Question

The probability that a blue-eyed person is left-handed is 1/7. The probability that a left-handed person is blue-eyed is 1/3. The probability that a person has neither of these attributes is 4/5. What is the probability that a person has both attributes?

Answer 1

the probability that someone has at least one attribute is\[P(at~least~one~attribute)=1-P(neither~attribute)\]let \(A=left-handed\) and \(B=right-handed\)
We can then write this as
\[P(A\cup B)=1-P(A^c\cap B^c)\]you can then apply the inclusion-exclusion principle to find the probability of the intersection \(P(A\cap B)\) (the probability of being both right-handed and left-handed):\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]