If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $
$P(\bar A)\,\,\,P(\bar B)$
$1 - P(A) - P(B)$
$P(A) + P(B) - P(A \cap B)$
$P(B) - P(A \cap B)$
(d) It is a fundamental concept.
If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is
If $A$ and $B$ are two events such that $P\,(A \cup B)\, + P\,(A \cap B) = \frac{7}{8}$ and $P\,(A) = 2\,P\,(B),$ then $P\,(A) = $
The odds against a certain event is $5 : 2$ and the odds in favour of another event is $6 : 5$. If both the events are independent, then the probability that at least one of the events will happen is
Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random, then the probability of the number to be divisible by $2$ or $3$, is
A die is thrown. Let $A$ be the event that the number obtained is greater than $3.$ Let $B$ be the event that the number obtained is less than $5.$ Then $P\left( {A \cup B} \right)$ is
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