When an automobile is stopped with a roving safety patrol, each tire is checked for tire wear, and each headlight is checkedto see whether it is properly aimed. Let X denote the numberof headlights that need adjustment, and let Y denote the numberof defective tires.

a) if X and Y are independent with Px(0) = .5, Px(1) =.3,Px(2) =.2 and Py(0) =.6, Py(1) =.1, Px(2) =Py(3) =..05, Py(4) =.2,display the joint probability distribution table.
b) compute P(X<= 1 and Y <= 1) fromthe joint probability distribution table and verify that it isequals to the product P(X <= 1) * P(Y<=1)
c) What is P(X+Y = 0) (theprobability of no violations)?
d) Compute P(X + Y <= 1).

a) Joint ptobability distribution

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

Then, the joint probabilities are:

X=0Y=0Px=0.5Px=0.6P(0,0)=0.3

X=0Y=1Px=0.5Px=0.1P(0,1)=0.05

X=0Y=2Px=0.5Px=0.05P(0,2)=0.025

X=0Y=3Px=0.5Px=0.05P(0,3)=0.025

X=0Y=4Px=0.5Px=0.2P(0,4)=0.1

X=1Y=0Px=0.3Px=0.6P(1,0)=0.18

X=1Y=1Px=0.3Px=0.1P(1,1)=0.03

X=1Y=2Px=0.3Px=0.05P(1,2)=0.015

X=1Y=3Px=0.3Px=0.05P(1,3)=0.015

X=1Y=4Px=0.3Px=0.2P(1,4)=0.06

X=2Y=0Px=0.2Px=0.6P(2,0)=0.12

X=2Y=1Px=0.2Px=0.1P(2,1)=0.02

X=2Y=2Px=0.2Px=0.05P(2,2)=0.01

X=2Y=3Px=0.2Px=0.05P(2,3)=0.01

X=2Y=4Px=0.2Px=0.2P(2,4)=0.04

We can write it in the form of a matrix:

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

We can calculate P(X<= 1) * P(Y<=1)

Both calculations give the same result.

c) Probability of no violations

d) P(X + Y <= 1)

b

step-by-step explanation:

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step-by-step explanation: