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A couple has 2 children. What is the probability that both are girls if the older of the two is a girl?

Short Answer

Expert verified
The probability that both children are girls given that the older child is a girl is \( \frac{1}{2} \), or 50%.

Step by step solution

01

Identify the events

Let's denote the following events: A - the event that the older child is a girl B - the event that both children are girls We want to find the probability P(B|A), which is the probability that both children are girls given that the older child is a girl.
02

List the possible outcomes

There are 4 possible outcomes when a couple has two children, consisting of the gender of each child (girl or boy) as shown below: 1. Older child: Girl, Younger child: Girl (GG) 2. Older child: Girl, Younger child: Boy (GB) 3. Older child: Boy, Younger child: Girl (BG) 4. Older child: Boy, Younger child: Boy (BB) In our problem, we know that the older child is a girl, so we can eliminate outcomes 3 and 4. This leaves us with only 2 possible outcomes: 1. Older child: Girl, Younger child: Girl (GG) 2. Older child: Girl, Younger child: Boy (GB)
03

Calculate the conditional probability

Now that we have listed the possible outcomes, we can calculate the conditional probability P(B|A) using the following formula: \[ P(B|A) = \frac{P(A \cap B)}{P(A)} \] Now, we need to determine P(A) and P(A ∩ B) to calculate the probability we are looking for. P(A) is the probability that the older child is a girl. Since there are 2 possibilities, GG and GB, each with an equal chance, P(A) = 1/2. P(A ∩ B) is the probability that both events A and B occur together. In other words, it is the probability that the older child is a girl and both children are girls. There is only 1 outcome that satisfies this requirement, which is GG. So, P(A ∩ B) = 1/4.
04

Calculate the final probability

Now that we have determined P(A) and P(A ∩ B), we can find P(B|A) by substituting the values into the formula: \[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{2} \] So, the probability that both children are girls given that the older child is a girl is 1/2, or 50%.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics concerned with the analysis of random events. The probability of an event is a measure of the likelihood that the event will occur, and it's usually expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

Within the realm of probability theory, conditional probability is a crucial concept. It refers to the probability of an event occurring given that another event has already occurred. In mathematical terms, the conditional probability of event B given event A is denoted by P(B|A).

To compute conditional probabilities, a basic understanding of how to identify and distinguish between independent events, mutually exclusive events, and complementary events is essential. The exercise highlighted involves such an understanding by dealing with the probability of both children being girls, given that the older child is already known to be a girl.
Combinatorics
Combinatorics is the area of mathematics that deals with counting, both as a means and an end in describing the structure of various objects. It is fundamental in the calculation of probability since outcomes and their likelihoods can be identified through combinatorial analysis.

In the example provided, combinatorics plays a role in listing the possible outcomes of the genders of two children. The four possible combinations GG, GB, BG, and BB can be considered exhaustive of all possibilities for this simple event space. The principles of combinatorics allow us to understand that, once the information that the older child is a girl is known, the number of relevant outcomes is reduced, which directly affects the conditional probability in question.

Understanding the basics of combinatorics can greatly simplify many probability problems by helping to clearly identify the sample space and the favorable outcomes, as seen in our textbook exercise.
Probability of Events
The probability of an event is a measure of how likely that event is to occur. It’s calculated by dividing the number of favorable outcomes by the total number of possible outcomes. When dealing with conditional probabilities, the set of possible outcomes may change based on the given conditions.

In our textbook problem, the event of interest (event B) was having both children as girls. Determining the probability involved identifying the reduced sample space once the condition (event A) that the older child is a girl was taken into account. From a combinatorial perspective, out of the two remaining scenarios (GG and GB), only one, GG, met the criteria for event B, leading to a conditional probability of 1/2.

Grasping the nuances of the probability of events, especially in conditional contexts, ensures that students can better tackle similar problems and comprehend the underlying mechanisms dictating the likelihood of various events.

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Most popular questions from this chapter

3.25. The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of people you encounter who are over \(50 .\) Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Comment on this method. Hint: Let \(p\) denote the proportion of people in the town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is the estimate approximately equal to \(p ?\)

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin that lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2} .\) Consider the following procedure for accomplishing our task: 1\. Flip the coin. 2\. Flip the coin again. 3\. If both flips land on heads or both land on tails, return to step 1. 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

In Laplace's rule of succession (Example \(5 \mathrm{e}\) ), are the outcomes of the successive flips independent? Explain.

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes \(1,2,\) or \(3 .\) Given that outcome 3 is the last of the three outcomes to occur, find the conditional probability that (a) the first trial results in outcome 1 ; (b) the first two trials both result in outcome \(1 .\)

There is a \(50-50\) chance that the queen carries the gene for hemophilia. If she is a carrier, then each prince has a \(50-50\) chance of having hemophilia. If the queen has had three princes without the disease, what is the probability that the queen is a carrier? If there is a fourth prince, what is the probability that he will have hemophilia?

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