An article in IEEE International Symposium on Electromagnetic Compatibility (Vol. 2, 2002, pp. 667-670) describes the quantifica
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Answer:
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=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3
X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05
X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025
X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025
X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1
X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18
X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03
X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015
X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015
X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06
X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12
X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02
X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01
X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01
X=2 Y=4 Px=0.2 Px=0.2 P(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)