Chance Experiment:

E: Woman gives birth to a child.

Possible outcomes:

= Boy

= Girl

= Girl

Sample Space (or Universal Set). This is the set of all possible outcomes of our chance experiment:

S = {
,
}

Assigned probabilities of each outcome occurring:

Pr() | = | Pr(Boy) | = | 1/2 | ||||

Pr() | = | Pr(Girl) | = | 1/2 |

Note the probabilities of all elements in the sample space add up to 1 as they should:

Pr() +
Pr()
=
Pr(Boy) +
Pr(Girl)
=
1/2 + 1/2 = 1

We now consider repeating this chance experiment E twice. This can be
thought of as a single 2-fold compound experiment, which can be
designated by:

E x E = E2 |

We perform the experiment once (woman gives birth to first child), then
we perform the experiment again (woman gives birth to second child).
The sample space of all possible outcomes for our
compound experiment E2 is the Cartesian product of the sample spaces
for each individual experiment E:

S2 = S x S = { | { | , | }, | = { | { | Boy, | Boy | }, | ||

{ | , | }, | { | Boy, | Girl | }, | ||||

{ | , | }, | { | Girl, | Boy | }, | ||||

{ | , | } } | { | Girl, | Girl | } } |

Compound Chance Experiment:

E2: Woman gives birth to two children (one at a time).

Set S2 above contains all the possible outcomes of our compound chance experiment E2.

We can now figure out the probabilities of each possible outcome in our sample space S2. The probability of any particular element in S2 is equal to the product of the probabilities of the corresponding elements from S which together comprise the element in S2. (This assumes that the outcome of the second chance experiment is independent of the outcome of the first chance experiment. That is, the gender of the first child does not affect the probabilities of the gender of the second child.)

Pr( {,} ) | = | Pr( {Boy,Boy} ) | = | ½ * ½ | = | ¼ |

Pr( {,} ) | = | Pr( {Boy,Girl} ) | = | ½ * ½ | = | ¼ |

Pr( {,} ) | = | Pr( {Girl,Boy} ) | = | ½ * ½ | = | ¼ |

Pr( {,} ) | = | Pr( {Girl,Girl} ) | = | ½ * ½ | = | ¼ |

Note again the probabilities of all elements in the sample space add up
to 1 as they should:

Pr( {Boy,Boy} ) + Pr( {Boy,Girl} ) + Pr( {Girl,Boy} ) + Pr( {Girl,Girl} )
=
¼ + ¼ + ¼ + ¼ = 1

So far we've set up our sample space and determined the probabilities for the initial sentence, "A mother has two children." Let's take an aside and ask at this point, "What is the probability that one of them is a daughter and the other child is a girl?" We get our answer by partitioning the sample space into the elements which satisfy the criteria "One of them is a daughter" AND "the other child is a girl" and the elements which do not satisfy the criteria.

Elements of the sample space which satisfy the criteria:

{Girl,Girl}

Pr({Girl,Girl}) = ¼

Pr({Girl,Girl}) = ¼

Elements of sample space which do not satisfy the criteria:

{Boy,Boy}, {Boy,Girl}, {Girl,Boy}

Pr({Boy,Boy} or {Boy,Girl} or {Girl,Boy}) = ¼ + ¼ + ¼ = ¾

Pr({Boy,Boy} or {Boy,Girl} or {Girl,Boy}) = ¼ + ¼ + ¼ = ¾

Note again the total probability adds up to 1 as it should:

¼ + ¾ = 1

However, this is not the question we were asked. The setup was, "A mother has two children. One of them is a daughter." The second sentence gives us additional information. We need to incorporate this additional information into our setup before we answer the question. This is where people go astray when they use their intuition instead of Bayes's Theorem. Whenever you are given a new piece of information, you have to use Bayes' Theorem to properly incorporate that new piece of information.

First we shall derive Bayes' Theorem.

Let *A* and *M* be two events. The probability of *A*
given *M* is denoted by Pr(*A*|*M*).

Pr(*A*|*M*) = {Probability of *A* given *M*}

This is the probability
that the outcome of our chance experiment lies within realm *A*,
given that the outcome is known to lie somewhere within realm *M*.
Here is a Venn diagram that may help as we now derive Bayes' Theorem.

