Problem 14E

For the lottery described in Exercise 2.4-13, find the smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than (a) 0.50; (b) 0.95.

Reference Exercise 2.4-13

It is claimed that for a particular lottery, 1/10 of the 50 million tickets will win a prize. What is the probability of winning at least one prize if you purchase (a) 10 tickets or (b) 15 tickets?

Solution 14E

Step1 of 3:

We have lottery game in that a particular lottery, 1/10 of the 50 million tickets will win a prize.

That is p =

= 0.1.

We need to find,

The smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than (a) 0.50; (b) 0.95.

Step2 of 3:

Let “X” be random variable which follows binomial distribution with parameters n and p.

That is X B(n, p)

The probability mass function of binomial distribution is given below

P(X) = nCx , x = 0,1,2,...,n.

Where,

X = random variable

n = sample size

p = probability of success(or proportion).

a).

Consider,

P(X1) > 0.50

1 - P(X1) > 0.50

1 - P(X = 0) > 0.50

1 - nC0 > 0.50

1 - [1(1)] > 0.50

1 - > 0.50

1 - 0.50 >

0.50 >

Now we need to solve for n for that take Ln(natural logarithm on both side)

Ln(0.50) > Ln[]

Ln(0.50) > n Ln(0.9)

n

n

n 6.57634

n 7

Hence, n7.

Therefore, minimum 7 tickets have to be purchase.

Step3 of 3:

b).

Consider,

P(X1) > 0.95

1 - P(X1) > 0.95

1 - P(X = 0) > 0.95

1 - nC0 > 0.95

1 - [1(1)] > 0.95

1 - > 0.95

1 - 0.95 >

0.05 >

Now we need to solve for n for that take Ln(natural logarithm on both side)

Ln(0.05) > Ln[]

Ln(0.05) > n Ln(0.9)

n

n

n 28.4225

n 29

Hence, n29.

Therefore, minimum 29 tickets have to be purchase.

Conclusion:

Therefore,The smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than

(a) 0.50 = 7 tickets

(b) 0.95 = 29 tickets.