Understanding the Probability Problem

The problem asks for the probability of selecting a specific combination of children (exactly 1 girl and 2 boys) when one child is chosen randomly from each of three different groups. To solve this, we need to consider the composition of each group and the possible ways to achieve the desired outcome.

Analyzing the Composition of Each Group

We have three distinct groups of children, each with a different number of girls and boys.

• Group 1: Contains 3 girls and 1 boy. Total children = $3 + 1 = 4$.
• Group 2: Contains 2 girls and 2 boys. Total children = $2 + 2 = 4$.
• Group 3: Contains 1 girl and 3 boys. Total children = $1 + 3 = 4$.

For each group, we can determine the probability of selecting a girl or a boy randomly.

Identifying Scenarios for 1 Girl and 2 Boys

We are selecting one child from each of the three groups. To end up with exactly 1 girl and 2 boys in total, there are three possible mutually exclusive scenarios:

• Scenario 1: Select a girl from Group 1, a boy from Group 2, and a boy from Group 3.
• Scenario 2: Select a boy from Group 1, a girl from Group 2, and a boy from Group 3.
• Scenario 3: Select a boy from Group 1, a boy from Group 2, and a girl from Group 3.

Since the selection from each group is an independent event, the probability of each scenario is the product of the probabilities of the individual selections.

Calculating Probability for Each Scenario

• Scenario 1 (Girl from G1, Boy from G2, Boy from G3):
  Probability = P(Girl from G1) $\times$ P(Boy from G2) $\times$ P(Boy from G3)
  Probability = $\frac{3}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{3 \times 1 \times 3}{4 \times 2 \times 4} = \frac{9}{32}$
• Scenario 2 (Boy from G1, Girl from G2, Boy from G3):
  Probability = P(Boy from G1) $\times$ P(Girl from G2) $\times$ P(Boy from G3)
  Probability = $\frac{1}{4} \times \frac{1}{2} \times \frac{3}{4} = \frac{1 \times 1 \times 3}{4 \times 2 \times 4} = \frac{3}{32}$
• Scenario 3 (Boy from G1, Boy from G2, Girl from G3):
  Probability = P(Boy from G1) $\times$ P(Boy from G2) $\times$ P(Girl from G3)
  Probability = $\frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} = \frac{1 \times 1 \times 1}{4 \times 2 \times 4} = \frac{1}{32}$

Calculating Total Probability

The total probability of selecting 1 girl and 2 boys is the sum of the probabilities of these three mutually exclusive scenarios:

Total Probability = P(Scenario 1) + P(Scenario 2) + P(Scenario 3)

Total Probability = $\frac{9}{32} + \frac{3}{32} + \frac{1}{32} = \frac{9 + 3 + 1}{32} = \frac{13}{32}$

Thus, the probability that the three selected children consist of 1 girl and 2 boys is $\frac{13}{32}$.

Revision Table: Probability Calculation Steps

Additional Information: Independent Events in Probability

In this probability problem, the selection of a child from one group is independent of the selection from another group. This means the outcome of choosing a child from Group 1 does not affect the outcome of choosing a child from Group 2 or Group 3. When events are independent, the probability that all of them occur is the product of their individual probabilities. This principle was used to calculate the probability of each scenario.

Also, the three scenarios identified (G1, B2, B3; B1, G2, B3; B1, B2, G3) are mutually exclusive. This means that if one scenario occurs, the others cannot occur at the same time. When calculating the total probability of any of these mutually exclusive events occurring, we simply add their individual probabilities.