# How to determine larger probability of drawing one element from a set when set size and number of draws with replacement vary?

Which is the better scenario in terms of probability with the following:

Scenario 1

• I have a hat with 5 tickets in it (I have one).
• There is one single draw.
• I have a 20% chance that my ticket is drawn.

Scenario 2

• I have a hat with 25 tickets in it (I have one).
• There are 5 draws.
• After each draw the ticket is returned before the next draw. i.e. potential chance of being drawn 5 times.
• I have a 4% chance each draw

Which has the better probability of being drawn at least once? Ideally I'd like to know how to work this out (and why) so I can learn something :-)

It is sometimes easier to work out the probability that an event did not occur rather than the probability that it did occur. On each draw, the probability that your ticket is not drawn is $\frac{24}{25} = 0.96$, and thus the probability that your ticket is not drawn on all $5$ draws is $(0.96)^5$. Thus, the probability that your ticket is drawn at least once is just $1 - (0.96)^5$. I will leave it to you to calculate the exact value and compare it to $0.20$.
• @love-stats $(0.04)^5$ is the probability that the ticket held by the OP is drawn on all $5$ draws, and so if the question was "What is the probability that your ticket is not drawn on at least one of the five draws?" the answer would be $1 - (0.04)^5$ which is very close to $1$; it is very highly likely that an event of probability $0.04$ will not occur at least once on $5$ trials. – Dilip Sarwate Oct 7 '11 at 1:52