Deriving probability of exactly one event occuring

Question

Show that for any events A and B, the probability that exactly one of them occur is Pr(A) + Pr(B) − 2 Pr(A ∩ B).

Solution: The probability that exactly one event occurs is $$ Pr ((A \cap B^c) \cup (A^c ∩ B)) = Pr(A ∩ B^c) + Pr(A^c ∩ B) \\ = (Pr(A) − Pr(A ∩ B)) + (Pr(B)− Pr(A ∩ B))\\ = Pr(A) + Pr(B) − 2 Pr(A ∩ B).$$

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I am going through this question and I don't understand how $$P(A \cap B^c) = (Pr(A) − Pr(A ∩ B))$$ and $$P(A^c \cap B) = (Pr(B) − Pr(A ∩ B))$$

If someone could explain this to me, I would be very happy.

Note: I understand why this works with a venn diagram, but I can't derive its formula...

Answer 1

Even without measure theory:

If $A$ has occurred, the also, either $B$ has occurred or or $B$ has not occurred. The probablility that $A$ has occurred and $B$ has not occurred is by definition $P(A\cap B^c)$ and similarly the probablility that $A$ has occurred and $B$ has
is $P(A\cap B)$.

So $$ P(A) = P(A\cap B)+P(A\cap B^c) \\ P(A\cap B^c) = P(A) - P(A\cap B) $$

Answer 2

Are you familiar with measure theory? If so, $P$ is a measure, and $P(A) = P(A \cap B) + P(A \cap B^{c})$ (since measures are countably additive as long as the sets are pairwise disjoint), which implies $P(A) - P(A \cap B) = P(A \cap B^{c})$.

You can apply a similar argument for $P(A^{c} \cap B)$.

Answer 3

You could think of it like $$P(B) \cup P(B^{c})=1$$ so $$P(A)=P(A)\cap(P(B) \cup P(B^{c}))$$ and therefore $$P(A \cap B^c) = (P(A) − P(A ∩ B))$$

I don't know if that is formally written correctly, but maybe conveys the right logic.

Answer 4

The reason for the identity

$P(A^c∩B)=(P(B)−P(A∩B))$

is that the two sets $A^c∩B$ and $A∩B$ are disjoint and make up $B$, that is

$(A^c∩B) \cup (A∩B) = B.$

The probability of the left hand side is thus equal to the probability on the right hand side, and since the union on the left hand side is disjoint, it is equal to the sum of the probabilities of each of the sets in the union.