York University
MATH 2030 3.0 (Elementary Probability) Fall 2011
Assignment 1 –Solutions, Sept 2011
§
1.1 No. 2
(a) We use an equally likely outcomes model with
Ω =
{
suppose, a, word, is, picked, at, random, from, this, sentence
}
.
This Ω has 10 elements. The event in question is
A
=
{
suppose, word, picked, random, from, this, sentence
}
which has 7 el-
ements. So
P
(
A
) = 7
/
10.
(b) Now we have an event
B
=
{
suppose, picked, random, sentence
}
, and
P
(
B
) = 4
/
10.
(c)
A
∩
B
=
B
in this particular case, so
P
(
A
∩
B
) =
P
(
B
) = 4
/
10.
§
1.3 No. 4
(a) Yes: the event is
{
0
,
1
}
(b) Yes, this is just another way of saying there was exactly one head: the
event is
{
1
}
.
(c) No, the event can’t be expressed in this model. This particular model lets
us keep track of the number of heads, but not the order they occur in.
(d) Yes: the event is
{
1
,
2
}
.
§
1.3 No. 6
(a) There are 10 words, each of which is equally likely to be picked. I’ll use an
equally likely outcomes model with sample space consisting of
{
suppose,
a, word, is, picked, at, random, from, this, sentence
}
.
Let
X
be the
random variable that counts the length of the word, eg
X
(suppose) = 7,
X
(word) = 4, etc. For each
x
we have to count the number of outcomes
(ie words in the sentence) of length
x
, and then divide by 10 to get the
desired probabilities. We compute that
x
1
2
3
4
5
6
7
8
P
(
X
=
x
)
1
10
2
10
0
3
10
0
2
10
1
10
1
10
(b) We have the same model, but a different random variable
Y
that counts the
number of vowels in a word. For example,
Y
(suppose) = 3,
Y
(sentence) =
3. Now for every
y
we have to count the number of words with
y
vowels in
order to compute probabilities. Doing this we get that
y
1
2
3
P
(
Y
=
y
)
6
10
2
10
2
10
1