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 favorable outcomes and divide it by the total number of possible outcomes.
First, let's determine the total number of possible outcomes. We are drawing 3 balls from an urn containing 10 numbered balls, so the total number of ways to choose 3 balls is given by the combination formula: 10C3 = (10!)/(3!(10-3)!) = 120.
Now, let's consider the favorable outcomes. To have a favorable outcome, the smallest number drawn must be 2. We have two cases to consider:
Case 1: The smallest number drawn is 2, and the other two numbers are greater than 2.
In this case, we have 8 balls numbered from 3 to 10 to choose the other two balls from. The number of ways to choose 2 balls from 8 is given by the combination formula: 8C2 = (8!)/(2!(8-2)!) = 28.
Case 2: The smallest number drawn is 2, and one of the other numbers is also 2.
In this case, we have one ball numbered 2 and seven balls numbered from 3 to 10 to choose the other ball from. The number of ways to choose 1 ball from 7 is given by the combination formula: 7C1 = (7!)/(1!(7-1)!) = 7.
Now, we can calculate the probability by summing up the number of favorable outcomes for both cases and dividing by the total number of possible outcomes:
P(lowest number drawn is 2) = (28 + 7)/120 = 35/120 = 7/24.
Therefore, the probability that the lowest number drawn is 2 is 7/24 or approximately 0.292.