5
$\begingroup$

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

I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and }B)$.

Is this correct?


If it is, then how can I solve the following problem taken from DeGroot's 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?

In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .5 + .65 - P(\text{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question.

Is my reasoning correct?


If it 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?

These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or } B)$, but how can I work out the probability of "exactly one of them"?

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ "Exactly one of A and B" means "Either A or B, but not both" which you can calculate as P(A or B) - P(A and B). $\endgroup$ – Chris Taylor Oct 14 '11 at 11:13
  • $\begingroup$ Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tutorial" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. $\endgroup$ – Chris Taylor Oct 14 '11 at 11:21
  • $\begingroup$ So if I download LaTeX and paste your notation then it displays it in a more readable form? $\endgroup$ – upabove Oct 14 '11 at 11:24
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Pr(A and not B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) $\endgroup$ – Petr Peller Apr 4 '15 at 17:44
  • $\begingroup$ Pr(A and not B) + Pr(B and not A) + Pr(A and B) is what you are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. $\endgroup$ – Petr Peller Apr 4 '15 at 18:06
  • $\begingroup$ $\Pr(A \text{ and not } B)+\Pr(B \text{ and not } A)$ is the answer to "but how can I work out the probability of exactly one of them?" which is what I meant by "your second question" $\endgroup$ – Henry Apr 4 '15 at 18:08
2
$\begingroup$

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

$\endgroup$
-2
$\begingroup$

probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.