If an integer is chosen at random from first | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{{33}}{{100}}$

Vedclass · Answer

(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}}$.

$P(A)$	$P(B)$	$P(A \cap B)$	$P (A \cup B)$
$\frac {1}{3}$	$\frac {1}{5}$	$\frac {1}{15}$	........

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

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