There is a large container which will hold 10 balls with two of the balls being purple. 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.
Initially, there are 2 purple balls out of 10 total balls, so the probability of selecting a purple ball on the first try is 2/10 or 1/5.
If you do not replace the first ball, there will be 9 balls left in the container, with only 1 purple ball remaining out of these 9 balls. Therefore, the probability of selecting a purple ball on the second try is 1/9.
To find the overall probability of selecting two purple balls in a row when not replacing the first ball, you multiply the probabilities together:
1/5 * 1/9 = 1/45
Now, if you replace the first ball, after selecting the first ball, there will still be 2 purple balls remaining out of 10 total balls. Therefore, the probability of selecting a purple ball on the second try is still 2/10 or 1/5.
Multiply this probability by the probability of selecting a purple ball on the first try:
1/5 * 1/5 = 1/25
Then, calculate the difference in probability of winning when replacing the first ball compared to not replacing it:
1/45 - 1/25 = 5/225 - 9/225 = -4/225
Therefore, when you are required to replace the first ball, you are 4/225 less likely to win than when not replacing it.