Chapter 13 · Probability

Example 2: Two children: both boys, given at least one is a boy

ExamplesConditional probability · Section 13.2

The question

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

Exam answer

Let $b$ stand for boy and $g$ for girl. The sample space is

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

with all four outcomes equally likely.

Let $E$ ⁠: both the children are boys, and $F$ ⁠: at least one of the children is a boy.

E = \{(b,b)\}

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

E\cap F = \{(b,b)\}

P(F) = \frac{3}{4},\quad P(E\cap F) = \frac{1}{4}

By the definition of conditional probability,

\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \dfrac{\;\frac{1}{4}\;}{\frac{3}{4}} = \frac{1}{4}\times\frac{4}{3} \end{aligned}

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

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

The one idea this problem is built on

"Given at least one boy" throws away only the family with two girls. Of the four equally likely families, three are left, and just one of those three has two boys. So the answer is $\frac{1}{3}$ ⁠, not $\frac{1}{2}$ ⁠. The definition of conditional probability turns this counting into a calculation:

P(E|F) = \frac{P(E \cap F)}{P(F)}.

Given, to find, and the plan

Given: A family with two children; each child is a boy or a girl, and the four possible families are equally likely. We are told that at least one child is a boy.
To find: The probability that both children are boys, given that at least one is a boy.
Plan: List the sample space with the children in order; name the two events and list their outcomes; find $P(F)$ and $P(E \cap F)$ by counting; divide.

Basics you need

1Events are sets of outcomes

An event is the set of outcomes in which it happens; "both" is the intersection $E \cap F$ ⁠. Basics: events as sets

2Equally likely outcomes

When all outcomes are equally likely, $P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: the classical definition

3Conditional probability

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ when $P(F) \ne 0$ (Definition 1, Section 13.2). Lesson: conditional probability

4Dividing fractions

Dividing by $\frac{3}{4}$ is multiplying by $\frac{4}{3}$ ⁠. Basics: dividing by a fraction

Solution

Write the sample space, in order

Let $b$ stand for boy and $g$ for girl, and write the elder child first. Then
$S = \{(b,b), (g,b), (b,g), (g,g)\},$ $n(S) = 4.$
Each child can be a boy or a girl, so there are $2 \times 2 = 4$ families. The order matters: $(g,b)$ (elder girl, younger boy) and $(b,g)$ (elder boy, younger girl) are different families. Keeping them apart is what makes the four outcomes equally likely; lumping them together as "one boy and one girl" would give an outcome twice as likely as the others.

Basics: multiply the choices
Name the events and list their outcomes

Let $E$ ⁠: both the children are boys, and $F$ ⁠: at least one of the children is a boy. Then
$E = \{(b,b)\},$ $F = \{(b,b), (g,b), (b,g)\},$ $E \cap F = \{(b,b)\}.$
$F$ contains every family with a boy in it, which is every family except $(g,g)$ ⁠. The only family in both $E$ and $F$ is $(b,b)$ ⁠. In fact $E$ lies inside $F$ (two boys certainly means at least one boy), so $E \cap F = E$ ⁠.

Basics: events as sets
Find the two probabilities by counting
$P(F) = \frac{n(F)}{n(S)} = \frac{3}{4},$ $P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{4}.$
The four outcomes are equally likely, so each probability is the number of outcomes in the event over 4.

Basics: the classical definition
Divide, using the definition
$\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} = \dfrac{\;\frac{1}{4}\;}{\frac{3}{4}} \\ &= \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}. \end{aligned}$
We are told that $F$ has happened, so $F$ becomes the whole world; the definition measures what share of $F$ is also in $E$ ⁠. We may divide because $P(F) = \frac{3}{4} \ne 0$ ⁠.

Definition 1, Section 13.2. Lesson: conditional probability
See it as a grid

The four equally likely families (elder child first). The shaded cells make $F$ ⁠; only the ringed cell $(b,b)$ is in $E$ ⁠. Given $F$ ⁠, one cell out of three: $\frac{1}{3}$ ⁠.

Once we know there is a boy, the cell $(g,g)$ is ruled out and the three shaded cells stay equally likely. One of the three has two boys.

Basics: picturing events

Result

$P(E|F) = \dfrac{1}{3}$ ⁠.

(Textbook answer: $\frac{1}{3}$ ⁠.)

Check it with numbers

Counting inside $F$ ⁠. $\frac{n(E \cap F)}{n(F)} = \frac{1}{3}$ ⁠, the same value, because with equally likely outcomes the $n(S)$ cancels.

Compare with a different condition. If instead we are told that the elder child is a boy, call this $F_1 = \{(b,b), (b,g)\}$ ⁠. Then $P(E|F_1) = \frac{1}{2}$ ⁠, because the younger child is still a boy or a girl. "At least one is a boy" is weaker information than "the elder is a boy", so it leaves more families, and the answer is smaller.

Decimals. $\frac{1}{3} \approx 0.3333$ ⁠. Without any information, $P(E) = \frac{1}{4} = 0.25$ ⁠; knowing there is a boy raises the chance of two boys from 0.25 to about 0.3333.

Common mistakes

Answering $\frac{1}{2}$ by thinking "the other child is a boy or a girl". We are not told which child is the boy, so there is no single "other child". That reasoning answers the question about the elder child (the check above).
Using three outcomes: two boys, one of each, two girls. These are not equally likely; "one of each" happens in two ways, $(g,b)$ and $(b,g)$ ⁠.
Answering $P(E) = \frac{1}{4}$ ⁠. That ignores the information that there is at least one boy.

What to take away

Write ordered outcomes so that they are equally likely.
The condition removes outcomes; the answer is the share of what is left.
Read the condition carefully: "at least one" and "the elder one" give different answers.

Every fact used, and where it comes from

Fact	Used in	Source
Two choices for each child: $2 \times 2 = 4$ families	Step 1	Counting principle, Class XI (Basics: multiply the choices)
Events as sets; "both" is the intersection	Step 2	Class XI, Probability (Basics: events as sets)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 3, Check	Class XI, Probability (Basics: the classical definition)
$P(E\|F) = \frac{P(E \cap F)}{P(F)}$ ⁠, $P(F) \ne 0$	Step 4, Check	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(E\|F) = \frac{n(E \cap F)}{n(F)}$	Check	Section 13.2, equation (1)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 2: Two children: both boys, given at least one is a boy

Basics you need

Solution

Write the sample space, in order

Name the events and list their outcomes

Find the two probabilities by counting

Divide, using the definition

See it as a grid

Every fact used, and where it comes from