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?

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

⇒  

⇒  

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

⇒  

⇒  

Correspondingly,

P (not getting cancer and testing effective or positive)

=

=

P (with a good test result)

=

=

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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