Problem 33
Question
A family has three children. Assuming a \(1: 1\) sex ratio, what is the probability that all of the children are girls?
Step-by-Step Solution
Verified Answer
The probability that all the children are girls is \(\frac{1}{8}\).
1Step 1: Determine Total Possible Outcomes
Each child can either be a boy (B) or a girl (G), leading to two possible outcomes for each child. Therefore, for three children, calculate the total number of combinations using \(2^3\): \[ \text{Total Outcomes} = 2^3 = 8 \]This means there are 8 total possible gender combinations for three children.
2Step 2: Identify Favorable Outcomes
To find the probability that all three children are girls, identify the specific outcome where each child is a girl: (G, G, G). This is one specific combination out of the total 8.
3Step 3: Calculate the Probability
Probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes. In this case:\[ P(\text{all girls}) = \frac{1}{8} \]This fraction \(\frac{1}{8}\) represents the probability that all three children are girls.
Key Concepts
Understanding Sex RatioExploring CombinatoricsCalculating Probability
Understanding Sex Ratio
In probability theory, when we talk about the sex ratio, we're often referring to the likelihood of a child being born as either a boy or a girl. The "1:1 sex ratio" simplifies this to mean that each child has an equal chance of being a boy or a girl. Just like flipping a fair coin, each gender has a probability of \( \frac{1}{2} \). This forms the basis for solving many problems in probability concerning bearings on gender, such as predicting outcomes for a family with multiple children.
When considering a family of three children, each individual child could be a boy or a girl, independent of what gender the previous child was. This results in a variety of potential combinations for the children's gender for the whole family.
When considering a family of three children, each individual child could be a boy or a girl, independent of what gender the previous child was. This results in a variety of potential combinations for the children's gender for the whole family.
Exploring Combinatorics
Combinatorics is the mathematical field dealing with counting, arrangement, and combination of objects. That's exactly the kind of theory used to figure out how many different ways the genders can be arranged in a family of three kids.
With two gender choices for each child — either boy (B) or girl (G) — the total number of combinations can be calculated using the formula \( 2^n \), where \( n \) represents the number of children. Here, \( n = 3 \), the result is \( 2^3 = 8 \), which indicates there are eight conceivable gender distributions for three children.
With two gender choices for each child — either boy (B) or girl (G) — the total number of combinations can be calculated using the formula \( 2^n \), where \( n \) represents the number of children. Here, \( n = 3 \), the result is \( 2^3 = 8 \), which indicates there are eight conceivable gender distributions for three children.
- (B, B, B)
- (B, B, G)
- (B, G, B)
- (B, G, G)
- (G, B, B)
- (G, B, G)
- (G, G, B)
- (G, G, G)
Calculating Probability
The core of probability calculation lies in determining how likely an event is to happen. It's expressed as the ratio of favorable outcomes to the total possible outcomes.
Here, we want to calculate the probability of all three children being girls (G, G, G). Step by step, we identify one favorable outcome — all children being girls — among the 8 possible combinations we previously calculated. Using the probability formula:
Here, we want to calculate the probability of all three children being girls (G, G, G). Step by step, we identify one favorable outcome — all children being girls — among the 8 possible combinations we previously calculated. Using the probability formula:
- Probability \( P(\text{all girls}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \ = \frac{1}{8} \)
Other exercises in this chapter
Problem 33
Toss a fair coin 10 times. Let \(X\) be the number of heads. Find (a) \(P(X=5)\). (b) \(P(X \geq 8)\). (c) \(P(X \leq 9)\).
View solution Problem 33
Suppose that \(X\) is standard normally distributed. Find \(E(|X|)\).
View solution Problem 33
A multiple-choice question has four choices, and a test has a total of 10 multiple-choice questions. A student passes the test only if he or she answers all que
View solution Problem 33
Let \(S=\\{a, b, c\\} .\) List all possible subsets, and argue that the total number of subsets is \(2^{3}=8\).
View solution