Suppose that a certain diagnostic screening programme has the probability of 0.023 ofmissing a positive diagnosis. In a sample of 100 cases independent of each other, whatis the probability that(i) Exactly 3 cases were undiagnosed?(ii) At least 2 cases were undiagnosed?
Question
Suppose that a certain diagnostic screening programme has the probability of 0.023 ofmissing a positive diagnosis. In a sample of 100 cases independent of each other, whatis the probability that(i) Exactly 3 cases were undiagnosed?(ii) At least 2 cases were undiagnosed?
Solution
This problem can be solved using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
(i) For exactly 3 cases undiagnosed:
Here, n=100 (the number of cases), k=3 (the number of undiagnosed cases we're interested in), and p=0.023 (the probability of a case being undiagnosed).
P(X=3) = C(100, 3) * (0.023^3) * ((1-0.023)^(100-3))
(ii) For at least 2 cases undiagnosed:
This is equivalent to 1 minus the probability of 0 or 1 case being undiagnosed.
P(X>=2) = 1 - P(X=0) - P(X=1)
where:
P(X=0) = C(100, 0) * (0.023^0) * ((1-0.023)^(100-0))
and
P(X=1) = C(100, 1) * (0.023^1) * ((1-0.023)^(100-1))
So, you can calculate these probabilities using the above formulas.
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