0
$\begingroup$

Sixty percent of the families in a certain community own their own car, thirty percent own their own home, and twenty percent own both their own car and their own home. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

I tried to solve this using conditional probability.

Let A = owns car or home

B = not both(car and home)

then P(A|B) = ?

P(A|B) = P(AB)/P(B)

P(A|B) = $\frac{P(B|A)*P(A)}{1-0.2}$

But i don't know how to get the value of $P(B|A)$. Is it the right way or i'm getting it wrong?

$\endgroup$

1 Answer 1

0
$\begingroup$

But the randomly selected person need not belongs to $B$.

We are required to compute

$$P(A \cup B)-P(A \cap B)=P(A)+(B)-2P(A \cap B)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .