Question

A lottery has a very large number of tickets, one in every 500 at which entitles the purchase to prize. Calculate the minimum number of tickets the agent must sell to have 90% chance of selling at least one prize winning ticket.

Answer 1

The probability that ticket entitle the purchaser to prize is

$$p=\dfrac{1}{500}=0.002$$

If purchaser buy N tickets, the probability that none of them will be winning according to the formula of probability of independent events intersection:

$$P(lose)=(1-p)^N$$

So, the probability that purchaser has at least one prize winning ticket is:

$$P(at\ least \ one\ prize)=1-P(lose)$$

$$=1-(1-p)^N$$

By condition, this value should be not less than

$$p_0=0.9$$

$$1-(1-p)^N\geq p_0$$

$$1-(1-0.002)^N\geq 0.9$$

$$(0.998)^N\leq 0.1$$

$$N\ln(0.998)\leq\ln(0.1)$$

$$\ln(0.998)<0: N\geq\dfrac{\ln(0.1)}{\ln(0.998)}$$

$$N\geq1151$$

So, the minimum no. of tickets the agent must sell to have 90% chance of selling at least one prize winning ticket is 1151.