ProbabilityPYQ May 25Question 4080 of 187

A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7.

Options

\displaystyle \frac{7}{20}

\displaystyle \frac{12}{37}

\displaystyle \frac{24}{67}

\displaystyle \frac{8}{20}

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Correct Answer

✅ Option d — $\displaystyle \frac{8}{20}$

All Options:

A $\displaystyle \frac{7}{20}$
B $\displaystyle \frac{12}{37}$
C $\displaystyle \frac{24}{67}$
D $\displaystyle \frac{8}{20}$

Detailed Solution & Explanation

Let the sample space be the first 20 natural numbers:

S = \{1, 2, 3, \dots, 20\} \implies N(S) = 20

Let

\displaystyle A

be the event that the selected number is divisible by 3:

A = \{3, 6, 9, 12, 15, 18\} \implies n(A) = 6

Let

\displaystyle B

be the event that the selected number is divisible by 7:

B = \{7, 14\} \implies n(B) = 2

Let

\displaystyle A \cap B

be the event that the selected number is divisible by both 3 and 7 (i.e., divisible by 21). Since the largest number is 20, there are no such numbers in

\displaystyle S

A \cap B = \emptyset \implies n(A \cap B) = 0

We want to find the probability of the number being divisible by 3 or 7 (

\displaystyle A \cup B

). By the addition theorem of probability:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cup B) = \frac{n(A)}{N(S)} + \frac{n(B)}{N(S)} - \frac{n(A \cap B)}{N(S)}

P(A \cup B) = \frac{6}{20} + \frac{2}{20} - \frac{0}{20} = \frac{8}{20}

Hence, **Option D** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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