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14.Probability
medium
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
$\frac{{41}}{{100}}$
B
$\frac{{33}}{{100}}$
C
$\frac{1}{{10}}$
D
None of these
Solution
(b) Let $A$ be the event to be multiple of $4$ and $B$ be the event to be multiple of $6$
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}}$.
Standard 11
Mathematics
