The probability that a particular screening test for breast cancer will be posit

Question

The probability that a particular screening test for breast cancer will be positive when, in fact, the patient does, in fact, have breast cancer is 0.9. You may assume that the outcome of any one test is independent of the outcome of any other. One morning five women, each of whom does have breast cancer, take the test. What is the probability that:

(a) All five of the tests are positive? [6 pts]

(b) The first test is negative and the others positive? [6 pts]

(c) at least one of the five tests is negative? [8 pts]

Explanation / Answer

Solution:-

a) The probability that  all five of the tests are positive is 0.5905.

p = 0.9

Probability of all five being positives = 0.9 × 0.9 × 0.9 × 0.9 × 0.9

Probability of all five being positives = 0.5905

b) The probability that the first test is negative and the others positive is 0.06561.

p = 0.9

Probability the first test is negative and the others positive = 0.10 × 0.9 × 0.9 × 0.9 × 0.9

Probability the first test is negative and the others positive = 0.06561

c) The probability that at least one of the five tests is negative is 0.40951.

Probability of being negative = 1 - p = 0.10

n = 5

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 1) = 0.40951

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