# Practice GMAT Questions

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**D**

- The probability of picking a 2, or any number, on the first try is 1/8.
- Since the picks are done with replacement, the probability of picking a 4, or any number other than that chosen on the first try, is 1/7.
- Since the picks are done with replacement, the probability of picking a 5, or any number other than those chosen on the first and second try, is 1/6.
- The probability of choosing 2, then 4, and finally 5 is:

(1/8)*(1/7)*(1/6) = 1/336 - However, since the order in which the balls are chosen does not matter in determining a winner, 1/336 is not the final answer and we are dealing with a combinatorics problem. The general formula is:

N!/(N-K)!

Where N is the total number of objects that can be selected

Where K is the number of objects that are selected - Since we have 3 total numbers that we care about (i.e., 2, 4, and 5) and can choose from in determining how many ways to select 3 numbers, N = 3. Note: N does not equal 8 in this instance since we only care about how many ways to select 2, 4, and 5.

Since we want to know how many ways we can select all three balls, K = 3 (i.e., we are selecting 3 balls). - Use the formula:

N!/(N-K)! = 3!/(3-3)! = 3!/0! = 3!/1 = 3! = 6 - Since there are 6 different permutations, the total probability is:

6*(1/336) = 6/336 = 1/56

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