An um contains balls numbered 1 through 10. A ball is chosen, retured to the ur, and a second ball is chosen. What is the probability that the first and second balls
Assuming that "um" refers to "urn" or "container," and "retured" refers to "returned," you are asking for the probability of selecting two specific numbers from a set of numbered balls.
The total number of possible outcomes is the total number of ways to select two balls from the urn, which can be calculated using the combination formula, nCr. In this case, n = 10 (total number of balls) and r = 2 (number of balls to be chosen).
nCr = n! / (r!(n-r)!)
10C2 = 10! / (2!(10-2)!)
= 10! / (2!8!)
= (10 x 9) / (2 x 1)
= 45
The number of favorable outcomes, in this case, is 1 (selecting a specific numbered ball twice).
Therefore, the probability of selecting a specific numbered ball twice is:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 1 / 45
≈ 0.0222 or 2.22%