To solve this probability problem, we can use the concept of combinations. The total number of ways to choose 4 people from a group of 3 men, 4 women, and 2 children is given by the combination formula:

nCr = n! / (r!(n-r)!)

where n is the total number of people (3 men + 4 women + 2 children = 9) and r is the number of people to be chosen (4).

Step 1: Find the total number of ways to choose 4 people from the group.
9C4 = 9! / (4!(9-4)!) = 9! / (4!5!) = (9*8*7*6) / (4*3*2*1) = 126

So, there are 126 different possible committees that can be formed.

Step 2: Find the number of ways to choose 1 man, 1 woman, and 1 child.
There are 3 men, 4 women, and 2 children. We need to choose 1 man, 1 woman, and 1 child, which can be done in the following ways:
- Choose 1 man from 3: 3C1 = 3
- Choose 1 woman from 4: 4C1 = 4
- Choose 1 child from 2: 2C1 = 2

The total number of ways to choose 1 man, 1 woman, and 1 child is given by the product of these combinations:
3C1 * 4C1 * 2C1 = 3 * 4 * 2 = 24

Step 3: Find the number of ways to choose the remaining person.
After choosing 1 man, 1 woman, and 1 child, there is 1 remaining spot in the committee, which can be filled by any of the remaining 3 men, 3 women, or 1 child. So, the number of ways to choose the remaining person is 3 + 3 + 1 = 7.

Step 4: Find the probability.
The probability of choosing 1 man, 1 woman, and 1 child in the committee is given by:
(Number of favorable outcomes) / (Total number of possible outcomes)
= (Number of ways to choose 1 man, 1 woman, and 1 child) * (Number of ways to choose the remaining person) / (Total number of ways to choose 4 people)

= 24 * 7 / 126
= 168 / 126
= 4/3

Therefore, the probability that the committee contains 1 man, 1 woman, and 1 child is 4/3.

However, the options provided do not include 4/3 as a possible answer. So, we need to simplify this fraction. Dividing the numerator and denominator by 3, we get:

4/3 = (4/3) / (3/3) = 4/9

Hence, the correct answer is 4/9, which matches with option B.