Question

A lottery has a very large number of tickets, one in every 500 of which entitles the purchaser to prize. Calculate the minimum no. of tickets the agent must sell to have 95% chance of selling at least one prize winning ticket.

Answer 1

The probability that ticket entitle the purchaser to prize is $p = \frac{1}{500} = 0.002$

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:

$p_{win} = 1 - p_{lose} = 1 - (1-p)^N$

By condition, this value should be not less than $p_0 = 0.95$ :

$p_0 \leq 1 - (1-p)^N => N \geq log_{1-p}(1 - p_0) = log_{0.998}0.05 = 1496.36$

So, the minimum no. of tickets the agent must sell to have 95% chance of selling at least one prize winning ticket is 1497.