Question

There are three groups of children having 3 girls and one boy, 2 girls and 2 boys, one girl and 3 boys respectively. One child is selected at random from each group. Find the probability that the three selected children have one girl and 2 boys.

Hint:

For probability problems involving multiple independent events and specific combined outcomes, it's essential to list all the mutually exclusive ways the outcome can occur. Calculate the probability for each way and then add them up.

Answer 1

Step 1: Understanding the Concept:
We need to find the probability of a specific outcome (1 girl and 2 boys) when making independent selections from three different groups. This can happen in several mutually exclusive ways (e.g., girl from group 1, boys from groups 2 & 3; or boy from group 1, girl from group 2, boy from group 3, etc.). The total probability is the sum of the probabilities of these individual cases.
Step 2: Key Formula or Approach:
1. Determine the probability of selecting a girl (G) and a boy (B) from each group.
2. Identify all possible combinations of selections that result in one girl and two boys.
3. Calculate the probability for each combination by multiplying the probabilities of the independent selections.
4. Sum the probabilities of all combinations.
Step 3: Detailed Explanation or Calculation:
Let's denote the probabilities for each group:
- Group 1: 3 Girls, 1 Boy (Total 4). \( P(G_1) = 3/4, P(B_1) = 1/4 \).
- Group 2: 2 Girls, 2 Boys (Total 4). \( P(G_2) = 2/4 = 1/2, P(B_2) = 2/4 = 1/2 \).
- Group 3: 1 Girl, 3 Boys (Total 4). \( P(G_3) = 1/4, P(B_3) = 3/4 \).
The three possible ways to get exactly one girl and two boys are:
- Case 1: Girl from Group 1, Boy from Group 2, Boy from Group 3 (G B B)
\( P(\text{Case 1}) = P(G_1) \times P(B_2) \times P(B_3) = \frac{3}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{9}{32} \)
- Case 2: Boy from Group 1, Girl from Group 2, Boy from Group 3 (B G B)
\( P(\text{Case 2}) = P(B_1) \times P(G_2) \times P(B_3) = \frac{1}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{32} \)
- Case 3: Boy from Group 1, Boy from Group 2, Girl from Group 3 (B B G)
\( P(\text{Case 3}) = P(B_1) \times P(B_2) \times P(G_3) = \frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{32} \)
The total probability is the sum of the probabilities of these mutually exclusive cases:
\[ P(\text{1 girl, 2 boys}) = P(\text{Case 1}) + P(\text{Case 2}) + P(\text{Case 3}) \] \[ = \frac{9}{32} + \frac{3}{32} + \frac{1}{32} = \frac{13}{32} \] Step 4: Final Answer:
The probability that the three selected children consist of one girl and two boys is \( \frac{13}{32} \).