An unbiased die is thrown twice. Let the event $A$ be 'odd number on the first throw' and $B$ the event 'odd number on the second throw '. Check the independence of the events $A$ and $B$.
If all the $36$ elementary events of the experiment are considered to be equally likely, we have
$P(A)=\frac{18}{36}=\frac{1}{2}$ and $P(B)=\frac{18}{36}=\frac{1}{2}$
Also $P(A \cap B)=P($ odd number on both throws $)$
$=\frac{9}{36}=\frac{1}{4}$
Now $\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
Clearly $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B})$
Thus, $A$ and $B$ are independent events
For any two independent events ${E_1}$ and ${E_2},$ $P\,\{ ({E_1} \cup {E_2}) \cap ({\bar E_1} \cap {\bar E_2})\} $ is
If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is
If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are independent, then $x =$
Three persons $P, Q$ and $R$ independently try to hit a target . If the probabilities of their hitting the target are $\frac{3}{4},\frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by $P$ or $Q$ but not by $R$ is
In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.