Miscellaneous · Q2: Two children: both males given at least one male, $\frac13$⁠; both females given the elder is female, $\frac12$

Basics you need

Part (i): both males, given at least one male

1. The sample space, elder child first

   Write M for male and F for female, with the elder child first. Then

   $$S = \{MM, MF, FM, FF\},$$

   and each outcome has probability $\frac14$⁠.

   Each child has 2 possibilities, so there are $2 \times 2 = 4$ families. MF (elder male, younger female) and FM (elder female, younger male) are different outcomes; merging them into one "one of each" outcome would give a sample space whose outcomes are not equally likely. Part (ii) needs the order anyway, because it mentions the elder child.
   Multiply the choices. Basics: counting

2. The two events as sets

   Let $E$⁠: both children are males, and $F$⁠: at least one child is male. Then

   $$E = \{MM\}, \quad F = \{MM, MF, FM\},$$ $$E \cap F = \{MM\}.$$

   So $P(F) = \frac34$ and $P(E \cap F) = \frac14$⁠.

   "At least one male" means one male or two males: every family except FF. Both males is the single family MM, which lies inside $F$⁠, so $E \cap F = E$⁠.
   An event is a set of outcomes. Basics: events as sets

3. Apply the definition

   $$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} \\ &= \frac14 \div \frac34 \\ &= \frac14 \times \frac43 = \frac13. \end{aligned}$$
   Knowing that at least one child is male shrinks the sample space to the three families MM, MF, FM. They stay equally likely, and only one of them has two males.
   Section 13.2, Definition 1. Lesson: conditional probability

4. Why the answer is not $\frac12$

   It is tempting to say "one child is a boy, the other is a boy or a girl, so $\frac12$". But "at least one is male" does not say which child is male. In MF it is the elder, in FM the younger, in MM both. Of these three equally likely families, one has a second boy: $\frac13$⁠.

   The $\frac12$ argument treats "the other child" as a fixed child, but the information does not single one child out. Part (ii) is different: there the information is about one named child (the elder).
   Knowing something shrinks the world. Examples 1 to 7

Part (ii): both females, given the elder is female

1. The two events as sets

   Use the same sample space as Part (i), Step 1. Let $A$⁠: both children are females, and $B$⁠: the elder child is female. Then

   $$A = \{FF\}, \quad B = \{FM, FF\},$$ $$A \cap B = \{FF\}.$$

   So $P(B) = \frac24$ and $P(A \cap B) = \frac14$⁠.

   The elder child is the first letter, so $B$ is the two families that start with F. Both females is FF, which is one of them.
   An event is a set of outcomes. Basics: events as sets

2. Apply the definition

   $$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac14 \div \frac24 \\ &= \frac14 \times \frac42 = \frac12. \end{aligned}$$
   Given that the elder is female, only FM and FF remain, and they are equally likely. Both children are females only in FF.
   Section 13.2, Definition 1. Lesson: conditional probability

3. A second way to see $\frac12$

   Once the elder is known to be female, both children are females exactly when the younger child is female. The younger child is male or female with probability $\frac12$ each, whatever the elder is, so the answer is $\frac12$⁠.

   The two children are independent, so information about the elder tells us nothing about the younger. In Part (i) the information did not name a child, which is why its answer is different.
   Independent events. Lesson: independent events

Every fact used, and where it comes from