Class 12 Maths Probability | Independent Events |

**Independent Events**:

Let E and F be two events associated with a sample space S. If theprobability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. Thus, two events E and F will be independent, if

**(a) P(F | E) = P(F), provided P (E) ≠ 0****(b) P(E | F) = P(E), provided P (F) ≠ 0**

Using the multiplication theorem on probability, we have

**(c) P(E ∩ F) = P(E) P(F)**

Three events A, B and C are said to be mutually independent if all the

following conditions hold:

**P(A ∩ B) = P(A) P(B)****P(A ∩ C) = P(A) P(C)****P(B ∩ C) = P(B) P(C)****and P(A ∩ B ∩ C) = P(A) P(B) P(C)**

*Example*: If you throw two coins, then probability of getting head or tail in second coin is independent of probability of getting head or tail in first coin.

*Example*: Let A and B be independent events with P(A)=0.3 and P(B) = 0.4.

Find(i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)*Solution*:

(i) we have to find the value of P(A ∩ B) and they gave P(A) =0.3 and P(B) = 0.4.

given A and B are independent events.

We know that if A and B are independent events then P(A ∩ B) = P(A) P(B)

=>P(A ∩ B) = 0.3×0.4 = 0.12

(ii)we have to find the value of P(A ∪ B)

the formula for P(A ∪ B) is P (A ∪B)=P(A)+P(B)−P( A∩B)

=> P(A ∪ B)=0.3+0.4-0.12 = 0.58

(iii) P(A|B)=?

the formula for P(A|B) is P(A|B)= P( A∩B)/ P(B)

=> P(A|B)= 0.12/ 0.4 = 0.3

(iv) P(B|A)=?

the formula for P(B|A) is P(B|A)= P( A∩B)/ P(A)

=> P(B|A)= 0.12/0.3 = 0.4

*Example:If* A and B are two events such that P(A)=1/4 ,P(B)=1/2 , ,P(A∩B)=1/8

find P(not A and not B).*Solution*:

we know that P(not A and not B)=P(A'∩B')

But we also know that A'∩B'=(A∪B)' => P(A'∩B')=P((A∪B))'

∴P(not A and not B)=P(A'∩B')

=P((A∪B))'

=1-P(A∪B) ( ∵ P(A)' = 1-P(A) )

=1- (P(A)+P(B)−P( A∩B)) (∵ P(A∪B)=P(A)+P(B)−P( A∩B) )

=1- ( 1/4 + 1/2 - 1/8 )

=1- 5/8

=3/8

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