Question

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?

Answer 1

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 `6`s
(b) two `6`s 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 `6`s 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`.