MCQ
If an integer is chosen at random from first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$, is
- A$\frac{{41}}{{100}}$
- ✓$\frac{{33}}{{100}}$
- C$\frac{1}{{10}}$
- DNone of these
So, $P(A) = \frac{{25}}{{100}},$ $P(B) = \frac{{16}}{{100}}$ and $P(A \cap B) = \frac{8}{{100}}$
Thus required probability is
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$ \Rightarrow P(A \cup B) = \frac{{25}}{{100}} + \frac{{16}}{{100}} - \frac{8}{{100}} = \frac{{33}}{{100}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $-2$:$\;\mathop \sum \limits_{r = 0}^n \left( {r + 1} \right)\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right){x^r}\; = {\left( {1 + x} \right)^n} + nx{\left( {1 + x} \right)^{n - 1}}$