Chapter 13 · Probability

Exercise 13.1 · Q12: Two children: both girls, given the youngest or at least one is a girl

Exercise 13.1Conditional probability · Section 13.2

The question

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Exam answer

Let b denote a boy and g a girl, with the elder child written first. The 4 outcomes are equally likely:

\begin{aligned} S = \{&(b, b), (b, g), \\ &(g, b), (g, g)\} \end{aligned}

Let $E$ ⁠: both are girls, so $E = \{(g, g)\}$ ⁠.

(i) Let $F$ ⁠: the youngest (the younger of the two) is a girl.

F = \{(b, g), (g, g)\}

so $E\cap F = \{(g, g)\}$ ⁠. By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{1}{4}\div\frac{2}{4} \end{aligned}

$P(E|F) = \frac{1}{2}$ ⁠.

(ii) Let $G$ ⁠: at least one is a girl.

G = \{(b, g), (g, b), (g, g)\}

so $E\cap G = \{(g, g)\}$ ⁠. By the definition of conditional probability,

\begin{aligned} P(E|G) &= \frac{P(E\cap G)}{P(G)} \\ &= \frac{1}{4}\div\frac{3}{4} \end{aligned}

$\therefore\ P(E|G) = \dfrac{1}{3}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

List the families as ordered pairs (elder, younger): there are 4 equally likely ones. The two conditions sound alike but keep different numbers of families. "The youngest is a girl" keeps 2 families; "at least one is a girl" keeps 3. In both cases only one family, $(g, g)$ ⁠, has two girls, so the answers are $\frac12$ and $\frac13$ ⁠.

Given, to find, and the plan

Given: A family with two children; each child is a boy (b) or a girl (g) with equal chance. We also take the two children to be independent of each other; the question assumes this without saying it, as the book's Example 2 does. With the elder child first, $S = \{(b, b), (b, g), (g, b), (g, g)\}$ ⁠, each with probability $\frac14$ ⁠. With two children, "the youngest" is the younger child, the second letter.
To find: $P(E|F)$ and $P(E|G)$ ⁠, where $E$ ⁠: both are girls; $F$ ⁠: the youngest is a girl; $G$ ⁠: at least one is a girl.
Plan: (1) Write $E$ ⁠, $F$ ⁠, $G$ as sets. (2) Intersect with $E$ ⁠. (3) Divide by $P(F)$ and by $P(G)$ ⁠.

Basics you need

1Ordered outcomes

An elder girl with a younger boy, $(g, b)$ ⁠, is a different family from $(b, g)$ ⁠. Each child has 2 possibilities, so there are $2\times 2 = 4$ families. Basics: multiply the choices

2Classical probability

With equally likely outcomes, $P(A) = \frac{n(A)}{n(S)}$ ⁠. Basics: P(E) = n(E)/n(S)

3Conditional probability

$P(E|F) = \dfrac{P(E\cap F)}{P(F)}$ ⁠, provided $P(F) \ne 0$ (Definition 1, Section 13.2). The book's Example 2 asks part (ii) with boys in place of girls. Lesson: conditional probability Example 2

Part (i): both girls, given the youngest is a girl

Write the events as sets
$E = \{(g, g)\},\qquad F = \{(b, g), (g, g)\},$ $E\cap F = \{(g, g)\}.$
$F$ fixes the second letter as g; the elder child can be b or g, so $F$ has 2 families. The only one with two girls is $(g, g)$ ⁠, which is in $F$ ⁠, so $E\cap F = E$ ⁠.

An event is the set of outcomes that make it happen. Basics: events as sets

The 4 families, elder child first, once for each part. Gold and blue tiles form the given event; blue is also "both girls". Given that the younger child is a girl, 1 of 2 families remain with two girls; given at least one girl, 1 of 3.
Divide
$P(E|F) = \frac{P(E\cap F)}{P(F)} = \frac{1/4}{2/4} = \frac{1}{2}.$
$P(F) = \frac24 \ne 0$ and $P(E\cap F) = \frac14$ ⁠. This matches common sense: once we know the younger child is a girl, both are girls exactly when the elder is a girl, and that has chance $\frac12$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability

Part (ii): both girls, given at least one is a girl

Write $G$ and $E\cap G$
$G = \{(b, g), (g, b), (g, g)\},$ $E\cap G = \{(g, g)\}.$
"At least one girl" leaves out only $(b, b)$ ⁠, so $G$ has 3 families. The families $(b, g)$ and $(g, b)$ are both in $G$ ⁠: the girl could be the elder or the younger child, and we are not told which.

"At least one" is everything except "none". Basics: events as sets
Divide
$P(E|G) = \frac{P(E\cap G)}{P(G)} = \frac{1/4}{3/4} = \frac{1}{3}.$
$P(G) = \frac34 \ne 0$ ⁠. Of the 3 equally likely families left, just 1 has two girls. The answer is smaller than in (i) because "at least one girl" is weaker information: it keeps the family $(g, b)$ ⁠, which (i) had ruled out.

Definition 1, Section 13.2. Basics: dividing by a fraction

Result

(i) $P(E|F) = \frac{1}{2}$ (ii) $P(E|G) = \frac{1}{3}$ ⁠.

(Textbook answer: (i) $\frac12$ (ii) $\frac13$ ⁠.)

Check it with numbers

Imagine 400 two-child families, about 100 of each type $(b, b)$ ⁠, $(b, g)$ ⁠, $(g, b)$ ⁠, $(g, g)$ ⁠. (i) The younger child is a girl in about 200 of them, and 100 of those have two girls: $\frac{100}{200} = 0.5$ ⁠. (ii) At least one girl: about 300 families, 100 with two girls: $\frac{100}{300} \approx 0.3333$ ⁠.

Without any condition, $P(E) = \frac14 = 0.25$ ⁠. Each piece of information raises the chance, the sharper one, (i), more.

Common mistakes

Answering $\frac12$ in (ii). That treats "at least one girl" as if we knew which child is the girl. The families $(b, g)$ and $(g, b)$ are different, so 3 families remain, not 2.
Merging $(b, g)$ and $(g, b)$ ⁠. With unordered families {two boys, one of each, two girls} the three cases are not equally likely: "one of each" is twice as likely as the others.
Dividing by 4. $\frac14$ is $P(E\cap F)$ ⁠; after conditioning, divide by $P(F)$ or $P(G)$ ⁠.

What to take away

Write families (or coins, or dice) as ordered pairs so that the outcomes are equally likely.
Two conditions that sound alike can keep different numbers of outcomes; list the given event before dividing.

Every fact used, and where it comes from

Fact	Used in	Source
Two children give $2\times 2 = 4$ equally likely ordered outcomes	Given	Section 13.2, Example 2 (the same sample space) (Basics: multiply the choices)
$P(A) = \frac{n(A)}{n(S)}$	Both parts	Class XI, Probability (Basics: classical probability)
$P(E\|F) = \frac{P(E\cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Both parts	Section 13.2, Definition 1 (Lesson: conditional probability)

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Exercise 13.1 · Q12: Two children: both girls, given the youngest or at least one is a girl

Basics you need

Part (i): both girls, given the youngest is a girl

Write the events as sets

Divide

Part (ii): both girls, given at least one is a girl

Write GGG and E∩GE\cap GE∩G

Divide

Every fact used, and where it comes from

Write $G$ and $E\cap G$