Question

13.36 a probability teaser. lets assume it is safe to say that people are either male or female at birth¹⁸ and that each child born is equally likely to be a boy or a girl and that the sexes of successive children are independent. if we let bg mean that the older child is a boy and the younger child is a girl, then each of the combinations bb, bg, gb, and gg has probability 0.25. ashley and brianna each have two children.

a. you know that at least one of ashleys children is a boy. what is the conditional probability that she has two boys?

b. you know that briannas older child is a boy. what is the conditional probability that she has two boys?

Explanation:

Step1: Define events for part a

Let \(A\) be the event that Ashley has at least one boy. The sample space for two - children is \(\{BB,BG,GB,GG\}\). \(n(A)=3\) (since \(A = \{BB,BG,GB\}\)). Let \(B\) be the event that Ashley has two boys. \(n(B) = 1\) (i.e., \(BB\)).
The formula for conditional probability is \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Since \(B\subseteq A\), \(A\cap B = B\).

Step2: Calculate conditional probability for part a

Using the formula \(P(B|A)=\frac{n(B)}{n(A)}\). Substituting \(n(B) = 1\) and \(n(A)=3\), we get \(P(B|A)=\frac{1}{3}\).

Step3: Define events for part b

Let \(C\) be the event that Brianna's older child is a boy. The sample - space for the condition is \(\{BB,BG\}\). Let \(D\) be the event that Brianna has two boys.
The formula for conditional probability \(P(D|C)=\frac{P(C\cap D)}{P(C)}\). Since \(D\subseteq C\) (if \(D = \{BB\}\) and \(C=\{BB,BG\}\)), \(C\cap D = D\).

Step4: Calculate conditional probability for part b

Using the formula \(P(D|C)=\frac{n(D)}{n(C)}\). Here, \(n(D) = 1\) and \(n(C)=2\). So \(P(D|C)=\frac{1}{2}\).

Answer:

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