A certain form of cancer is known to be found in women over 60 with probability of 0.07. A blood test exists for the detection of the disease, but the test is not infallible. In fact, it is known that 10% of the time the test gives a false negative (i.e. the test incorrectly gives negative result when the woman actually has the cancer) and 5% of the time the test gives false positive (i.e. incorrectly gives a positive result when the woman actually does not have the cancer). What is the probability that a randomly selected woman over 60 will test positive?

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SerenaBochenek SerenaBochenek · Answer 1 · 2020-09-29T19:52:59+02:00

Answer:

Throughout the segment below, the definition including its particular question is mentioned.

Step-by-step explanation:

The probability of developing cancer

= 0.07

The probability of someone not getting cancer

= 1 - 0.07

= 0.93

Provided that if women have cancer, the risk of someone not testing positive is:

= 0.10

Therefore, if a female requires cancer, the risk of testing positive

= 1 - 0.10

= 0.90

The Probability of positive test

= 0.055 (whenever a woman does not have cancer)

Therefore, whenever a woman does not have cancer, the risk of not testing positive will be:

= 1 - 0.05

= 0.955

Now,

By using the law of conditional probability, we get

⇒ [tex]P(\frac{B}{A} ) = \frac{P(A \ and \ B)}{P(A)}[/tex]

⇒ [tex]P (A \ and \ B) = P(A)\times P(B)[/tex]

⇒ P (having cancer as well as positive tests) = P(having cancer) × P(Effective results, because she has cancer)

⇒ [tex]0.07\times 0.90[/tex]

⇒ [tex]0.063[/tex]

Correspondingly,

P (not getting cancer and testing effective or positive)

= [tex]0.93\times 0.05[/tex]

= [tex]0.04655[/tex]

P (with a good test result)

= [tex]0.063 + 0.0465[/tex]

= [tex]0.1095[/tex]