A number is selected at random from the first 100 natural numbers. What is the probability that it would be a multiple of 3 or 7 ?

Option d: 43/100. Solution: Probability of Selecting a Multiple of 3 or 7 Total numbers = 100 (from 1 to 100) Let A be the event of selecting a multiple of 3. n(A) = 100/3 = 33 Let B be the event of selecting a multiple of 7. n(B) = 100/7 = 14 Let A B be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21). n(A B) = 100/21 = 4 By the Addition Theorem of Probability: n(A B) = n(A) + n(B) - n(A B) = 33 + 14 - 4 = 43 Thus, the probability is: P(A B) = n(A B)/100 = 43/100 Hence, Option D is the correct answer.

A number is selected at random from the first $100$ natural numbers. What is the probability that it would be a multiple of $3$ or $7$?

The correct answer is Option d: $43/100$. **Probability of Selecting a Multiple of 3 or 7** Total numbers = 100 (from 1 to 100) Let $A$ be the event of selecting a multiple of 3. $$n(A) = \lfloor \frac{100}{3} \rfloor = 33$$ Let $B$ be the event of selecting a multiple of 7. $$n(B) = \lfloor \frac{100}{7} \rfloor = 14$$ Let $A \cap B$ be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21). $$n(A \cap B) = \lfloor \frac{100}{21} \rfloor = 4$$ By the Addition Theorem of Probability: $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 14 - 4 = 43$$ Thus, the probability is: $$P(A \cup B) = \frac{n(A \cup B)}{100} = \frac{43}{100}$$ Hence, **Option D** is the correct answer.

ProbabilityMCQPYQ Dec 23Question 3339 of 187

All Questions

A number is selected at random from the first $\displaystyle 100$ natural numbers. What is the probability that it would be a multiple of $\displaystyle 3$ or $\displaystyle 7$ ?

Options

\displaystyle 33/100

\displaystyle 100/3

\displaystyle 21/100

\displaystyle 43/100

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Correct Answer

✅ Option d — $\displaystyle 43/100$

All Options:

A $\displaystyle 33/100$
B $\displaystyle 100/3$
C $\displaystyle 21/100$
D $\displaystyle 43/100$

Detailed Solution & Explanation

**Probability of Selecting a Multiple of 3 or 7** Total numbers = 100 (from 1 to 100) Let

\displaystyle A

be the event of selecting a multiple of 3.

n(A) = \lfloor \frac{100}{3} \rfloor = 33

Let

\displaystyle B

be the event of selecting a multiple of 7.

n(B) = \lfloor \frac{100}{7} \rfloor = 14

Let

\displaystyle A \cap B

be the event of selecting a multiple of both 3 and 7 (i.e., a multiple of 21).

n(A \cap B) = \lfloor \frac{100}{21} \rfloor = 4

By the Addition Theorem of Probability:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 14 - 4 = 43

Thus, the probability is:

P(A \cup B) = \frac{n(A \cup B)}{100} = \frac{43}{100}

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.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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A number is selected at random from the first 100\displaystyle 100100 natural numbers. What is the probability that it would be a multiple of 3\displaystyle 33 or 7\displaystyle 77?