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

(a) at least once

(b) exactly once

(c) at least twice?

.: Let $p$ be the probability of success that is chance of winning a prize

$= \cfrac{1}{{100}} \Rightarrow q = 1 - p = 1 - \cfrac{1}{{100}} = \cfrac{{99}}{{100}}$

and let $X$ has a binomial distribution with
$n = 50,p = \cfrac{1}{{100}},q = \cfrac{{99}}{{100}}$

$\therefore$ $P(X = r){ = ^n}{C_r}{(q)^{n - r}}{p^r}$

(a) $P$ (the probability of person wins a prize at least once) $= P(X \ge 1)$

$= 1 - P(X = 0) = 1{ - ^{50}}{C_0}{q^{50}}{p^0} = 1 - (1){\left( {\cfrac{{99}}{{100}}} \right)^{50}} = 1 - {\left( {\cfrac{{99}}{{100}}} \right)^{50}}$

(b) $P$ (the probability of person winning a prize exactly once)

$= P(X = 1)$

${ = ^{50}}{C_1}{q^{49}}{p^1} = 50{\left( {\cfrac{{99}}{{100}}} \right)^{49}}\left( {\cfrac{1}{{100}}} \right) = \cfrac{1}{2}{\left( {\cfrac{{99}}{{100}}} \right)^{49}}$

(c) $P$ (the probability of person wins a prize at least twice)

$= P(X \ge 2)$

$= 1 - (P(0) + P(1)) = 1 - {(^{50}}{C_0}{q^{50}}{p^0}{ + ^{50}}C{q^{49}}{p^1})$

$= 1 - \left( {(1){q^{50}}(1) + 50{q^{49}}p} \right) = 1 - {q^{49}}(q + 50p)$

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