Question

A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $1/100$. Then the probability that he will win a prize exactly once is:

• A. ${\left(\frac{99}{100}\right)}^{49}$
• B. $\frac{1}{2}$
• C. $\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$

Let $X$ represents the number of prizes winning in $50$ lotteries and the trials are Bernoulli trials
Here clearly, we have $X$ is a binomial distribution where $n=50$ and $p=1/100$
Thus, $q=1-p=1-1/100=99/100$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x{=}^{50}{C}_x{\left(\frac{99}{100}\right)}^{50-x}.{\left(\frac{1}{100}\right)}^x$
Probability of winning in lottery exactly once$=P(X=1)$
${=}^{50}{C}_1{\left(\frac{99}{100}\right)}^{49}.{\left(\frac{1}{100}\right)}^1$
$=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}$
$=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}$