Question

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\dfrac{1}{{100}}$. What is the probability that he will win a prize?
$
  (a){\text{ at least one}} \\
  (b){\text{ exactly one}} \\
  (c){\text{ at least twice}} \\
$

Answer 1

Hint – In this question use the concept of Bernoulli trials. Let Y be the random variable representing the number of prizes in 50 lotteries. So in the first case find $P\left( {Y \geqslant 1} \right)$, in the second case find $P\left( {Y = 1} \right)$ and in the last case find $P\left( {Y \geqslant 2} \right)$.

Complete step-by-step answer:
Let Y represent the number of winning prizes in 50 lotteries.
The trials are Bernoulli trials.
Clearly, Y has a Binomial distribution with n = 50 (given) and p = (1/100) (given).
So the chance of not winning the prize (q) = 1 – p = 1 – (1/100) = (99/100).
So here we use the Bernoulli probability formula
$\therefore P = {}^n{C_Y}{\left( p \right)^Y}{\left( q \right)^{n - Y}}$..................... (1)
$\left( a \right)$ At least once.
$ \Rightarrow P\left( {Y \geqslant 1} \right)$
$ \Rightarrow 1 - P\left( {Y < 1} \right)$
$ \Rightarrow 1 - P\left( {Y = 0} \right)$
So from equation (1) we have,
$ \Rightarrow 1 - {}^{50}{C_0}{\left( {\dfrac{1}{{100}}} \right)^0}{\left( {\dfrac{{99}}{{100}}} \right)^{50 - 0}}$, $\left[ {\because {}^n{C_Y} = \dfrac{{n!}}{{Y!\left( {n - Y} \right)!}},{}^{50}{C_0} = 1} \right]$
$ \Rightarrow 1 - 1.1{\left( {\dfrac{{99}}{{100}}} \right)^{50}} = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{50}}$
$\left( b \right)$ Exactly once
$ \Rightarrow P\left( {Y = 1} \right)$
So from equation (1) we have,
$ \Rightarrow {}^{50}{C_1}{\left( {\dfrac{1}{{100}}} \right)^1}{\left( {\dfrac{{99}}{{100}}} \right)^{50 - 1}}$, $\left[ {\because {}^n{C_Y} = \dfrac{{n!}}{{Y!\left( {n - Y} \right)!}},{}^{50}{C_1} = 50} \right]$
$ \Rightarrow 50\left( {\dfrac{1}{{100}}} \right){\left( {\dfrac{{99}}{{100}}} \right)^{49}} = \dfrac{1}{2}{\left( {\dfrac{{99}}{{100}}} \right)^{49}}$
$\left( c \right)$ At least twice
$ \Rightarrow P\left( {Y \geqslant 2} \right)$
$ \Rightarrow 1 - P\left( {Y < 2} \right)$
$ \Rightarrow 1 - \left[ {P\left( {Y = 0} \right) + P\left( {Y = 1} \right)} \right]$
So from equation (1) we have,
$ \Rightarrow 1 - \left[ {{}^{50}{C_0}{{\left( {\dfrac{1}{{100}}} \right)}^0}{{\left( {\dfrac{{99}}{{100}}} \right)}^{50 - 0}} + {}^{50}{C_1}{{\left( {\dfrac{1}{{100}}} \right)}^1}{{\left( {\dfrac{{99}}{{100}}} \right)}^{50 - 1}}} \right]$
$ \Rightarrow 1 - \left[ {1.1{{\left( {\dfrac{{99}}{{100}}} \right)}^{50}} + 50\left( {\dfrac{1}{{100}}} \right){{\left( {\dfrac{{99}}{{100}}} \right)}^{49}}} \right] = 1 - \left[ {{{\left( {\dfrac{{99}}{{100}}} \right)}^{50}} + \dfrac{1}{2}{{\left( {\dfrac{{99}}{{100}}} \right)}^{49}}} \right]$
$ \Rightarrow 1 - \left[ {{{\left( {\dfrac{{99}}{{100}}} \right)}^{50}} + \dfrac{1}{2}{{\left( {\dfrac{{99}}{{100}}} \right)}^{49}}} \right] = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{49}}\left[ {\dfrac{{99}}{{100}} + \dfrac{1}{2}} \right] = 1 - {\left( {\dfrac{{99}}{{100}}} \right)^{49}}\left[ {\dfrac{{149}}{{100}}} \right]$
So this is the required answer.

Note – In the problems related to probability, a Bernoulli trial is the random experiment with exactly two possible outcome success or failure. Key point here is we are eligible to use Bernoulli trial if and only if the probability of winning remains same all throughout the event or it is the same every time the experiment is conducted.