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

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 $np=50×\frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{â„™}(Xâ‰\yen 1)$

$=1-\mathrm{â„™}(X=0)$

$=1-e^{-0.5}$

$=0.39$.

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

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

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 $np=50×\frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{â„™}(X=1)$

$=\frac{e^{-0.5}·0.5^1}{1!}$

$=0.30$.

Probability of winning precisely once is $0.3$.

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

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 $np=50×\frac{1}{100}=\frac{1}{2}$. So we have:

$\mathrm{â„™}(Xâ‰\yen 2)=1-\mathrm{â„™}(X=0)-\mathrm{â„™}(X=1)$

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

$=1-e^{-0.5}-0.5·e^{-0.5}=0.09$.

Probability of winning at least twice is $0.09$.