1}{100}$
$q=1-p=1-\frac{1}{100}=\frac{99}{100}$
$Hence\phantom{\rule{0.167em}{0ex}}P\left(X=x\right){=}^{50}{C}_{x}\left(\frac{1}{100}{\right)}^{x}\left(\frac{99}{100}{\right)}^{50-x}$
a) Probability that he wins the lottery atleast once
$P\left(\text{at least once}\right)=P\left(X\ge 1\right)$
$=1-P\left(0\right)$
$=1{-}^{50}{C}_{0}\left(\frac{1}{100}{\right)}^{0}\left(\frac{99}{100}{\right)}^{50-0}$
$=1-1×1×\left(\frac{99}{100}{\right)}^{50}$
$=1-\left(\frac{99}{100}{\right)}^{50}$
b) Probability that he wins the lottery exactly once
$P\left(\text{exactly once}\right)=P\left(X=1\right)$
${=}^{50}{C}_{1}\left(\frac{1}{100}{\right)}^{1}\left(\frac{99}{100}{\right)}^{50-1}$
$=50×\frac{1}{100}×\left(\frac{99}{100}{\right)}^{49}$
$=\frac{1}{2}\left(\frac{99}{100}{\right)}^{49}$
c) $P\left(\text{at least twice}\right)=P\left(X\ge 2\right)$
$=1-\left[P\left(X=0\right)+P\left(X=1\right)\right]$
$=1-{\left[}^{50}{C}_{0}\left(\frac{1}{100}{\right)}^{0}\left(\frac{99}{100}{\right)}^{50-0}{+}^{50}{C}_{1}\left(\frac{1}{100}{\right)}^{1}\left(\frac{99}{100}{\right)}^{50-1}\right]$
$=1-\left[\left(\frac{99}{100}{\right)}^{50}+\frac{1}{2}\left(\frac{99}{100}{\right)}^{49}\right]$
$=1-\left(\frac{99}{100}{\right)}^{49}\left[\frac{99}{100}+\frac{1}{2}\right]$
$=1-\left(\frac{99}{100}{\right)}^{49}\left[\frac{99+50}{100}\right]$
$=1-\frac{149}{100}\left(\frac{99}{100}{\right)}^{49}$