A die marked $1,\,2,\,3$ in red and $4,\,5,\,6$ in green is tossed. Let $A$ be the event, $'$ the number is even,$'$ and $B$ be the event, 'the number is red'. Are $A$ and $B$ independent?
When a die is thrown, the sample space ( $S$ ) is
$\mathrm{S}=\{1,2,3,4,5,6\}$
Let $A:$ the number is even $=\{2,4,6\}$
$\Rightarrow P(A)=\frac{3}{6}=\frac{1}{2}$
$B:$ the number is red $=\{1,2,3\}$
$\Rightarrow P(B)=\frac{3}{6}=\frac{1}{2}$
$\therefore $ $A \cap B=\{2\}$
$P(A B)=P(A \cap B)=\frac{1}{6}$
$P(A) P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6}$
$\Rightarrow $ $P(A) \cdot P(B) \neq P(A B)$
Therefore, $A$ bad $B$ are not independent.
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$
$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $
Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and $\mathrm{P}$ $($ not $ \mathrm{A}$ or not $\mathrm{B})=\frac{1}{4} .$ State whether $\mathrm{A}$ and $\mathrm{B}$ are independent?
$\mathrm{A}$ die is thrown. If $\mathrm{E}$ is the event $'$ the number appearing is a multiple of $3'$ and $F$ be the event $'$ the number appearing is even $^{\prime}$ then find whether $E$ and $F$ are independent ?
If the probability of a horse $A$ winning a race is $1/4$ and the probability of a horse $B$ winning the same race is $1/5$, then the probability that either of them will win the race is