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Using the axioms of probability, prove the following:

a. For any event $A, P\left(A^{c}\right)=1-P(A)$.
b. The probability of the empty set is zero, i.e., $P(\emptyset)=0$.
c. For any event $A$, $P(A) \leq 1$.
d. $P(A-B)=P(A)-P(A \cap B)$.
e. $P(A \cup B)=P(A)+P(B)-P(A \cap B)$, (inclusion-exclusion principle for $n=2$ ).
f. If $A \subset B$ then $P(A) \leq P(B)$.
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a. This states that the probability that $A$ does not occur is $1-P(A)$. To prove it using the axioms, we can write
\begin{aligned} 1 &=P(S) \\ &=P\left(A \cup A^{c}\right) \end{aligned}
$($ axiom 2$)$ (definition of complement) $=P(A)+P\left(A^{c}\right) \quad$ (since $A$ and $A^{c}$ are disjoint $)$
b. Since $\emptyset=S^{c}$, we can use part (a) to see that $P(\emptyset)=1-P(S)=0$. Note that this makes sense as by definition: an event happens if the outcome of the random experiment belongs to that event. Since the empty set does not have any element, the outcome of the experiment never belongs to the empty set.
c. From part (a), $P(A)=1-P\left(A^{c}\right)$ and since $P\left(A^{c}\right) \geq 0$ (the first axiom), we have $P(A) \leq 1$
d. We show that $P(A)=P(A \cap B)+P(A-B)$. Note that the two sets $A \cap B$ and $A-B$ are disjoint and their union is $A$ (Figure $1.17)$. Thus, by the third axiom of probability
$$\begin{array}{rlr} P(A) & =P((A \cap B) \cup(A-B)) & (\text { since } A=(A \cap B) \cup(A-B)) \\ & =P(A \cap B)+P(A-B) & (\text { since } A \cap B \text { and } A-B \text { are disjoint }) . \end{array}$$

Note that since $A-B=A \cap B^{c}$, we have shown
$$P(A)=P(A \cap B)+P\left(A \cap B^{c}\right)$$
Note also that the two sets $B$ and $B^{c}$ form a partition of the sample space (since they are disjoint and their union is the whole sample space). This is a simple form of law of total probability that we will discuss shortly and is a very useful rule in finding probability of some
events.
e. Note that $A$ and $B-A$ are disjoint sets and their union is $A \cup B$. Thus,
\begin{aligned} P(A \cup B) &=P(A \cup(B-A)) & &(A \cup B=A \cup(B-A)) \\ &=P(A)+P(B-A) & &(\text { since } A \text { and } B-A \text { are disjoint) } \end{aligned}
$$=P(A)+P(B)-P(A \cap B) \quad \text { (by part }(\mathrm{d}))$$
f. Note that $A \subset B$ means that whenever $A$ occurs $B$ occurs, too. Thus intuitively we expect that $P(A) \leq P(B)$. Again the proof is similar as before. If $A \subset B$, then $A \cap B=A$. Thus,
$$P(B)=P(A \cap B)+P(B-A)$$
(by part (d))
$$\begin{array}{ll} =P(A)+P(B-A) & (\text { since } A=A \cap B) \\ \geq P(A) & (\text { by axiom } 1) \end{array}$$
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