If $A$ and $B$ are arbitrary events, then
$P(A \cap B) \ge P(A) + P(B)$
$P(A \cup B) \le P(A) + P(B)$
$P(A \cap B) = P(A) + P(B)$
None of these
If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is
Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, $B$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then
Probability that a student will succeed in $IIT$ entrance test is $0.2$ and that he will succeed in Roorkee entrance test is $0.5$. If the probability that he will be successful at both the places is $0.3$, then the probability that he does not succeed at both the places is
Let $A$ and $B$ be events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals
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 $NSS$.