If P,({A₁} cup {A₂}) = 1 - P(A₁^c),P(A₂^c) where c stands for complement, then events {A₁}

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14.Probability

easy

If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are

A

Mutually exclusive

B

Independent

C

Equally likely

D

None of these

Solution

(b) $P({A_1} \cup {A_2}) = 1 – [1 – P({A_1})]\,[1 – P({A_2})]$

$ = P({A_1}) + P({A_2}) – P({A_1})\,.\,P({A_2})$.

Standard 11

Mathematics

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