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

If $A, B, C$ are three events associated with a random experiment, prove that

$P ( A \cup B \cup C ) $ $= P ( A )+ P ( B )+ P ( C )- $ $P ( A \cap B )- P ( A \cap C ) $ $- P ( B \cap C )+ $ $P ( A \cap B \cap C )$

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 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

Two aeroplanes $I$ and $II$ bomb a target in succession. The probabilities of $l$ and $II$ scoring a hit correctlyare $0.3$ and $0.2,$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

- [AIEEE 2007]

Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.4$, $P ( A \cap B )=0.8$