1) If the joint probability distribution of X and Y is given by f(x, y) = x + y

Question

1) If the joint probability distribution of X and Y is given by f(x, y) = x + y 30 , for x = 0, 1, 2, 3; y = 0, 1, 2, find (a) P(X 2, Y = 1); (b) P(X > 2, Y 1); (c) P(X>Y ); (d) P(X + Y = 4).

2) If the joint probability distribution of X and Y is given by f(x, y) = x + y 30 , for x = 0, 1, 2, 3; y = 0, 1, 2, find (a) the marginal distribution of X; (b) the marginal distribution of Y .

3) An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by f(x) = 3x4, x> 1, 0, elsewhere. (a) Verify that this is a valid density function. (b) Evaluate F(x). (c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

Explanation / Answer

Question 1:

we are given here that f(x,y) = (x+y)/30

a) P(X 2, Y = 1) = P(X=0, Y=1 ) + P(X = 1,Y=1) + P(X=2,Y=1 ) = (1/30) ( 1 +2 + 3) = 0.2

Therefore 0.2 is the required probability here.

b) P(X > 2, Y 1) = P(X=3, Y=0) + P(X=3, Y=1) = (1/30) (3 + 4) = 7/30

Therefore 7/30 is the required probability here.

c) P(X>Y )= P(Y=0, X=1) + P(Y=0, X=2) + P(Y=0, X=3) + P(Y=1, X=2) + P(Y=1, X=3) + P(Y=2, X=3) = (1/30)(1 + 2 + 3 + 3 + 4 + 5) = 18/30 = 0.6

Therefore 0.6 is the required probability here.

d) P(X + Y = 4) = P(X=2,Y=2) + P(X=3,Y=1) = (1/30)(4 + 4) = 0.2667

Therefore 0.2667 is the required probability here.

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