A new test has been developed to detect a particular type of cancer. The test must be evaluated before it is put into use. A medical researcher selects a random sample of 2,000 adults and finds (by other means) that 1% have this type of cancer. Each of the 2,000 adults is given the test, and it is found that the test indicates cancer in 99% of those who have it and in 2% of those who do not.
a) What is the probability of a randomly chosen person having cancer given that the test indicates cancer?
b)What is the probability of a person having cancer given that the test does not indicate cancer

Question

A new test has been developed to detect a particular type of cancer. The test must be evaluated before it is put into use. A medical researcher selects a random sample of 2,000 adults and finds (by other means) that 1% have this type of cancer. Each of the 2,000 adults is given the test, and it is found that the test indicates cancer in 99% of those who have it and in 2% of those who do not.
a) What is the probability of a randomly chosen person having cancer given that the test indicates cancer?
b)What is the probability of a person having cancer given that the test does not indicate cancer

kropot72 · Answer

|dw:1407818368798:dw|
Does the probability tree make sense to you?

anonymous · Answer

yeah that makes sense!

anonymous · Answer

oh wait so is a) just 99%?

anonymous · Answer

or 0.99

kropot72 · Answer

The probability that a randomly chosen person has cancer and tests positive is:$$\large P(cancer+test\ positive)=0.01\times0.99=0.0099$$
The probability that a randomly chosen person does not have cancer and tests positive is:
$$\large P(no\ cancer+test\ positive)=0.99\times0.02=0.0198$$
Given that a randomly chosen person tests positive, the probability that they have cancer is:
$$\large \frac{P(cancer+test\ positive)}{P(cancer+test\ positive)+P(no\ cancer+test\ positive)}$$

anonymous · Answer

i got 1/3 for that

dumbcow · Answer

you can also use conditional probability here:
$$P(A|B) = \frac{P(A and B)}{P(B)}$$
P(A|B) means prob of A given B

kropot72 · Answer

Correct!
b) The probability that a person has cancer, although the test is negative is:
$$\large \frac{0.01\times0.01}{(0.01\times0.01)+(0.99\times0.98)}$$

anonymous · Answer

hmm is that one supposed to be 1.030609?

kropot72 · Answer

How did you get that value, bearing in mind that the numerator comes to 0.0001 and the denominator is close to 1?

anonymous · Answer

i just plugged it into my calculator and thats what came up!

anonymous · Answer

I'm getting 1E-4 for (0.01*0.01)

kropot72 · Answer

Correct! Now divide by 0.9703.

anonymous · Answer

same answer!

anonymous · Answer

is it supposed to be 0.99*0.02?

kropot72 · Answer

Correct again. The more exact answer is 0.000103061.
"is it supposed to be 0.99*0.02"
No it is 0.99 * 0.98, which is the probability of no cancer and testing negative.

anonymous · Answer

how did you come up with 0.000103061? everytime i try it i get 1.030609

anonymous · Answer

ohhh wait because its -4 do the zeros go infront?

kropot72 · Answer

The 3 zeros surely go after the decimal point.

anonymous · Answer

okay thats where i was confused! thank you :) @kropot72

kropot72 · Answer

You're welcome :)