Problem 63 - Algebra and Trigonometry

Question

Mrs. Smith says, “I have two children—the older one is named William.” Mrs. Jones replies, “One of my two children is also named William.” For each woman, list the sample space for the genders of her children, and calculate the probability that her other child is also a son. Explain why these two probabilities are different.

Step-by-Step Solution

Answer

Mrs. Smith's probability: 1/2; Mrs. Jones's probability: 2/3. The difference is due to initial information constraints.

Step 1: Understanding the problem

We have two women, each with two children. Mrs. Smith's older child is named William, while Mrs. Jones says that one of her children is named William. We need to list the sample spaces for possible gender combinations of their children and find the probability that the other child is a son for each woman.

Step 2: List sample space for Mrs. Smith

Since we know her older child is William (a son), the sample space for the gender of the two children is limited to the possibilities where the first child is a son: 1. Older son, Younger son (SS) 2. Older son, Younger daughter (SD) Thus, Mrs. Smith's sample space is SS and SD.

Step 3: Calculate probability for Mrs. Smith

Out of the two possibilities in her sample space (SS and SD), only one possibility (SS) results in the other child being a son. Therefore, the probability that the other child is a son for Mrs. Smith is given by: \[P(\text{Other child is a son}) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{1}{2}\]

Step 4: List sample space for Mrs. Jones

Mrs. Jones has one son named William, but we don't know the order. Her sample space includes: 1. Older son, Younger son (SS) 2. Older son, Younger daughter (SD) 3. Older daughter, Younger son (DS) This includes cases where William could be the older or younger child.

Step 5: Calculate probability for Mrs. Jones

Out of the three possibilities in her sample space (SS, SD, and DS), two possibilities (SS and DS) result in the other child being a son. Therefore, the probability for Mrs. Jones is:\[P(\text{Other child is a son}) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{2}{3}\]

Step 6: Explain difference in probabilities

The difference arises because knowing that Mrs. Smith's older child is William restricts possible combinations to those that begin with a son, affecting the sample space. In Mrs. Jones's case, we only know that one child is William, allowing more possibilities.

Key Concepts

Sample Space

In probability theory, a sample space encompasses all the possible outcomes of a particular situation. When dealing with children and their genders in families, the sample space helps us frame all potential gender combinations.

For Mrs. Smith, whose older child is definitely William, we know that the first child is a boy. Thus, the sample space is narrowed down considerably. The potential combinations are:

• Older son, Younger son (SS)
• Older son, Younger daughter (SD)

This list represents all possible gender scenarios under the given conditions.

For Mrs. Jones, things are different. She mentions having a child named William, but doesn't specify their order among her two children. Therefore, the scenarios expand to include:

• Older son, Younger son (SS)
• Older son, Younger daughter (SD)
• Older daughter, Younger son (DS)

Each of these possibilities acknowledges that William can be either the older or the younger child. Understanding sample space is crucial for accurately calculating probabilities.

Gender Combinations

Gender combinations in a family context refer to the different sequences of sons and daughters possible among siblings. For any given pair of children, if not specified, they can be either a, boy or girl - resulting in these combinations:

• Older son, Younger son (SS)
• Older son, Younger daughter (SD)
• Older daughter, Younger son (DS)
• Older daughter, Younger daughter (DD)

Because of specific information provided in this problem about the children named William, not all combinations are applicable.

For Mrs. Smith, her older child being William limits the sequence to SS or SD. In contrast, Mrs. Jones's description allows three distinct possibilities – SS, SD, and DS – since we don't know if William is first or second.

Breaking down gender combinations in such manner helps clarify how probabilities are derived from restricted sample spaces.

Conditional Probability

Conditional probability is the likelihood of an event occurring based on a given condition or event happening first. In these family scenarios, we're given certain information (a child named William) that impacts our probability calculations.

For Mrs. Smith, the condition stated is that the older child is William. Thus, the probability that the other child is also a son depends on this fixed scenario: out of two possibilities (SS and SD), only SS results in the second child being a son. Therefore, the probability is \(\frac{1}{2}\).

For Mrs. Jones, the condition is less specific—one child is named William, not specifying which one. With her sample space being SS, SD, and DS, we look at all scenarios where the unmentioned child is a son (SS or DS). Therefore, the probability calculation becomes \(\frac{2}{3}\).

Understanding these probabilities lies in recognizing how the initial condition changes the likelihood by limiting possible outcomes.