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Question

(A) Let $A$ be the event that the chosen integer is divisible by $6$,and $B$ be the event that the chosen integer is divisible by $8$.The total number

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) Let $A$ be the event that the chosen integer is divisible by $6$,and $B$ be the event that the chosen integer is divisible by $8$.The total number of outcomes is $n(S) = 200$.The number of integers divisible by $6$ is $n(A) = \lfloor \frac{200}{6} floor = 33$.The number of integers divisible by $8$ is $n(B) = \lfloor \frac{200}{8} floor = 25$.The integers divisible by both $6$ and $8$ are divisible by $	ext{lcm}(6, 8) = 24$. The number of such integers is $n(A \cap B) = \lfloor \frac{200}{24} floor = 8$.Using the inclusion-exclusion principle,the number of integers divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 25 - 8 = 50$.The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{50}{200} = \frac{1}{4}$.

An integer is chosen at random from the first $200$ natural numbers. What is the probability that the integer chosen is divisible by $6$ or $8$?

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