# What is the probability that either $A$ or $B$ but not both will occur?

Events $A$ and $B$ are defined on a sample space $S$ such that $P((A\cup B)^c)=0.5$ and $P(A\cap B) = 0.2$. What is the probability that either $A$ or $B$ but not both will occur?

I have tried using Venn diagrams to solve the problem but I have been unsuccessful so far. How do I solve this problem?

• Hint: $P(A\cup B) = P(A\cap B^c) + P(A^c \cap B) + P(A\cap B)$ while $$P(\mathrm{exactly~one~of }~A~\mathrm{and}~B~\mathrm{occurs}) = P(A\cap B^c) + P(A^c \cap B).$$ – Dilip Sarwate Apr 7 '13 at 3:16

Hint: You are looking for the probability of an event $(A\cup B)\setminus(A\cap B)$.

Sum of events $A\cup B$ contains all outcomes for which at least one of $A,B$ occurs while the intersection $A \cap B$ - those for which both $A$ and $B$ do.

Since the difference of $E$ and $F$ is a set of outcomes for which (at the same time) $E$ occurs but $F$ does not (i.e. $E\setminus F= E\cap F^c$), the set for which at least one $A$ or $B$ but not both hold is exactly $(A\cup B)\setminus(A\cap B)$.

• ok so that would be .5/.2. but what is the significance of that event, can someone explain? – notamathwiz Apr 6 '13 at 23:47
• Let me ask if I understand your calculation correctly: what is the probability of $\mathbb{P}(E\setminus F)$ in terms of $\mathbb{P}(E)$ and $\mathbb{P}(F)$, whenever $F$ is a subset of $E$? – Kuba Helsztyński Apr 6 '13 at 23:54
• P(E/F) would be different from P(E)/P(F) – notamathwiz Apr 7 '13 at 0:05
• Notice that $0.5/0.2$ exceeds $1$, in particular it can't be the probability of any event. I suggest you to look at the Venn diagram of $E$ and $F$ when $F \subset E$ and to think of a probability as a measure of those sets. One should expect $\mathbb{P}(E)=\mathbb{P}(F)+\mathbb{P}(E\setminus F)$ to hold in this case. And indeed it can be easily derived from Kolmogorov axioms. – Kuba Helsztyński Apr 7 '13 at 0:32
• it would be 0.3 right? – notamathwiz Apr 7 '13 at 3:12

The Venn diagram method will work fine.

Draw the usual two intersecting ovals, and label them $A$ and $B$.

You can see $A\cap B$ in the picture. It is the region in common between $A$ and $B$. We are told that $\Pr(A\cap B)=0.2$. Write $0.2$ in the region in common between $A$ and $B$.

The event (set) $(A\cup B)^c$ is the complement of the union of $A$ and $B$. So it is the region which is outside both $a$ and $B$, it is the rest of the world. The probability of that is $0.5$. Write $0.5$ somewhere in that region.

We want the probability of being in $A$ or $B$ but not in both. That region is made up of two parts, neither of which has yet been counted in. Call the probability of being in that split region $x$. Then $$0.2+0.5+x=1,$$ because the probability of landing somewhere is $1$. Now we can find $x$.

Another way: Because the probability of the "rest of the world" is $0.5$, the probability of $A\cupB$ is $1-0.5$, which is $0.5$.

The part in both $A$ and $B$ has probability $0.2$. So the probability of being in $A$ or $B$ but not both is $0.5-0.2$.

I'm not really great with probability notations but If I may share a very handy formula for problems like this.

Let P(A), P(B) be proabability representing occurence of events A and B, respectively.
P(A or B) is the probability that either or both of A and B occur.
P(A and B), both A and B occur.
P(A or B)', neither of A and B occurs. This is just the complement of P(A or B).
P(A xor B), probability that either A or B will occur but not both!

First basic equation:

P(A or B) = P(A) $+$ P(B) $-$ P(A and B)

$1$ $-$ P(A or B)' = P(A) $+$ P(B) $-$ P(A and B)
$1$ $-$ $0.5$ = P(A) $+$ P(B) $-$ $0.2$
$\therefore$ P(A) $+$ P(B) = $0.7$

Second basic equation:

P(A xor B) = P(A) $+$ P(B) $-$ $2\times$P(A and B)

P(A xor B) = $0.7 - 2\times0.2$
$\therefore$ P(A xor B) = $0.3$

answer: $0.3$