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Probability Formulas

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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Probability Formulas

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List of Important Probability Formulas:
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 {E1, E2, ..., En) is nothing but a division of any given sample space and also let each of the E1, E2 , ..., En has non-zero probability. Let A be any event related with S, then
a.    P(A) = P(E1) P (A|E1) + P (E2) P (A|E2) + .............. + P (En) P(A| En)
8.    Bayes' theorem: If E1, E2, ............, En are events which constitute a partition of sample space S, i.e. E1, E2, ..., En are pairwise disjoint and E1 union E2 union ... union En = S and A be any event with nonzero probability, then P(Ei | A) = P(Ei ) P(A|Ei ) / ∑ P(Ej ) P(A|Ej ), where j varying from 1 to n.

Solved Examples:
Example 1: Find the probability that an even number is obtained, when a die is rolled.
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) = $\frac{n(E)}{n(S)}$  = $\frac{3}{6}$ = $\frac{1}{2}$ .

Example 2: Find the probability of obtaining the two heads, when two coins are tossed.
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) = $\frac{n(E)}{n(S)}$ = $\frac{1}{4}$.

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