The probability of choosing at random a numbe | vedclass

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(D) Let $A$ be the set of numbers divisible by $6$ and $B$ be the set of numbers divisible by $8$ in the range $1$ to $90$.Total numbers $n(S) = 90$.N

Vedclass · Accepted Answer

$\frac{23}{90}$

Vedclass · Answer

(D) Let $A$ be the set of numbers divisible by $6$ and $B$ be the set of numbers divisible by $8$ in the range $1$ to $90$.Total numbers $n(S) = 90$.Number of elements divisible by $6$ is $n(A) = \lfloor \frac{90}{6} floor = 15$.Number of elements divisible by $8$ is $n(B) = \lfloor \frac{90}{8} floor = 11$.Numbers divisible by both $6$ and $8$ are divisible by $	ext{lcm}(6, 8) = 24$.Number of elements divisible by $24$ is $n(A \cap B) = \lfloor \frac{90}{24} floor = 3$.Using the inclusion-exclusion principle,the number of elements divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 15 + 11 - 3 = 23$.The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{23}{90}$.

The probability of choosing at random a number that is divisible by $6$ or $8$ from among $1$ to $90$ is equal to

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