A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is

Solution:

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, $\mathrm{X}$ has a binomial distribution with $n=50$ and $p=\frac{1}{100}$

$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}$

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-x} p^{x}={ }^{50} \mathrm{C}_{x}\left(\frac{99}{100}\right)^{50-x} \cdot\left(\frac{1}{100}\right)^{2}$

(a) P (winning at least once) = P (X ≥ 1)

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

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

$=1-{ }^{50} \mathrm{C}_{0}\left(\frac{99}{100}\right)^{50}$

$=1-1 \cdot\left(\frac{99}{100}\right)^{50}$

$=1-\left(\frac{99}{100}\right)^{30}$

(b) P (winning exactly once) = P(X = 1)

$={ }^{50} \mathrm{C}_{1}\left(\frac{99}{100}\right)^{49} \cdot\left(\frac{1}{100}\right)^{1}$

$=50\left(\frac{1}{100}\right)\left(\frac{99}{100}\right)^{49}$

$=\frac{1}{2}\left(\frac{99}{100}\right)^{49}$

(c) P (at least twice) = P(X ≥ 2)

$=1-P(X<2)$

$=1-P(X \leq 1)$

$=1-[P(X=0)+P(X=1)]$

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

$=1-\left(\frac{99}{100}\right)^{50}-\frac{1}{2} \cdot\left(\frac{99}{100}\right)^{49}$

$=1-\left(\frac{99}{100}\right)^{49} \cdot\left[\frac{99}{100}+\frac{1}{2}\right]$

$=1-\left(\frac{99}{100}\right)^{49} \cdot\left(\frac{149}{100}\right)$

$=1-\left(\frac{149}{100}\right)\left(\frac{99}{100}\right)^{49}$