# 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?

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.

$1 - {\left(1 - p\right)}^{n} \ge 0.5$
$1 - {0.9}^{n} \ge 0.5$
${0.9}^{n} \le 0.5$.

Put n=1,2,… to the inequation you will find:
${0.9}^{6} = 0.531441 \ge 0.5$
${0.9}^{7} = 0.4782969 \le 0.5$
Thus, the minimum integer that meets ${0.9}^{n} \le 0.5$ 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]$\textcolor{g r e e n}{\text{What is the odds to have at least one 6 when we roll three dices?}}$
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 $1 - \frac{1}{6} = \frac{5}{6}$.
(2) When three dices are rolled, they are independent. So the probability of having no $6$s is ${\left(\frac{5}{6}\right)}^{3} = \frac{125}{216}$.
(3) Having at least one 6 is $\textcolor{red}{\text{complement}}$ to (2) and the probability is $1 - \frac{125}{216} = \frac{91}{216}$.