There is a large container which will hold 10 balls.

Question

The balls are all the same size and texture. You are required to select two balls.
If both of the balls that you select are purple then you win.

How much less likely are you to win if you are required to replace the first ball? Show all of your work and explain your steps.

Answer 1

When selecting two balls without replacement, there are a total of 10 choose 2 = 45 different ways to select two balls from a container of 10 balls.

To calculate the probability of winning with the balls being purple, we need to first determine the total number of ways to select two purple balls, which can be done as follows:

There are 5 purple balls out of 10, so the number of ways to choose 2 purple balls is 5 choose 2 = 10.

Therefore, the probability of winning without replacement is 10/45 = 2/9.

Now, if we replace the first ball after selecting it, the probability of selecting a purple ball on the second draw is still 5 out of 10, as the composition of the remaining balls in the container remains the same.

Therefore, the probability of winning with replacement is (5/10) * (5/10) = 1/4.

The comparison between the two probabilities shows that the probability of winning without replacement (2/9) is greater than the probability of winning with replacement (1/4).

Therefore, you are less likely to win if you replace the first ball when selecting two balls from a container of 10 balls.