What is the probability that at least one red ball is drawn when three balls are drawn from a bag with $5$ red and $5$ yellow balls?

Question

I am having a tough time trying to understand the concept of probability and figuring out which is the right way to solve this problem. So I would really appreciate a lot if you can help me with this question

Problem: There are 5 red balls and 5 yellow balls. If we make 3 draws from this set, without replacing the ball drawn in each set, what is the probability that in the end we have drawn at least one red ball.

Method 1:

Calculate the probability of not drawing any red ball in all three draws and subtract that from 1.

The probability of drawing 3 yellow balls in succession : 5/10 * 4/9 * 3/8

So then the probability of drawing at least one red ball is 1-(5/10 * 4/9 * 3/8) = 11/12

Method 2: Create the possible color combinations that the 3 draws can produce. I can come up with only 8 such combinations:

RRR, YYY, YRR, YRY, YYR, RRY, RYR, RYY

So based on these combinations I see that there is only 1 combination that has no red ball. So isthe probability of drawing a red ball then 7/8?

Which is the right answer and the right way of solving this problem.

Answer 1

The first is correct. In your second method you haven’t taken into account the fact that the eight different outcomes have different probabilities.

Answer 2

Your first method is correct. Let's correct your second method.

There are $$\binom{10}{3}$$ ways to select three of the ten balls.

The number of selections with at least one red ball is $$\binom{5}{3} + \binom{5}{2}\binom{5}{1} + \binom{5}{1}\binom{5}{2}$$ where the first term counts the number of ways of selecting three red balls, the second term counts the number of ways of selecting two red and one yellow ball, and the third term counts the number of ways of selecting one red and two yellow balls. Observe that $$\frac{\dbinom{5}{3} + \dbinom{5}{2}\dbinom{5}{1} + \dbinom{5}{1}\dbinom{5}{2}}{\dbinom{10}{3}} = \frac{11}{12}$$ The problem with your second method is that the eight events are not equally likely.