$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is

An experiment has $10$ equally likely outcomes. Let $\mathrm{A}$ and $\mathrm{B}$ be two non-empty events of the experiment. If $\mathrm{A}$ consists of $4$ outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent, is

Urn $A$ contains $6$ red and $4$ black balls and urn $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from urn $A$ and placed in urn $B$. Then one ball is drawn at random from urn $B$ and placed in urn $A$. If one ball is now drawn at random from urn $A$, the probability that it is found to be red, is

If $A$ and $B$ are two independent events, then $P\,(A + B) = $

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 $NSS$ but not $NCC$.

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