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 $B$.
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
Similar Questions
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
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
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
- [IIT 1991]
Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.
Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $
If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $