A lottery claims that 10% of tickets win a prize. How many tickets should you purchase to be more than 50% sure of winning a prize?

1 Answer
Oct 2, 2017

You have to buy at least 7 tickets.

Explanation:

Let p=0.1, which is the probability for winning a prize.

In this lottery tickets, the probability for NOT winning a prize is
1-p=1-0.1=0.9.

If you want more than 50% sure of winning at least one prize,
the number of tickets you have to buy n satisfies the following inequation.

1-(1-p)^n>=0.5
1-0.9^n>=0.5
0.9^n<=0.5.

Put n=1,2,… to the inequation you will find:
0.9^6=0.531441>=0.5
0.9^7=0.4782969<=0.5
Thus, the minimum integer that meets 0.9^n<=0.5 is n=7.

[What is the point?]
The concept I applied to this question is called complementary event.
https://en.wikipedia.org/wiki/Complementary_event

This idea is useful to evaluate the probability for having at least one event.
Consider a simpler case and you will notice the convenience.

[Example]color(green) "What is the odds to have at least one 6 when we roll three dices?"
If you solve this from the front, you need to consider three cases.

(a) three 6s
(b) two 6s and another number
(c) a 6 and two other numbers

It will be complicated. Insted, you can solve it from the back door.

(1) If a dice is rolled, the probability of not having a 6 is 1-1/6=5/6.
(2) When three dices are rolled, they are independent. So the probability of having no 6s is (5/6)^3=125/216.
(3) Having at least one 6 is color(red)"complement" to (2) and the probability is 1-125/216=91/216.