Question

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table.

p(x, y)		0	5	10
x	0	0.03	0.08	0.09
	5	0.09	0.20	0.20
	10	0.02	0.15	0.16

(a) what is the probability that a randomly selected student scores 5 on both parts?
(b) what is the probability that a randomly selected student scores at least 5 on part 1 and no points on part 2?

(c) find the marginal PMFs of X and Y (find PMF of X and PMF of Y).

(d) If the score recorded in the grade book is the toal number of points earned in two parts, what is the expected recorded score E(X+Y)?

Answer 1

(Note that : total probability is always equal to 1. So please check the given probabilities).

Here total of the probabilities = 1.02

Let's take P(X = 10, Y = 10 ) = 0.14 instead of 0.16 so we get total probability = 1 .

(a) what is the probability that a randomly selected student scores 5 on both parts?

From the joint probability mass function of X and Y we get P( X = 5, Y = 5) = 0.20

(b) what is the probability that a randomly selected student scores at least 5 on part 1 and no points on part 2?

P( X >= 5 , Y = 0 ) = P( X = 5, Y = 0) + P( X = 10, Y = 0) = 0.09 + 0.02 = 0.11

(c) find the marginal PMFs of X and Y (find PMF of X and PMF of Y).

P(X = 0) = 0.03+0.08+0.09 =0.20

P(X = 5) = 0.09+0.20+0.20 = 0.49

P( X = 10) = 0.02+0.15+0.14 = 0.31

X	0	5	10	Total
P(X = x)	0.20	0.49	0.31	1

Similarly, the pmf of Y is as follows:

Y	0	5	10	Total
P(Y = y)	0.14	0.43	0.43	1

E(X + Y) = 0*0.03 + (0+5)*0.08 + (0+10)*0.09 + (5+0)*0.09 + (5+5)*0.20 + (5+10)*0.20 + (10+0)*0.02 + (10+5)*0.15 + (10+10)*0.14 = 12