1. Suppose that the number of typographical errors in a new text is Poisson dist
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Question
1. Suppose that the number of typographical errors in a new text is Poisson distributed with mean . Two proofreaders independently read the text. Suppose that each error is independently found by proofreader i with probability pi, i = 1, 2. Let X1 denote the number of errors that are found by proofreader 1 but not by proofreader 2. Let X2 denote the number of errors found by proofreader 2 but not by proofreader 1. Let X3 be the number of errors found by both proofreaders. Finally, let X4 denote the number of errors found by neither proofreader.
(a) What is the expected value of Xi (i=1, 2, 3, 4)?
(b) What is the expected value of XiXj when ij (i, j=1, 2, 3, 4)?
I have the solution which is:
E[X1] = p1(1-p2), E[X2] = p2(1-p1), E[X3] = p2p1, E[X4] = (1-p2)(1-p1)
The last two make sense to me but I don't understand why the first two aren't just E[X1] = p1 and E[X2] = p2
Explanation / Answer
see, look at the expected value as lambda * respective probability of 1st proofreader * respective probability of 2nd proof reader. It's the multiplication of two probabilities along with lambda.
Now, as per you expected value of x would be lambda * p1.. Wrong? it pertains to the error detected by proofreader 1 but are they also detected by proofreader! You have no information about that. p2 is the probabilty that error is detected by proofreader2.
Look, this way, A hits the target with probability p1, B hits the target with probability p2.
A doesn't hits the target with probability=1-p1, B doesn't hits the target with probabilty=1-p2
q1 a hits the target, b does not == p1*(1-p2)]]
q2 a doesnt hits the target, b hits =(1-p1)*p2
q3 both hit the target=p1*p2
q4 noone hits the target=(1-p1)*(1-p2)
Aren't these questions equivalent? just hitting the target is equivalent to findingg the error. That's proofreader hit ratio. Hope it helps.All the answers are straightforward i guess.
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