<span>If two events are independent, the occurrence of one event does not affect the other.
That is if two events are independent, then P(Aâ©B)=P(A)P(B)
Let A be the even getting a sum of 6 in a single roll of two dice.
Sample space of A ={(1,5)(5,1)(2,4)(4,2)(3,3)}
n(A)=5; n(S)=36
Therefore P(A) =n(A)/n(S) =5/36 ---------(1)
Let B be the event of rolling doubles.
Sample space for B ={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}
n(B)=6;n(S)=36
P(B) = n(B)/n(S) = 6/36 --------------(2)
Aâ©B is the event of getting a sum of 6 and rolling doubles.
Therefore n(Aâ©B)=1
P(Aâ©B)=1/36 ------(3)
Multiplying equation (1) and (2)
(5/36)*(6/36)=5/216
but P(Aâ©B)=1/36
P(Aâ©B) ≠P(A)P(B)
Therefore, the events are not independent.</span>
