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?

Question

Hunter Aldridge · Accepted Answer

You have to buy at least #7# tickets.

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://tutor.hix.ai 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#.