2. A copy machine in a certain o'ice breaks down every so often. The ofce workers try to x it when it breaks. If they can't x it, they call a technician. In a particular month, let X be the number of times the machine breaks, and let Y be the number of times the technician is called. Assume that the PMF of X is P(X = o) = 0.3 P(X = 1) = 0.4 P(X = 2) = 0.2 P(X = 3) = 0.1 Also assume that when the machine breaks, the workers can x it 60% of the time and they have to call the technician 40% of the time. (a) Draw a table for the joint distribution of X and Y. Solution: If X = 0, then certainly Y = 0. Thus we have P(X=0,Y=0)=0.3, P(X=0,Y=1)=0, P(X=0,Y=2)=0, P(X=0,Y=3)=O. Given that X = n, for n = 1,2,3, we know that Y is between 0 and n. In fact, the distribution of Y is Binomial with 11 trials and p = 0.4. The precise statement is that for n = 1,2,3 and 0 5 k 5 n, P(Y : k | X : n) = (E) 11.4" - nan-k. To get the joint PMF, we use that P(X=nY= k): P(X=n)P(Y= k|X=n). P(X:3,Y:0 :01 0.63:0.0216 P(X=3,Y=1 :01 3-0.4-0.62=0.0432 P(X=3,Y=2 :01 3-0.42-0.6=0.0288 So, P(X : 1, Y: 0): 0.6 : 0.24 P(X= 1, Y: 1): 0.4=0.16 P(X : 2, Y: 0): 0 52 : 11.072 P(X=2, Y: 1):o.2 2-0.4-0.6=0.096 P(X:2, Y:2)=U.2 0.42:0.032 ) ) )