Let A , B and C be three events such that Pleft( {A cap bar B cap bar C} right) = 0.6 ,

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14.Probability

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Let $A$,$B$ and $C$ be three events such that $P\left( {A \cap \bar B \cap \bar C} \right) = 0.6$, $P\left( A \right) = 0.8$ and $P\left( {\bar A \cap B \cap C} \right) = 0.1$, then the value of $P$(atleast two among $A$,$B$ and $C$ ) equals

A

$0.1$

B

$0.2$

C

$0.3$

D

$0.5$

Solution

$P$ (at least two among $ \mathrm{A}, \mathrm{B} $ and$\mathrm{C}$)

$=0.8+0.1-0.6=0.3$

Standard 11

Mathematics

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