STAT 311 Practice Questions
Q1. We are drawing balls from the urn with 2 white balls and 3 red balls with
replacement. X be the number of red balls drawn in the first two draw, then X=0, 1, 2. Y
can go on forever
until
you get the
first
red ball in your draw. (That means, Y has a
geometric distribution, and Y=1,2,3,
…
..)
(a)
Joint probability mass function of p
X,Y
(x, y)
In this kind of example, (When things do not look that straightforward), trying
to figure out the exact kind of distribution may be rather futile. Instead calmly
think about all the possible cases! First fixate on X. When X=0 what would Y be?
Then when X=1, what would be the corresponding Y? Then when X=2?
Joint probability mass function is what assigns the probability to each possible
pair of X and Y.
Remember, the value of what Y can take is restricted with the condition,
“
first
red ball is drawn
”
, therefore, bounded by the value of X.

(X=0, Y=1): No such thing can happen

(X=0, Y=2): No such thing can happen

(X=0, Y>2): White balls are drawn in the first two draw, and then whatever
happens after that does not matter. (WW)

(X=1, Y=1): When the drawn balls are RW

(X=1, Y=2): When the drawn balls are WR

(X=1, Y>2): No such thing can happen

(X=2, Y=1): When the drawn balls are RR

(X=2, Y=2): No such thing can happen

(X=2, Y>2): No such thing can happen
P
X = 0, Y = 1
= 0
P
X = 0, Y = 2
= 0
P
X = 0, Y > 2
=
2
5
∗
2
5
=
4
25
P
X = 1, Y = 1
=
3
5
∗
2
5
=
6
25
P
X = 1, Y = 2
=
2
5
∗
3
5
=
6
25
P
X = 1, Y > 2
= 0
P
X = 2, Y = 1
=
3
5
∗
3
5
=
9
25
P
X = 2, Y = 2
= 0
P
X = 2, Y > 2
= 0
To make sure, add up all the probabilities and see if it is equal to 1.
(b)
Probability mass function of X and Y assigns probabilities to the possible values of X
and Y
P
X = 0) = P(X = 0, Y = 1
+ P
X = 0, Y = 2
+ P
X = 0, Y > 2
= 0 + 0 +
4
25
=
4
25
P
X = 1) = P(X = 1, Y = 1
+ P
X = 1, Y = 2
+ P
X = 1, Y > 2
= 0 +
6
25
+
6
25
=
12
25
P
X = 2) = P(X = 2, Y = 1
+ P
X = 2, Y = 2
+ P
X = 2, Y > 2
=
9
25
+ 0 + 0 =
9
25
Again, make sure that the sum is equal to 1
P
Y = 1) = P(X = 0, Y = 1
+ P
X = 1, Y = 1
+ P
X = 2, Y = 1
= 0 +
6
25
+
9
25
=
15
25