A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 / 100. What is the probability that he will win a prize
(a) At least once
(b) Exactly once
(c) At least twice?
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 / 100. What is the probability that he will win a prize
(a) At least once
(b) Exactly once
(c) At least twice?

Solution:

(a) Let X be the number of prizes won in 50 lottery drawings, and the trials be Bernoulli trials.
Here clearly, we have \mathrm{X} is a binomial distribution where \mathrm{n}=50 and \mathrm{p}=1 / 100 Thus, q=1-p=1-1 / 100=99 / 100

\therefore \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \mathrm{p}^{\mathrm{x}}

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

Hence, probability of winning in lottery at least once =\mathrm{P}(\mathrm{X} \geq 1)

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

=1-\mathrm{P}(\mathrm{X}=0)
=1-{ }^{50} C_{0}\left(\frac{99}{100}\right)^{50}
=1-1 \cdot\left(\frac{99}{100}\right)^{50}
=1-\left(\frac{99}{100}\right)^{50}

(b) Probability of winning in lottery exactly once =\mathrm{P}(\mathrm{X}=1)
={ }^{50} 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) Probability of winning in lottery at least twice =\mathrm{P}(\mathrm{x} \geq 2)

=1-\mathrm{P}(\mathrm{X}<2)

=1-P(X \leq 1)

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

=[1-P(X=0)]-P(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}