# A few basic probability question

• Dec 9th 2009, 12:27 PM
Danneedshelp
A few basic probability question
Q: A large group of people is to be checked for two common symptoms of a certain disease. It is thought that 20% of the people possess symptom A alone, 30% possess symptom B alone, 10% possess both symptoms, and the remainder have neither symptom. For one person chosen at random from this group, find these probabilitys:

a) The person has neighter symptom.

b) The person has at least one symptom.

c) The person has both symptoms, given that he has symptom B.

A: I am assuming these events are independent.

a) $P(A\cup\\B)^{c}=1-P(A\cup\\B)=1-[P(A)+P(B)-P(A\cap\\B)]=
1-[.2+.3-.1]=1-.4=.6$

b) $P(A\cup\\B)=P(A)+P(B)-P(A\cap\\B)=.2+.3-.1=.4$

c) $P(A\cap\\B|B)=\frac{P[(A\cap\\B)\cap\\B]}{P(B)}
=\frac{P[(A\cap\\B)\cap\\(B\cap\\B)]}{P(B)}
=\frac{P(A\cap\\B)P(B)}{P(B)}=.33333$

The book says teh answers are:

a).4
b).6
c).25

I am not sure what I am doing wrong.

Thanks
• Dec 9th 2009, 01:28 PM
Plato
Quote:

Originally Posted by Danneedshelp
Q: A large group of people is to be checked for two common symptoms of a certain disease. It is thought that 20% of the people possess symptom A alone, 30% possess symptom B alone, 10% possess both symptoms, and the remainder have neither symptom. For one person chosen at random from this group, find these probabilitys:

The point is alone: $P(A\cap B^c)=0.2,~P(B\cap A^c)=0.3~\&~P(A\cap B)=0.1$.
So $P(A)=0.3~\&~P(B)=0.4$.