Calculating probability of 'at least one event occurring'

Question

If I know the probability of A and the probability of B how can I calculate the probability of "at least one of them" occurring?

I was thinking that this is P(A or B) = P(A) + P(B) - P(AB). Is this correct?

If it is, then how can I solve the following problem taken from 'DeGroot - Probability and Statistics':

If 50 percent of families in a certain city subscribe to the morning newspaper, 65 percent of the families subscribe to the afternoon newspaper, and 85 percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?

Here the question is P(morning)=.5, P(afternoon)=.65 P(morning OR afternoon)=.85

P(morning OR afternoon) = .5 + .65 - .3 = .85

P(morning AND afternoon) = P(morning) + P(afternoon) - P(morning OR afternoon)

So the answer to the question is .3

Is my reasoning correct?

EDIT.

If this reasoning is correct:

How can I calculate the following:

If the probability that student A will fail a certain statistics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1,what is the probability that exactly one of the two students will fail the examination?

So this questions highlights the difference between !at least one of them! !or exactly one of them!

I understand that at least one of them = P(A or B)

but how can I work out the probability of exactly one of them?

Answer 1

You are correct.

To expand a little: if $A$ and $B$ are any two events then

$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$

or, written in more set-theoretical language,

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

In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.

Answer 2

For your second question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together.

Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.

Answer 3

For the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\cap B)$$