Question

If you buy a lottery ticket in $50$lotteries, in each of which your chance of winning a prize is role="math" localid="1646465220038" $\frac{1}{100}$, what is the (approximate) probability that you will win a prize

(a) at least once?

(b) exactly once?

(c) at least twice?

Short answer

$\left(a\right)$The probability of winning prize at least once is $0.39$.

$\left(b\right)$The probability of winning prize exactly once is $0.30$.

$\left(c\right)$The probability of winning prize is$0.09$.

Step-by-step solution

Given Information (Part a)

If you buy a lottery ticket in $50$lotteries, in each of which your chance of winning a prize is $\frac{1}{100}$.

Calculation (Part a)

We buy lottery tickets in $50$lotteries, in each of which chance of a winning prize is $\frac{1}{100}$.

Find the probability of winning the prize at least once.

We use the fact that for large $n$and small $p$Poisson distribution approximates binomial distribution. Obviously $np=50\times \frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{\mathbb{P}}(X\ge 1)$

$=1-\mathrm{\mathbb{P}}(X=0)$

$=1-e^{-0.5}$

$=0.39$.

Final answer (Part a)

Probability of winning the lottery at least once is $0.39$.

Given Information (Part b)

If you buy a lottery ticket in $50$lotteries, in each of which your chance of winning a prize is $\frac{1}{100}$.

Calculation (Part b)

We buy lottery tickets in $50$lotteries, in each of which chance of a winning prize is $\frac{1}{100}$.

Find the probability of winning the prize exactly once.

We use the fact that for large $n$and small $p$Poisson distribution approximates binomial distribution. Obviously $np=50\times \frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{\mathbb{P}}(X=1)$

$=\frac{e^{-0.5}·0.5^1}{1!}$

$=0.30$.

Final answer (Part b)

Probability of winning precisely once is $0.3$.

Given Information (Part c)

If you buy a lottery ticket in $50$ lotteries, in each of which your chance of winning a prize is$\frac{1}{100}$.

Calculation (Part c)

We buy lottery tickets in $50$lotteries, in each of which chance of a winning prize is $\frac{1}{100}$.

Find the probability of winning the prize at least twice.

We use the fact that for large $n$and small $p$Poisson distribution approximates binomial distribution. Obviously $np=50\times \frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{\mathbb{P}}(X\ge 2)=1-\mathrm{\mathbb{P}}(X=0)-\mathrm{\mathbb{P}}(X=1)$

$=1-\frac{e^{-0.5}·0.5^0}{0!}-\frac{e^{-0.5}·0.5}{1!}$

$=1-e^{-0.5}-0.5·e^{-0.5}=0.09$.

Final answer (Part c)

Probability of winning at least twice is $0.09$.