Word problem of probability In a group of people, 21% own both phones and computers, 49% own just phones and 7% own just computers. One person is randomly chosen, what are the odds that they own a phone if we know that they own a computer?
Let $A$ represent the event that the person owns a cellphone, and let $B$ represent the event that a person owns computers. 

Note that it is given that $P(A \cap B) = 0.21, P(A - B) = 0.49$, and $P(B - A) = 0.07$. We then need to find $P(A)$ and $P(B)$. However, one can see that $P(A) = P(A \cap B) + P(A-B)$ and similarly $P(B) = P(A \cap B) + P(B-A)$. We then have $P(A \cap B) = 0.21$, $P(A) = 0.7$, and $P(B) = 0.28$. By Bayes, we know that $P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.21}{0.28} = 3/4$.
Let $A$ represent the event that the person owns a cellphone, and let $B$ represent the event that a person owns computers. 

Note that it is given that $P(A \cap B) = 0.21, P(A - B) = 0.49$, and $P(B - A) = 0.07$. We then need to find $P(A)$ and $P(B)$. However, one can see that $P(A) = P(A \cap B) + P(A-B)$ and similarly $P(B) = P(A \cap B) + P(B-A)$. We then have $P(A \cap B) = 0.21$, $P(A) = 0.7$, and $P(B) = 0.28$. By Bayes, we know that $P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.21}{0.28} = 3/4$.