Suppose an urn contains 10 balls numbered from 1 to 10. Someone draws 3 balls simultaneously from the urn (i.e. without replacement.) What is the probability that the lowest number drawn is 2?
(8C3)/(10C3) but I know that is wrong. I also tried ((8C3)+(7C3)+(6C3))/(10C3)
To calculate the probability that the lowest number drawn is 2, we need to consider the number of ways we can choose 3 balls from the urn such that the lowest numbered ball is 2, divided by the total number of ways to choose 3 balls from the urn.
First, let's find the total number of ways to choose 3 balls from the urn. This can be calculated using the combination formula, denoted as (nCr), where nCr represents the number of ways to choose r items from a group of n.
Using the combination formula, we calculate:
Total number of ways to choose 3 balls from 10: 10C3 = 120
Next, let's find the number of ways to choose 3 balls from the urn where the lowest numbered ball is 2. In this case, we fix the ball numbered 2 as the lowest, and we need to choose 2 more balls from the remaining 8 balls numbered 3 to 10.
Using the combination formula, we calculate:
Number of ways to choose 2 balls from 8: 8C2 = 28
Therefore, the probability that the lowest number drawn is 2 is:
Probability = Number of ways to choose 2 balls from 8 / Total number of ways to choose 3 balls from 10
Probability = 28 / 120 = 7 / 30
So, the correct answer is 7/30 or approximately 0.2333.