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?

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#.