Question

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

Answer 1

It is given that a person has a chance of a winning a prize is 1 100 .

Let the number of winning prizes in 50 lotteries is represented by X.

The probability of winning a prize is,

p= 1 100

So,

q=1−p =1− 1 100 = 99 100

Then, X has a binomial distribution with n=50 and p= 1 100 .

(a)

The probability of x successes P( X=x ) is,

P( X=x )= C n x q n−x p x = C 50 x ( 99 100 ) 50−x ( 1 100 ) x

Where x=0,1,…,n.

The probability of winning at least once is,

P( X≥1 )=1−P( X<1 ) =1−P( X=0 )

Substitute the value in above expression,

P( X≥1 )=1− C 50 0 ( 99 100 ) 50 =1−1⋅ ( 99 100 ) 50 =1− ( 99 100 ) 50

Therefore, the probability of winning a prize at least once is 1− ( 99 100 ) 50 .

(b)

The probability of winning exactly once is,

P( X=1 )= C 50 1 ( 99 100 ) 49 ( 1 100 ) 1 =50( 1 100 ) ( 99 100 ) 49 = 1 2 ( 99 100 ) 49

Therefore, the probability of winning a prize exactly once is 1 2 ( 99 100 ) 49 .

(c)

The probability of winning at least twice is,

P( X≥2 )=1−P( X<2 ) =1−[ P( X=0 )+P( X=1 ) ]

So,

P( X≥2 )=1−[ ( 99 100 ) 50 + 1 2 ( 99 100 ) 49 ] =1− ( 99 100 ) 49 [ 99 100 + 1 2 ] =1− ( 99 100 ) 49 [ 149 100 ] =1− 149 100 ( 99 100 ) 49

Therefore, the probability of winning a prize at least twice is 1− 149 100 ( 99 100 ) 49 .