- Home
- Standard 11
- Mathematics
14.Probability
normal
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