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You are given the following information about events A, B, and C P(A)0.35, P (B)-0.3, P(C) 0.51 Events A and B are independent. The probability of at least two of these events occurring is 0.27. The probability of at exactly two of these events occurring is 0.2 Find P(4jc) 0.3698 0.3489 0.3384 0.3279 0.3593

It is known that 2.6% of the population has a certain disease. A new test is developed to screen for the disease. A study has shown that the test returns a positive result for 18% of all individuals, and returns a positive result for 92% of individuals who do have the disease If a person tests positively for the disease under this test, what is the probability that they actually have the disease? 0.1276 0.1329 0.1435 0.1223 O 0.1382

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Answer #1

2)
A - actually have the disease
X - test positive

P(A) = 0.026
P(X) = 0.18
P(X|A) = 0.92

P(A|X) = P(A and X) /P(X)

P(A and X)= 0.92 * 0.026 = 0.02392

hence
answer is 0.02392/0.18
= 0.1328888

option B) 0.1329 is correct

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