Question

Studies show that about five women in seven (approximately 71.4%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 25% of the time. Also, suppose that in the general population of women, the test for breast cancer is negative about 45% of the time. Let B=women develops breast cancer, P=tests positive, and N=tests negative. Suppose one woman is selected at random. Round your answer to three decimal places. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P|B) = 1-P(N|B). What is the probability that a woman develops breast cancer and tests positive. Find P(B AND P) = P(P|B)P(B). What is the probability that a woman does not develop breast cancer. Find P(B’) = 1-P(B). What is the probability that a woman tests positive for breast cancer. Find P(P)=1-P(N).

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Fred

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Let's define the events:

B : Woman develops breast cancer

P : Test result is positive

N : Test result is negative

Given probabilities:

P(B) = 5/7

P(N|B) = 0.25

P(N) = 0.45

1. Probability a woman who develops breast cancer tests positive:P(P|B) = 1 - P(N|B)

P(P|B) = 1 - P(N|B) = 1 - 0.25 = 0.75

2. Probability a woman develops breast cancer and tests positive:P(B and P) = P(P|B) \cdot P(B)

P(B and P) = P(P|B) \cdot P(B) = 0.75 \cdot 5/7 = 0.5357

3. Probability a woman does not develop breast cancer:P(B') = 1 - P(B)

P(B') = 1 - P(B) = 1 - 5/7 = 2/7 \approx 0.286

4. Probability a woman tests positive for breast cancer:P(P) = 1 - P(N)

P(P) = 1 - P(N) = 1 - 0.45 = 0.55

**Answer:**

1.P(P|B) = 0.75

2.P(B and P) = 0.5357

3.P(B') = 0.286

4.P(P) = 0.55

Given probabilities:

1. Probability a woman who develops breast cancer tests positive:

2. Probability a woman develops breast cancer and tests positive:

3. Probability a woman does not develop breast cancer:

4. Probability a woman tests positive for breast cancer:

**Answer:**

1.

2.

3.

4.

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