Four persons are chosen from a group containing 3 men, 2 women and 4 children. Show thatthe chance that exactly two of them will be children is 10/21.

To solve this problem, we need to use the concept of combinations in probability.

Step 1: Calculate the total number of ways to choose 4 persons from a group of 9 (3 men, 2 women, and 4 children). This can be done using the combination formula C(n, r) = n! / [(n-r)!r!], where n is the total number of items, and r is the number of items to choose.

So, the total number of ways to choose 4 persons from 9 is C(9, 4) = 9! / [(9-4)!4!] = 126.

Step 2: Calculate the number of ways to choose exactly 2 children from 4. This is C(4, 2) = 4! / [(4-2)!2!] = 6.

Step 3: Calculate the number of ways to choose the remaining 2 persons from the 5 adults (3 men and 2 women). This is C(5, 2) = 5! / [(5-2)!2!] = 10.

Step 4: Multiply the results of steps 2 and 3 to get the number of ways to choose exactly 2 children and 2 adults. This is 6 * 10 = 60.

Step 5: The probability that exactly two of the chosen persons will be children is the number of ways to choose exactly 2 children and 2 adults (from step 4) divided by the total number of ways to choose 4 persons (from step 1). This is 60 / 126 = 10 / 21.

Therefore, the chance that exactly two of the chosen persons will be children is 10/21.