Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys: 1 girl and 3 boys. One child is selected at random from each group. The probability that the three selected consist of 1 girl and 2 boys is
13/32
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.
We have three distinct groups of children, each with a different number of girls and boys.
For each group, we can determine the probability of selecting a girl or a boy randomly.
| Group | Number of Girls | Number of Boys | Total Children | P(Select Girl) | P(Select Boy) |
|---|---|---|---|---|---|
| Group 1 | 3 | 1 | 4 | $\frac{3}{4}$ |
$\frac{1}{4}$ |
| Group 2 | 2 | 2 | 4 | $\frac{2}{4} = \frac{1}{2}$ |
$\frac{2}{4} = \frac{1}{2}$ |
| Group 3 | 1 | 3 | 4 | $\frac{1}{4}$ |
$\frac{3}{4}$ |
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:
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.
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 $\frac{13}{32}$.
| Step | Description | Calculation/Result |
|---|---|---|
| 1 | Determine group compositions and individual selection probabilities. | P(G1)=3/4, P(B1)=1/4, P(G2)=1/2, P(B2)=1/2, P(G3)=1/4, P(B3)=3/4 |
| 2 | Identify scenarios for 1 girl and 2 boys. | (G1, B2, B3), (B1, G2, B3), (B1, B2, G3) |
| 3 | Calculate probability of each scenario. | P(G1,B2,B3) = 9/32 P(B1,G2,B3) = 3/32 P(B1,B2,G3) = 1/32 |
| 4 | Sum scenario probabilities for total probability. | 9/32 + 3/32 + 1/32 = 13/32 |
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.
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