P(E)= 1−P(E′)
P(ϕ)= 0
If A⊆B ,then P(A)≤P(B)
For any event E, 0≤P(E)≤1
For A and B, P(A∪B)= P(A)+P(B)−P(A∩B)
P(A∪B∪C)= P(A)+P(B)+P(C)−P(AB)−P(AC)−P(BC)+P(ABC)
P(A∩B)= P(A)*P(B|A)
P(A∩B∩C)= P(A)*P(B|A)*P(C|A∩B)

(Bayes’ Theorem) P(A|B)=

Let S be the same space, then the probability of occurrence of an event E is denoted by P(E) and is defined as

Sample Space :
In tossing a coin, S = {H, T}
If two coins are tossed, the S = {HH, HT, TH, TT}
In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}

Example: Two coins are tossed, find the probability that two heads are obtained.

Explanation:The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}

Let E be the event “two heads are obtained”. ‘
E = {(H,H)}

We use the formula of the classical probability.
P(E) = n(E) / n(S) = 1 / 4
Hence the probability of two heads obtained is ¼.

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