**BAYES' THEOREM:**

Let *A* and *M* be two events.
The outcomes common to two events *A* and *M* form their intersection
*A* ∩ *M*. The conditional probability of event *A*, given event
*M*, is defined as:

Pr(A|M) = |
Pr(A∩M) |

Pr(M) |

Then by this definition of conditional probability,

Pr(

whence

Pr(A|M) = |
Pr(M|A)*Pr(A) | (Bayes' Theorem) |

Pr(M) |

We now must apply Bayes' Theorem to every element in our sample space S2 to incorporate our new piece of information, "One of them is a daughter," to determine the new probabilities of each outcome occurring.

__First element of S2__:

Let *A* = {Boy,Boy}

Let*M* = "One of them is a daughter."

Apply Bayes' Theorem:

Let

Apply Bayes' Theorem:

Pr( {Boy,Boy} | {"One of them is a daughter"} ) | = | Pr( {"One of them is a daughter"} | {Boy,Boy} ) * Pr( {Boy,Boy} ) |
||

Pr( {"One of them is a daughter"} ) | ||||

= | 0 * ¼ |
= 0 | ||

¾ |

__Second element of S2__:

Let *A* = {Boy,Girl}

Let*M* = "One of them is a daughter."

Apply Bayes' Theorem:

Let

Apply Bayes' Theorem:

Pr( {Boy,Girl} | {"One of them is a daughter"} ) | = | Pr( {"One of them is a daughter"} | {Boy,Girl} ) * Pr( {Boy,Girl} ) |
||

Pr( {"One of them is a daughter"} ) | ||||

= | 1 * ¼ |
= 1/3 | ||

¾ |

__Third element of S2__:

Let *A* = {Girl,Boy}

Let*M* = "One of them is a daughter."

Apply Bayes' Theorem:

Let

Apply Bayes' Theorem:

Pr( {Girl,Boy} | {"One of them is a daughter"} ) | = | Pr( {"One of them is a daughter"} | {Girl,Boy} ) * Pr( {Girl,Boy} ) |
||

Pr( {"One of them is a daughter"} ) | ||||

= | 1 * ¼ |
= 1/3 | ||

¾ |

__Fourth element of S2__:

Let *A* = {Girl,Girl}

Let*M* = "One of them is a daughter."

Apply Bayes' Theorem:

Let

Apply Bayes' Theorem:

Pr( {Girl,Girl} | {"One of them is a daughter"} ) | = | Pr( {"One of them is a daughter"} | {Girl,Girl} ) * Pr( {Girl,Girl} ) |
||

Pr( {"One of them is a daughter"} ) | ||||

= | 1 * ¼ |
= 1/3 | ||

¾ |

We now have the probabilites for each possible outcome
of "A mother has two children" given "One of them is a daughter."

Pr( {Boy,Boy} | {"One of them is a daughter"} ) | = 0 |

Pr( {Boy,Girl} | {"One of them is a daughter"} ) | = 1/3 |

Pr( {Girl,Boy} | {"One of them is a daughter"} ) | = 1/3 |

Pr( {Girl,Girl} | {"One of them is a daughter"} ) | = 1/3 |

Since Pr({Boy,Boy}) = 0 we can remove that element from our sample space.

Pr( {Boy,Girl} | {"One of them is a daughter"} ) | = 1/3 |

Pr( {Girl,Boy} | {"One of them is a daughter"} ) | = 1/3 |

Pr( {Girl,Girl} | {"One of them is a daughter"} ) | = 1/3 |

We can now answer the question, "What is the probability that the other child is a girl?" given the setup, "A woman has two children", AND "one of them is a daughter".

We do this by partitioning the sample space into the elements which satisfy the criteria "One of them is a daughter" AND "the other child is a girl" and the elements which do not satisfy the criteria.

Elements of the sample space which satisfy the criteria:

{Girl,Girl}

Pr({Girl,Girl}) = 1/3

Pr({Girl,Girl}) = 1/3

Elements of the sample space which do not satisfy the criteria:

{Boy,Boy}, {Boy,Girl}, {Girl,Boy}

Pr({Boy,Girl} or {Girl,Boy}) = 1/3 + 1/3 = 2/3

Pr({Boy,Girl} or {Girl,Boy}) = 1/3 + 1/3 = 2/3

Note again the total probability adds up to 1 as it should:

1/3 + 2/3 = 1

Hence the answer to our question *A mother has two children. One
of them is a daughter. What is the probability that the other child is
a girl?* is 1/3, as {Girl,Girl} is the only outcome which satisfies
the criteria "One of them is a daughter" AND "the other child is a
girl", and the probability of {Girl,Girl}, given the setup "A mother
has two children." AND "One of them is a daughter", is 1/3.