Math 135, Spring 2011: HW 3 Solutions
September 21, 2011
This homework is due at the beginning of class on
Thursday, February 3
. You are free to talk with each
other and get help. However, you should write up your own answers and understand everything you write.
Problem 1.
A woman has 2 children, one of whom is a boy born on a Tuesday. What is the probability
that both children are boys? (You may assume that boys and girls are equally likely and independent, and
that children are equally likely to born on any given day of the week.)
SOLUTION. In this problem, one of the first questions that comes to mind is, “Why does Tuesday matter?”
To determine whether this information matters, let us work from the definitions.
We are asked to find the probability that the woman has two boys,
given
that one of her two children is a
boy born on a Tuesday. The problem does
not
tell us which of the two children is a boy. Let the days of the
week be numbered 1
,
2
,
3
,
· · ·
,
7, where 1=Mon, 2=Tues, 7=Sunday. Since the woman has two children, the
sample space Ω consists of ordered pairs (
Bi, Bj
)
,
(
Bi, Gj
)
,
(
Gi, Bj
)
,
(
Gi, Gj
) where
B
or
G
represents the
gender, and
i
and
j
each take on any integer value between 1 and 7 and represent the day of the week that
the child is born. Furthermore, (
Bi, Gj
) represents the event that the firstborn child is a boy born on day
i, and the secondborn is a girl born on day j, etc. Note that all of these outcomes are equally likely.
Since we know that one of the children was a boy born on a Tuesday, however, we can restrict ourselves
to considering the following outcomes in our sample space: the 7 outcomes (
B
2
, Gj
), where
j
= 1
,
2
,
· · ·
,
7;
the 7 outcomes (
Gj, B
2) where
j
= 1
,
2
,
· · ·
,
7, the 7 outcomes (
B
2
, Bj
), where
j
= 1
,
2
· · ·
,
7, and the
six
outcomes (
Bj, B
2) where
j
= 1
,
3
,
· · ·
,
7, because we already counted the pair (
B
2
, B
2). Hence the event
C
=
{
The woman has two children, one of whom is a boy born on a Tuesday
}
has a total of 27 outcomes.
The event
D
=
{
both children are boys
}
has 49 outcomes, but the event
D
∩
C
=
{
both children are boys, one of whom was born on a Tuesday
}
has only 13 outcomes. Hence
P
(woman has two boys

woman has two children, one of whom is Tuesdayborn boy)
=
P
(
D

C
) =
P
(
D
∩
C
)
P
(
C
)
=
13
27
which is slightly less than 1/2.
One way to think about this is that if you only know that the woman has two children, then the probability
that both are boys is 1/4; if you know that one of them is a boy, the probability that both are boys rises to
1/3; and if you know that one of them is a boy born on a Tuesday, the probability that both are boys rises
to nearly 1/2.
Problem 2
(
The Matching Problem).
Suppose there are
n
people at a party, and everyone has put their
coats in a dark coat room. As the guests leave, they grab their coats from the coat room at random, since
they cannot see.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Staff
 Math, Normal Distribution, Probability, Standard Deviation, Probability theory, Binomial distribution

Click to edit the document details