Diagnostic Tests and Expected Values

I have this question from my textbook I'm not sure how to answer. I got the first part but the second part is a bit confusing.

It goes something like this:

"Two percent of the population has a certain condition for which there are two diagnostic tests. Test A which costs \$1 per person, gives positive results for 80% of persons with the condition and for 5% of persons without the condition. Test B, which costs$100 per person , gives positive results for all persons with the condition and negative results for all persons without it."

Suppose that 2000 persons are given test A, and then only those who test positive are given test B. Show that the expected cost is \$15,000 and the expected number of cases. I know the expected value of random variable X is defined as: $$E[X]=x_1p_1 + x_2p_2\dots+x_kp_k$$ - 2 Answers 2000 people. 2% have the condition so an expected 40 have the condition and 1960 do not. Test A gives positive results for 80% of persons with the condition (an expected 32 positive and 8 negative) and for 5% of persons without the condition (98 positive and 1862 negative), so in total an expected 130 positive and 1870 negative test results. So Test A is applied to 2000 people costing$\$1$ each and Test B to an expected 130 people costing $\$100\$ each, for a total expected cost of $$2000 \times \1 + 130 \times \100 = \15000.$$

With this you identify an expected 32 actual cases and miss 8 actual cases.

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You can solve this by working out the expected cost per person, and then multiply that by 2000, since each person's cost is independent of all others.

There are four possible cases to consider: (a) a person has the condition and tests negative for it with test A; (b) a person does not have the condition and tests negative for it with test A; (c) a person has the condition and tests positive for it with test A, and so is given test B; (c) a person does not have the condition and tests positive for it with test A, and so is given test B.

The probability of case (a) occurring is (0.02)(0.2).

The probability of case (b) occurring is (0.98)(0.95).

The probability of case (c) occurring is (0.02)(0.8).

The probability of case (d) occurring is (0.98)(0.05).

The cost of case (a) and (b) is one dollar, while the cost of case (c) and (d) is 100 dollars.

Hence, the expected cost per person is $$1((0.02)(0.2)+(0.98)(0.95))+100( (0.02)(0.8) + (0.98)(0.05)) = 7.5.$$ With 2000 people, the expected cost is thus 15000 dollars.

The number of cases can be worked out in a similar way, but it is simpler.

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