30% of brain tumors are malignant. A new test is 90% accurate. That is, if a tum

Question

30% of brain tumors are malignant. A new test is 90% accurate. That is, if a tumor is malignant, it will indicate a positive with 90% of the time. If a tumor is non-malignant, it will be falsely positive with 10% of the time.

a)Construct a probability tree for this problem.

b)What is the probability of a tumor non-malignant if the test is negative?

c)Bob is tested non-malignant for his tumor. He is happy but wants to be sure. So he tested again and it is again negative. Given that he has two negative tests, and assuming the two tests are independent, what is the probability that Bob

Explanation / Answer

Let M represent having a malignant tumor, and N represent a negative test.
We are given P(M) = 0.3, P(not M) = 1 - 0.3 = 0.7, P(N|M) = 1 - 0.9 = 0.1, and P(N|not M) = 1 - 0.1 = 0.9.

Because the tests are independent (given the actual condition of the tumor), we also have
P(two N's|M) = (0.1)^2 = 0.01 and P(two N's|not M) = (0.9)^2 = 0.81.

From Bayes's Theorem, the probability that the tumor is non-malignant if the test is negative is
P(not M|N) = P((not M) and N)/P(N)
= P(N|not M)P(not M) / [P(N|not M)P(not M) + P(N|M)P(M)]
= 0.9(0.7) / [0.9(0.7) + 0.1(0.3)]
= 0.63/0.66
= 21/22.

From Bayes's Theorem, the probability that the tumor is non-malignant if the test is negative twice is
P(not M|two N's) = P((not M) and two N's)/P(two N's)
= P(two N's|not M)P(not M) / [P(two N's|not M)P(not M) + P(two N's|M)P(M)]
= 0.81(0.7) / [0.81(0.7) + 0.01(0.3)]
= 0.567/0.57
= 189/190.

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