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)]=<br />
1-[.2+.3-.1]=1-.4=.6](http://latex.codecogs.com/png.latex?P(A\cup\\B)^{c}=1-P(A\cup\\B)=1-[P(A)+P(B)-P(A\cap\\B)]=<br />
1-[.2+.3-.1]=1-.4=.6)
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)}<br />
=\frac{P[(A\cap\\B)\cap\\(B\cap\\B)]}{P(B)}<br />
=\frac{P(A\cap\\B)P(B)}{P(B)}=.33333](http://latex.codecogs.com/png.latex?P(A\cap\\B|B)=\frac{P[(A\cap\\B)\cap\\B]}{P(B)}<br />
=\frac{P[(A\cap\\B)\cap\\(B\cap\\B)]}{P(B)}<br />
=\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
