A number is chosen from the first 100 natural | vedclass

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(D) Let $S$ be the set of the first $100$ natural numbers,so $n(S) = 100$.Let $A$ be the event that the number is even. The even numbers are $2, 4, 6,

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(D) Let $S$ be the set of the first $100$ natural numbers,so $n(S) = 100$.Let $A$ be the event that the number is even. The even numbers are $2, 4, 6, \dots, 100$. Thus,$n(A) = 50$.Let $B$ be the event that the number is divisible by $5$. The numbers divisible by $5$ are $5, 10, 15, \dots, 100$. Thus,$n(B) = 20$.The intersection $A \cap B$ represents numbers that are both even and divisible by $5$,which means they are divisible by $10$. These are $10, 20, 30, \dots, 100$. Thus,$n(A \cap B) = 10$.Using the inclusion-exclusion principle,the number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 50 + 20 - 10 = 60$.The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{60}{100} = \frac{3}{5}$.

$A$ number is chosen from the first $100$ natural numbers. The probability that the number is even or divisible by $5$ is:

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