Problem:
Three numbers are chosen at random without replacement from the set {1, 2, 3, ....8}. The probability that their minimum is 3, given that their maximum is 6, is:

a) 1/4
b) 2/5
c) 3/8
d) 1/5

Solution:
To solve this problem, we need to find the probability that the minimum of the three chosen numbers is 3, given that the maximum is 6. Let's break down the problem into smaller steps to find the solution.

Step 1: Determine the total number of ways to choose three numbers from the given set.
The set {1, 2, 3, ....8} contains 8 elements. Therefore, the total number of ways to choose three numbers from this set without replacement is given by the combination formula:

C(8, 3) = 8! / (3! * (8-3)!) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56

Step 2: Determine the number of ways to choose three numbers such that the maximum is 6.
In order to have the maximum as 6, we need to choose at least one number from the set {1, 2, 3, 4, 5, 6}. The remaining two numbers can be chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}.

Number of ways to choose one number from {1, 2, 3, 4, 5, 6} = C(6, 1) = 6
Number of ways to choose two numbers from {1, 2, 3, 4, 5, 6, 7, 8} = C(8, 2) = 8! / (2! * (8-2)!) = 8! / (2! * 6!) = (8 * 7) / (2 * 1) = 28

Therefore, the total number of ways to choose three numbers such that the maximum is 6 is given by:

6 * 28 = 168

Step 3: Determine the number of ways to choose three numbers such that the minimum is 3 and the maximum is 6.
In order to have the minimum as 3 and the maximum as 6, we need to choose one number from the set {3, 4, 5, 6}. The remaining two numbers can be chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}.

Number of ways to choose one number from {3, 4, 5, 6} = C(4, 1) = 4
Number of ways to choose two numbers from {1, 2, 3, 4, 5, 6, 7, 8} = C(8, 2) = 28

Therefore, the total number of ways to choose three numbers such that the minimum is 3 and