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(B) Let $A$ be the event that the chosen number is a multiple of $4$,and $B$ be the event that the chosen number is a multiple of $6$.The number of mu

Vedclass · Accepted Answer

$\frac{33}{100}$

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(B) Let $A$ be the event that the chosen number is a multiple of $4$,and $B$ be the event that the chosen number is a multiple of $6$.The number of multiples of $4$ in the first $100$ integers is $\frac{100}{4} = 25$. So,$P(A) = \frac{25}{100}$.The number of multiples of $6$ in the first $100$ integers is $\lfloor \frac{100}{6} floor = 16$. So,$P(B) = \frac{16}{100}$.The multiples of both $4$ and $6$ are multiples of $	ext{lcm}(4, 6) = 12$. The number of multiples of $12$ in the first $100$ integers is $\lfloor \frac{100}{12} floor = 8$. So,$P(A \cap B) = \frac{8}{100}$.Using the addition theorem of probability,the required probability is:$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$P(A \cup B) = \frac{25}{100} + \frac{16}{100} - \frac{8}{100} = \frac{33}{100}$.

If an integer is chosen at random from the first $100$ positive integers,then the probability that the chosen number is a multiple of $4$ or $6$ is:

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