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What is the probability that a family of two children has (a) two boys given that it has at least one boy? (b) two boys given that the first child is a boy?

Short Answer

Expert verified
(a) \( \frac{1}{3} \); (b) \( \frac{1}{2} \).

Step by step solution

01

Understanding the Problem

We need to calculate the probabilities under two different conditions for a family with two children. We'll focus on all possible gender combinations for two children: Boy-Boy (BB), Boy-Girl (BG), Girl-Boy (GB), and Girl-Girl (GG).
02

Determining the Probability for Part (a)

For part (a), if a family has at least one boy, we must consider the combinations BB, BG, and GB, excluding GG since it contains no boys. The probability is the number of favorable outcomes (BB) divided by all possible outcomes (BB, BG, GB).
03

Calculating the Probability for Part (a)

There are three scenarios where there is at least one boy: BB, BG, GB. Among these, only BB has two boys. So, the probability is: \[ P(\text{2 boys} \mid \text{at least 1 boy}) = \frac{1}{3} \]
04

Determining the Probability for Part (b)

For part (b), given that the first child is a boy, we only consider the combinations BB and BG. The probability is the number of favorable outcomes (BB) divided by these possible outcomes (BB, BG).
05

Calculating the Probability for Part (b)

With the first child being a boy, there are two outcomes: BB and BG. Out of these, one has two boys (BB). So, the probability is: \[ P(\text{2 boys} \mid \text{first is a boy}) = \frac{1}{2} \]

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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 that studies randomness and uncertainty. It deals with calculating the likelihood of events in a probabilistic space. In our context of a family with two children, we use probability theory to determine the chance of having specific gender combinations, such as two boys.

When calculating probabilities, we assign numerical values between 0 and 1, where 0 means an event will not happen, and 1 means it is certain to happen. For example, if you flip a fair coin, the probability of landing on heads is 0.5, since there are two equally likely outcomes: heads or tails.

The step-by-step solution uses these fundamental principles of probability by identifying all possible outcomes for the children's genders and considering the situations where specific conditions (like having at least one boy) are met. By focusing only on the relevant outcomes, we effectively address each part of the problem.
Combinatorics
Combinatorics is the area of mathematics concerned with counting, arrangement, and combination of sets of elements. In probability questions, it helps us determine the number of possible outcomes and combinations in a given scenario.

For our exercise, there's a need to count the different possible gender combinations in a two-child family: Boy-Boy (BB), Boy-Girl (BG), Girl-Boy (GB), and Girl-Girl (GG). This is like asking how you can arrange two items when each could be either a boy or a girl.

Combinatorics helps in identifying subsets of these combinations by applying conditions, like ignoring the GG outcome when considering families with at least one boy. This precise counting ensures the accuracy of the probability calculation by analyzing only those scenarios that meet the conditions of the problem. Through combinatorial methods, we simplify the task of counting outcomes that fulfill particular conditions.
Bayes' Theorem
Bayes' theorem is a fundamental concept in probability theory allowing us to update the probability of a hypothesis based on new evidence. While it's not explicitly used in the simple calculations of this exercise, understanding its foundation enriches the comprehension of conditional probabilities.

Bayes’ theorem describes the probability of an event, based on prior knowledge of conditions that might be related to the event. For instance, in finding the probability of two boys given at least one boy, we implicitly use the idea of Bayes’ theorem by revising our probability with the information that excludes one possible event (Girl-Girl).

Formally, Bayes’ theorem is expressed as:
  • \[ P(A \,|\, B) = \frac{P(B \,|\, A) \, P(A)}{P(B)} \]
Where:
  • \(P(A \,|\, B)\) is the probability of event A given B is true.
  • \(P(B \,|\, A)\) is the probability of event B given A is true.
  • \(P(A)\) is the probability of A occurring.
  • \(P(B)\) is the probability of B occurring.
In our exercise, although we've used straightforward conditional probability steps, Bayes' theorem provides the theoretical background that supports these calculations by showing how probabilities are updated as conditions change.

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Most popular questions from this chapter

In the problem of points, discussed in the historical remarks in Section 3.2, two players, A and B, play a series of points in a game with player A winning each point with probability \(p\) and player \(\mathrm{B}\) winning each point with probability \(q=1-p\). The first player to win \(N\) points wins the game. Assume that \(N=3 .\) Let \(X\) be a random variable that has the value 1 if player A wins the series and 0 otherwise. Let \(Y\) be a random variable with value the number of points played in a game. Find the distribution of \(X\) and \(Y\) when \(p=1 / 2\). Are \(X\) and \(Y\) independent in this case? Answer the same questions for the case \(p=2 / 3\).

A coin is tossed three times. Consider the following events A: Heads on the first toss. B: Tails on the second. \(C:\) Heads on the third toss. \(D:\) All three outcomes the same \((\mathrm{HHH}\) or TTT). \(E:\) Exactly one head turns up. (a) Which of the following pairs of these events are independent? (1) \(A, B\) (2) \(A, D\) (3) \(A, E\) (4) \(D, E\) (b) Which of the following triples of these events are independent? (1) \(A, B, C\) (2) \(A, B, D\) (3) \(C, D, E\)

A coin is in one of \(n\) boxes. The probability that it is in the \(i\) th box is \(p_{i}\). If you search in the \(i\) th box and it is there, you find it with probability \(a_{i}\). Show that the probability \(p\) that the coin is in the \(j\) th box, given that you have looked in the \(i\) th box and not found it, is $$ p=\left\\{\begin{array}{cc} p_{j} /\left(1-a_{i} p_{i}\right), & \text { if } j \neq i \\ \left(1-a_{i}\right) p_{i} /\left(1-a_{i} p_{i}\right), & \text { if } j=i \end{array}\right. $$

It is desired to find the probability that in a bridge deal each player receives an ace. A student argues as follows. It does not matter where the first ace goes. The second ace must go to one of the other three players and this occurs with probability \(3 / 4 .\) Then the next must go to one of two, an event of probability \(1 / 2,\) and finally the last ace must go to the player who does not have an ace. This occurs with probability \(1 / 4 .\) The probability that all these events occur is the product \((3 / 4)(1 / 2)(1 / 4)=3 / 32 .\) Is this argument correct?

A coin is tossed three times. What is the probability that exactly two heads occur, given that (a) the first outcome was a head? (b) the first outcome was a tail? (c) the first two outcomes were heads? (d) the first two outcomes were tails? (e) the first outcome was a head and the third outcome was a head?

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