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Chapter 4: Problem 11

Consider a family with 3 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also \(0.5,\) and that the events "boy" and "girl" are independent. (a) List the equally likely events for the gender of the 3 children, from oldest to youngest. (b) What is the probability that all 3 children are male? Notice that the complement of the event "all three children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female.

Short Answer

Expert verified
Sample space: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG. \( P(\text{BBB}) = 0.125 \). \( P(\text{at least one girl}) = 0.875 \).

Step by step solution

01

Define the Sample Space

Each child can either be a boy (B) or a girl (G). Thus, for 3 children, we list all possible combinations of boys and girls: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG. These 8 combinations make up the sample space, and each is equally likely.
02

Calculate the Probability of All Boys

The probability for each child to be a boy is 0.5, and since the events are independent, the probability of all 3 being boys (BBB) is calculated by multiplying each probability: \( P(\text{BBB}) = 0.5 \times 0.5 \times 0.5 = 0.125 \).
03

Determine the Complement Event

The complementary event to all 3 children being boys is that at least one child is a girl (not all are boys). Since these are complementary, their probabilities sum to 1. Calculate the probability of the complement by subtracting the probability of "all boys" from 1.
04

Calculate Probability of at Least One Girl

Using the complement principle: \( P(\text{at least one girl}) = 1 - P(\text{BBB}) = 1 - 0.125 = 0.875 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Space
In probability theory, a sample space is an essential concept that represents all possible outcomes of an experiment. For instance, consider a family with three children. Each child can be either a boy (B) or a girl (G). So, to construct the sample space, we must list every combination of B's and G's for the three children. This will help us visualize all potential outcomes.
  • BBB
  • BBG
  • BGB
  • BGG
  • GBB
  • GBG
  • GGB
  • GGG
Each combination signifies one unique outcome where the order corresponds to the children from oldest to youngest. For example, "BGB" signifies that the oldest and the youngest are boys, and the middle child is a girl. Since each child has an independent and equal chance, these eight combinations form the sample space of the problem, with each being equally likely.
Complement Rule
The complement rule is a handy tool in probability that helps us calculate the likelihood of an event not occurring by using its counterpart, the complement. In the context of the problem, if we want to find the probability that at least one child is a girl, it's easier to first find the probability that all three children are boys and then use the complement rule.
Let's consider the event that all three children are boys (denoted as BBB). We previously calculated its probability as 0.125. Since the only alternative to all children being boys is for at least one to be a girl, these two scenarios are complements of each other.
Using the complement rule, we denote the event of at least one child being a girl as "not BBB." We find the probability of this event, "not BBB," by simply subtracting the probability of BBB from 1:
\[P( ext{at least one girl}) = 1 - P( ext{all boys}) = 1 - 0.125 = 0.875\]This approach effectively leverages the fact that the probabilities of complementary events sum to 1, making calculations simpler when dealing with many possible outcomes.
Independent Events
Understanding independent events is crucial in probability theory, as it describes situations where the occurrence of one event does not affect another. In this scenario, the gender of one child does not influence the gender of any siblings. Thus, each event鈥攂eing a boy (B) or a girl (G)鈥攈appens with an equal probability of 0.5, independently of others.
In our specific task, to anchor the idea of independence, consider calculating the probability of the three children all being boys. Since each child's gender does not affect the others, we multiply the probabilities for each child. This gives us:
\[P( ext{all boys}) = 0.5 imes 0.5 imes 0.5 = 0.125\]This multiplication works because the events are independent. If the occurrence of one outcome affected another, we couldn't simply multiply the probabilities. Recognizing independence allows simplifying complex probability calculations, especially when dealing with multiple events simultaneously.

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