The following three axioms

1. P(E) $\geq$ 0 for all E $\in$ C

2. P(S) = 1

3. let E_1, E_2,.... be pairwise disjoint events then

$P(\bigcup_{i=1}^{\infty} E_i)=\sum_{i=1}^{\infty} P(E_i)$

a,b,c are disjoint events

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

$P(A \cup B \cup C)=P(A)+P(B)+P(C)$

Let $A$ and $B$ be any two events. Then

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

Proof: $~$ We can write

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

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

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

$E= F \cup(E-F)$

$P(E)= P(F)+P(E-F)$

So $P(E-F)=P(E)-P(F) \geq 0$

$\Rightarrow P(E) \geq P(F)$

i.e probability is a monotone function.

3) $E \cup E^c=S$

$P(E)+P\left(E^c\right)=P(S)=1$

$\Rightarrow P\left(E^c\right)=1-P(E)$

Extension to 3 events: $A, B, C$

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

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

$\quad-P((\underbrace{A \cap C)} \cup(B \cap C))$

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

$\quad-\quad(P(A \cap C)+P(B \cap C)-P((A \cap C) \cap(B \cap C)))$

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

$\quad-P(B \cap C)+P(A \cap B \cap C)$

Let $A_1, A_2, \ldots, A_n$ be any events. Then

$P(\bigcup_{i=1}^n A_i)=\sum_{i=1}^n P(A_i)-\underset{i<j}{\sum \sum} P(A_i \cap A_j)$

$+ (\underset{i<j<k}){\sum\sum \sum}_k P(A_i \cap A_j \cap A_k)-\cdots$

$+(-1)^{n+1} P(\bigcap_{i=1}^n A_i)$

Proof: We will prove the relation (1) using the principle of mathematical induction.

P(1)

$P(K) \rightarrow P(K+1)$

For $n=1$, the statement (1) becomes

$P(A_1)=P(A_1)$ which always holds true.

Next we assume the statement (1) to be true for $n=k$. Then we prove it for $n=k+1$

$P(\bigcup_{i=1}^{k+1} A_i)=P((\underbrace {\bigcup_{i=1}^k A_i}) \cup A_{k+1})$

$= P(\bigcup_{i=1}^k A_i)+P(A_{k+1})-P((\bigcup_{i=1}^k A_i) \cap A_{k+1})$

(by addition rule for two events)

$=\sum_{i=1}^k P(A_i)-\underset{i<j}{\sum ^k \sum^k} P(A_i \cap A_j)+\underset{i<j<m}{\sum ^k \sum ^k \sum ^k} P(A_i \cap A_j \cap A_m$

$-\cdots+(-1)^{k+1} P(\bigcap_{i=1}^k A_i)+P(A_{k+1})$

$-P(\bigcup_{i=1}^k(A_i \cap A_{k+1}))$

$= \sum_{i=1}^{k+1} P(A_i)- \underset{i<j}{\sum ^{k+1} \sum ^{k+1}} P (A_i \cap A_j)$

$+ \underset{i<j<m} {\sum ^{k+1} \sum ^{k+1} \sum ^{k+1}} P (A_i \cap A_j \cap A_m) -\cdots$

$+(-1)^{k+2} P(\bigcap_{i=1}^{k+2} A_i )$

this shows that statement (1) is true for $n=k+1$.

Hence by the Principle of mathematical Induction the General addition rule holds for all $n$ ( $n$ is a positive integer).

(Using the general addition ).

Let A denote the event that the set of six cards contains at least one card of each suit .

Then $A^c \rightarrow$ at least one suit does not appear in the set $\eta$ six cards.

Let,

$B_1 \rightarrow$ hearts do not appear

$B_2 \rightarrow$ spades do not appear

$B_3 \rightarrow$ clubs do not appear

$B_4 \rightarrow$ diamonds do not appear

Then $A^c=\bigcup_{i=1}^4 B_i$

By general addition rule

$P(A^c) = P(\bigcup_{i=1}^4 B_i)$

$= \sum_{i=1}^4 P(B_i) - \underset{i<j}{\sum ^4 \sum ^4} P(B_i \cap B_j)$

$+ \underset{i<j<k}{\sum ^4 \sum^4 \sum ^4} P (B_i \cap B_j \cap B_k)$

$- P (\bigcap_{i=1}^4 B_i)$

$P(B_i)=\frac{3}{4} \times \frac{3}{4} \times \cdots \times \frac{3}{4}=(\frac{3}{4})^6$

$P(B_i)=(\frac{3}{4})^6, \quad i=1,2,3,4$

$P(B_1 \cap B_2)=(\frac{1}{2})^6, \quad P(B_i \cap B_j)=(\frac{1}{2})^6$ 6 terms

$P(B_i \cap B_j \cap B_k) = (\frac{1}{4})^6, \quad (1 \leq i<j<k \leq 4)$ 4 terms

$P(\bigcap_{i=1}^4 B_i)=0$

So,

$P(A^c) = 4 \times (\frac{3}{4})^6 - 6 \times (\frac{1}{2})^6 + 4 (\frac{1}{4})^6 = \frac{317}{512} \approx 0.62$

$P(A) = 1 -P(A) = \frac{195}{512} \approx 0.38$