There is nothing different about these questions at all.

For part (c), let be the number of defective items found.There are trials and success is defined as finding a defective item so that the probability of success is . Indeed we could write to denote this.

Now for part (i) we are looking for the probability that is less than two ( here means OR):

.

I trust that you can finish this question off.

For part (d) (I think this is Q.3 on P.103 so barring silly typos ye have the answers), first you split into those who have the disease and those who do not have the disease (you can use your own notation of course), and with 0.01 on the arrow to and the rest of the probability, 0.99, on the arrow to .

Then each of these split again. A notation I would use is for a positive diagnosis and for a negative diagnosis (you can do this using correct/incorrect diagnoses also). Now the splits into and with respective probabilities 0.97 and 0.03. Similarly the splits into and with respective probabilities 0.05 and 0.95.

Now part (ii) is looking for . Now we need the formula for conditional probability. They way I do this is write

(the probability of AND is the probability of times the probability of given that is true).

This implies that

.

Adapted to our case we have

.

Now is fairly straightforward: it is equal to (0.01)(0.97). However there are TWO ways of getting a positive diagnosis:

.

I trust ye can finish this off.

For part (iii) you are looking for the probability of a correct diagnosis; that is and or and :

.

Again I trust that ye can finish this off.

Regards,

J.P.

Sorry for bothering you so late. We are struggling with part (c) and (d) of question 1 of the summer 2013 exam. Anyway you could forward on worked solutions to give us a clue as they seem different from the worked example in the book.

Thanks.

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