Boy/Girl Paradox
A family has two children. Given that one of the children is a boy, and that he was born on a Tuesday, what is the probability that both children are boys?
Assume that the probability of being born on a particular day of the week is 7 1 and is independent of whether the child is a boy or a girl.
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3 solutions
Discussions for this problem are now closed
So, here's my solution:
-
Knowing that there are 7 days in a week, let's first consider every day that's not a Tuesday. On these days, you can either get a boy or a girl as the first child, but the next child would have to be a boy born on a Tuesday. Therefore you have 6 instances of each (other child being boy or girl) and just add this to a little chart categorised into boy and girl (so 6 for boy and 6 for girl).
-
Now we consider Tuesday itself, there are two cases; boy or girl. If the first child is a girl, then the next child would have to be a boy born on Tuesday so there is one instance of a girl being born here (Now it's 6 for boy and 7 for girl). If the first child is a boy, then it doesn't matter what day it is or whether it's a boy or girl because the condition of a Tuesday-born boy has already been fulfilled. Therefore, there are 7 instances of both boy and girl here. Now we have 14 instances of girl and 13 instances of boy. Therefore, the probability of the other child being a boy is 13/27.
I couldn't believe that the answer is 13/27 but look at this:
I will explicitly list all combinations where one child is boy born on Tuesday
(I use numbers for days of the week.
0 Sunday, 1 Monday, 2 Tuesday, etc
and B for boy and G for girl.
So read G1B2 as "girl born on Monday and boy burn on Tuesday".)
1) First child is boy born on Tue, second child is girl (7 combinations):
B2G0 B2G1 B2G2 B2G3 B2G4 B2G5 B2G6
2) First child is girl, second child is boy born on Tue (7 combinations):
G0B2 G1B2 G2B2 G3B2 G4B2 G5B2 G6B2
3) First child is boy born on Tue, second child is boy (7 combinations):
B2B0 B2B1 B2B2 B2B3 B2B4 B2B5 B2B6
4) First child is boy, second child is boy born on Tue (7 combinations)
B0B2 B1B2 B2B2 B3B2 B4B2 B5B2 B6B2
Total number of unique combinations is 27 ( 28 minus 1 because B2B2 has been counted twice).
Out of those 27 the number of combinations where both children are boys is 13.
So the answer IS 13/27 !
Perfect!
A: Both children were boys.
B: One boy was born on Tuesday.
P ( B A ) = P ( B ) P ( A ∩ B ) = P ( B ) P ( A ) P ( B ) [Since A and B are independent] = P ( A ) = 2 1
Where am I mistaken?
How did you conclude that A and B are independent?
See the wiki page for the equation that you should be using for dependent events.
The question states: "Assume that the probability of being born on a particular day of the week is 7 1 and is independent of whether the child is a boy or a girl."
Doesn't that mean A and B independent?
You mean P ( A ∣ B ) = P ( B ) P ( A ∩ B ) ??
@Pranjal Jain – P(A|B) is the probability of occurence of A when B has occurred while P(B) is probability of occurrence of B. So P ( A ∣ B ) P ( B ) = P ( A ∩ B )
@Pranjal Jain – Sorry, I meant "how did you conclude that A and B are independent"? I have updated the original comment.
The equation P ( A ∣ B ) = P ( B ) P ( A ∩ B ) is correct.
@Calvin Lin – It says "Knowing the day of the week a child is born on can’t possibly give you additional information, right?"
Exactly! Thats where I am confused!
@Pranjal Jain – Read on further. It goes on to explain what is going on.
The events are:
A - both are boys.
B- One of the children is a boy that is born on Tuesday.
Consider the following scenarios:
B
:
We know that one of the children
is
a boy born on Tuesday.
B
c
:
We know that one of the children
is not
a boy born on Tuesday.
In which scenario does it make it more or less likely that both are boys?
In the first scenario, we already are guaranteed that one of the children is a boy. In the second scenario, we only have a
2
7
1
3
chance that the child is a boy. It seems unlikely that these events are independent.
Here is the calculation explanation for why the events A and B are not independent.
2 events are independent if
P
(
A
)
×
p
(
B
)
=
P
(
A
∩
B
)
. Let's calculate these probabilities.
P
(
A
)
=
4
1
, which is clear.
To calculate
P
(
B
)
, let
I
1
be the indicator variable that the first child is a boy born on Tuesday and
I
2
be the indicator variable that the second child is a boy born on Tuesday. Then,
B
=
I
1
∪
I
2
. Hence, by the principle of inclusion and exclusion, we have:
P
(
B
)
=
P
(
I
1
∪
I
2
)
=
P
(
I
1
)
+
P
(
I
2
)
−
P
(
I
1
∩
I
2
)
=
1
4
1
+
1
4
1
−
1
9
6
1
=
1
9
6
2
7
.
To calculate
P
(
A
∩
B
)
, we use the same idea, and see that
P
(
A
∩
B
)
=
P
(
I
1
∩
second child is boy
)
+
P
(
first child is boy
∩
I
2
)
+
P
(
I
1
∩
I
2
)
.
Now, by independence, this probability is equal to
1
4
1
×
2
1
+
2
1
×
1
4
1
+
1
4
1
×
1
4
1
=
1
9
6
1
3
.
Hence, P ( A ∩ B ) = P ( A ) × P ( B ) , so these events are not independent.
@Calvin Lin – The question is worded terribly then. It doesn't clearly state the boy born on a Tuesday is the only boy born on a Tuesday, which is what you have assumed.
@Raphael Nasif – No, I have not assumed that is the only boy born on a Tuesday.
The event I 1 ∩ I 2 accounts for those scenarios where both boys are born on a Tuesday. The reason why it appears that we have subtracted that is due to the Principle of Inclusion and Exclusion, because we have actually double counted this event in both I 1 and I 2 .
If one child is a boy (given) with no assumptions as to the probability of a girl vs a boy. How is the question any different than is the second child a boy or girl? It's 50:50
All that you know is "at least one of the two children is a boy born on Tuesday".
Think about the more basic case first. Suppose that you are given that at least one of the children is a boy, what is the probability that both are boys? Well, there are 3 cases where at least one child is a boy, namely BB, BG, GB. Out of these, there is only 1 case where the child is a boy, namely BB. Hence, the probability of (both are boys given that at least one is a boy) is 3 1 .
Then is it 1/3?
I don't like the way this question is worded. This question assumes that the children cannot be twins, without explicitly stating that the children cannot be born on the same day. Not only that, but children could've been born during different weeks. The assumption that the second child cannot be born on a Tuesday is ridiculous.
EDIT: Never mind, I see it now.
Can you explain why it must be 2 7 1 3 ?
Read the wiki. There is a complete solution posted there in the examples.
You mentioned its independent, so your answer is incorrect.
Read the question carefully. Make sure you define the events properly. Certain events are independent, certain events are not.
Read the wiki for more information.
I used Baye's Theorem.
We are given "one" is a boy, that means 3 cases. Bg, Gb, Bb. B is a boy and G is a girl. CAPS means elder child.
Thus, we need to find out P(A/B) where A=both children are boys, B = one boy was born on a Tuesday. Thus, Bb becomes a vivid case. Bb, when 1. only B was born on Tuesday, 2. only b was born on Tuesday and 3. Both B and b were born on tuesday. that makes it (1/7 * 6/7) + (1/7 * 6/7) + (1/7 * 1/7) = (13/49)
thus it's the case to be looked for. And rest is easy as chocolate :-D
Detailed solution can be seen in the pic I uploaded. (edit) I can't upload the .jpeg image, some error is encountered in brilliant.org
Probability is most favourite topic ^_^