# Conditional probability problem

• May 15th 2011, 11:58 AM
paupsers
Conditional probability problem
Diseases I and II are prevalent among people in a certain population. It is assumed
that 10% of the population will contract disease I sometimes during their lifetime, 15% will contract disease II eventually, and 3% will contract both diseases.

a. Find the probability that a randomly chosen person from this population will contract at least one disease in his/her lifetime.

b. Find the conditional probability that a randomly selected person from this population will contract both diseases, given that he or she has contracted at least one disease in his/her lifetime.

For a, I did P(I)+P(II)-P(I or II)=0.1+0.15-0.03=0.22

I don't know how to do b, though. Any ideas?
• May 15th 2011, 12:09 PM
Plato
Quote:

Originally Posted by paupsers
Diseases I and II are prevalent among people in a certain population. It is assumed that 10% of the population will contract disease I sometimes during their lifetime, 15% will contract disease II eventually, and 3% will contract both diseases.
a. Find the probability that a randomly chosen person from this population will contract at least one disease in his/her lifetime.

b. Find the conditional probability that a randomly selected person from this population will contract both diseases, given that he or she has contracted at least one disease in his/her lifetime.

For a, I did P(I)+P(II)-P(I or II)=0.1+0.15-0.03=0.22

Let's use $III$ to mean has both $I~\&~II$.
We want to find $\mathcal{P}(III|I\cup II)$.
What would $III\cap(I\cup II)=~?$