Question

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

Atleast once ?

exactly once ?

atleast twice ?

Answer 1

Let X denotes the number of times that the person wins the prize, Here, it is a case of Bernoulli trails with n=50, p=1/100 and q=1-p=1-$\frac{1}{100}$=$\frac{99}{100}$.

$\therefore$ P(X=r)= ${}^n{C}_r{p}^r{q}^{n-r}={}^{50}{C}_r{\left(\frac{1}{100}\right)}^r.{\left(\frac{99}{100}\right)}^{50-r}$

(a) P(he wins a prize atleast once) =1- P(0)

= $1-{}^{50}{C}_0{p}^0{q}^{50}=1-{\left(\frac{99}{100}\right)}^{50}$

Let X denotes the number of times that the person wins the prize, Here, it is a case of Bernoulli trails with n=50, p=1/100 and q=1-p=1-$\frac{1}{100}$=$\frac{99}{100}$.
$\therefore$ P(X=r)= ${}^n{C}_r{p}^r{q}^{n-r}={}^{50}{C}_r{\left(\frac{1}{100}\right)}^r.{\left(\frac{99}{100}\right)}^{50-r}$

P (he wins a prize exactly once)

=${}^{50}{C}_1{p}^1{q}^{49}=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$

Let X denotes the number of times that the person wins the prize, Here, it is a case of Bernoulli trails with n=50, p=1/100 and q=1-p=1-$\frac{1}{100}=\frac{99}{100}$.
$\therefore P(X=r)={}^n{C}_r{p}^r{q}^{n-r}={}^{50}{C}_r{\left(\frac{1}{100}\right)}^r.{\left(\frac{99}{100}\right)}^{50-r}$

P (he wins atleast twice)= $P(X\ge 2)=1-P(X=0,1)=1-\left[P\right(0)+P(1\left)\right]$
=$1-({}^{50}{C}_0{p}^0{q}^{50}+{}^{50}{C}_1{p}^1{q}^{49})=1-(q^{50}+50p{q}^{49})$

=$1-q^{49}(p+50p)=1-{\left(\frac{99}{100}\right)}^{49}(\frac{99}{100}+50\times \frac{1}{100})=1-\frac{149}{100}{\left(\frac{99}{100}\right)}^{49}$