Remarks:

  • If P(A) = 0 => for event ‘B’, 0 ≤ P(A∩B) ≤ P(A).
    => P(A∩B) = 0, thus P(A∩B) = P(A)×P(B) = 0. Hence an impossible event would be independent of any other event.
  • Distinction between independent and mutually exclusive events must be carefully made.Independenceis a property of probability whereas the mutual exclusion is a set-theoretic concept. If A and B are two mutually exclusive and possible events of sample space ‘S’  then P(A) > 0, P(B) > 0 and P(A∩B) = 0 ≤ P(A)×P(B) so that A and B can’t be independent. Infact P(A/B) = 0 similarly P(B/A) = 0. Consequently, mutually exclusive events are strongly dependent.
  • Two events A and B are independent if and only if A and B’ are independent or A¢ and B are independent or A’ and B’ are independent.

We have P(A∩B)     = P(A)×P(B)

Now P(A∩B’)                        = P(A) – P(A∩B) = P(A) – P(A)×P(B) = P(A) (1-P(B)) = P(A)×P(B’)

Thus A and B’ are independent.

Similarly P(A’∩B)     = P(B) – P(A∩B) = P(B)×P(A’)

Finally P(A’ ∩ B’)     = 1 – P(A∪B) = 1 – P(A) – P(B) + P(A∩B)

= 1 – P(A) – P(B) + P(A)×P(B) = (1-P(A)) – P(B) (1-P(A))

= (1-P(A)) (1-P(B)) = P(A’)×P(B’)

Thus A’ and B’ are also independent.

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CBSE Class 12 Maths Probability All Topic Notes CBSE Class 12 Maths All Chapters Notes

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