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?

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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?

CBSE | Maths | NCERT | Probability

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 $=\mathrm{P}(\mathrm{x} \geq 2)$

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

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

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

$=[1-P(X=0)]-P(X=1)$
$=1-\left(\frac{99}{100}\right)^{50}-\frac{1}{2} \cdot\left(\frac{99}{100}\right)^{49}$
$=1-\left(\frac{99}{100}\right)^{49} \cdot\left[\frac{99}{100}+\frac{1}{2}\right]$