Probability of an event, in simple words is defined as, the number of favorable cases upon the total number or exhaustive cases.

For example, if an even E can happen in h ways out of a total number or exhaustive cases of n possible ways, then probability of occurrence of the event (called as its success) is denoted as p = P[E] = $\frac{h}{n}$.

And the probability of nonoccurrence of the event (called as its failure) is denoted by q = P [not E] = 1 - P[E] = 1- ($\frac{h}{n}$). Thus, p + q = 1.Thus, the probability of occurrence and nonoccurrence of an event always equals to one.

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1. The conditional probability of an event E, given the occurrence of the event F is given by P(E | F) = $\frac{P(E∩ F)}{P(F)}$, provided that P(F) not equal to 0.

2. 0 ≤ P (E|F) ≤ 1, and P (E′ | F) = 1 – P (E | F)

3. P ((E union F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G)

4. P (E ∩ F) = P (E) * P (F|E), P (E) ≠ 0

5. P (E ∩ F) = P (F) * P (E|F), P (F) ≠ 0

6. If E and F are independent events, then

a. P (E ∩ F) = P (E) * P (F)

b. P (E|F) = P (E), P (F) ≠ 0

c. P (F|E) = P (F), P(E) ≠ 0

7. Theorem of total probability: Let {E

a. P(A) = P(E

8. Bayes' theorem: If E

Solution: Sample space of the said experiment is S = {1, 2, 3, 4, 5, 6}

Let E be considered as the event of obtaining the even number as E = {2, 4, 6}

Using the formula of probability, we get, P (E) = $\frac{n(E)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$ .

Solution: Here, we could have two possibilities of outcomes which are H and T which results in the sample space S for this experiment as given by S = {(H,T), (H,H), (T,H), (T,T)}

If E be the event of obtaining two heads then E = {(H, H)}

Using the formula of probability, we get P(E) = $\frac{n(E)}{n(S)}$ = $\frac{1}{4}$.