# Proabability Bayes theorem cancer question

Cancer is present in 22% of a population and is not present in the remaining 78%. An imperfect clinical test successfully detects the disease and with probability 0.70. Thus, if a person has the disease in the serious form, the probability is 0.70 that the test will be positive and it is 0.30 if the test is negative. Moreover among the unaffected persons, the probability that the test will be positive is 0.05. A person selected at random from the population is given the test and the result is positive. What is the probability that this person has the cancer ?

I have tried it, Please correct me if I am doing something wrong or there is any other easy way of doing that.

p(A) be the probability population has cancer = 0.22 p(A) be the probability population don't have cancer = 0.78

E1 be the event of probability that test is positive = 0.70 E2 be the event of probability that test is negative = 0.30

p(person has cancer) = p(E1|A)*E1 / [p(E1|A)*E1] * [p(E2|A)*E2]

p(person has cancer) = 0.22*0.70/0.22*0.70+0.78*0.30

is this correct ?? or I am doing something wrong ?

Thanks . Help is appreciated

Close ... but not close enough!

What you are asked to figure out is $P(A|E1)$, and Bayes' formula for that is:

$$P(A|E1)=\frac{P(E1|A)\cdot P(A)}{P(E1)}$$

So note that in the numerator you get $P(E1|A) \cdot P(A)$ rather than $P(E1|A) \cdot P(E1)$

Also note that what you know is:

$$P(E1|A)=0.7$$

(rather than your $P(E1)=0.7$)

and we have that:

$$P(E1|A')=0.05$$

Oh, and of course:

$$P(A)=0.22$$

and

$$P(A')=0.78$$

Finally:

$$P(E1)=P(E1|A) \cdot P(A) + P(E1|A') \cdot P(A')$$

(whereas you brought in $E2$ in your equations)

• can you please solve this question ? As from formulae it's difficult to understand what are you saying. – himanshu chawla Dec 11 '17 at 18:11
• so there will no use of 22% and 78% ? – himanshu chawla Dec 11 '17 at 18:16
• @himanshuchawla Hmmm, I used the same variables as you defined, and I indicated exactly what values you are given. From there, it really is just plugging them into the formula. And yes, you do use the 22%; $P(A)=0.22$ and $P(A')=0.78$. You'll need those to calculate $P(E1)$ and your eventual $P(A|E1)$ – Bram28 Dec 11 '17 at 18:16
• personally, i am requesting you to please pen down the answer as it's becoming more cumbersome for me to understand. Thanks a lot – himanshu chawla Dec 11 '17 at 18:19
• @himanshuchawla Really, it's all there: the formulas, and the values to plug into those formulas! – Bram28 Dec 11 '17 at 18:22