A number is selected at random from the set 1, 2, ..., 99 . The probability that it is divisible by 9 or 11 is

Option b: 19/99. Solution: Probability of Divisibility by 9 or 11 Total numbers in the set 1, 2, ..., 99 is: n(S) = 99 Let A be the event that the number is divisible by 9: n(A) = 99/9 = 11 Let B be the event that the number is divisible by 11: n(B) = 99/11 = 9 The event A B represents numbers divisible by both 9 and 11 (i.e., divisible by 99): n(A B) = 99/99 = 1 Using the Addition Theorem of Probability: n(A B) = n(A) + n(B) - n(A B) = 11 + 9 - 1 = 19 Thus, the probability of selecting a number divisible by 9 or 11 is: P(A B) = 19/99 Hence, Option B is the correct answer.

A number is selected at random from the set ${1, 2, ..., 99}$. The probability that it is divisible by $9$ or $11$ is

The correct answer is Option b: $19/99$. **Probability of Divisibility by 9 or 11** Total numbers in the set $\{1, 2, \dots, 99\}$ is: $$n(S) = 99$$ Let $A$ be the event that the number is divisible by 9: $$n(A) = \lfloor \frac{99}{9} \rfloor = 11$$ Let $B$ be the event that the number is divisible by 11: $$n(B) = \lfloor \frac{99}{11} \rfloor = 9$$ The event $A \cap B$ represents numbers divisible by both 9 and 11 (i.e., divisible by 99): $$n(A \cap B) = \lfloor \frac{99}{99} \rfloor = 1$$ Using the Addition Theorem of Probability: $$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 11 + 9 - 1 = 19$$ Thus, the probability of selecting a number divisible by 9 or 11 is: $$P(A \cup B) = \frac{19}{99}$$ Hence, **Option B** is the correct answer.

ProbabilityMCQMTP Sep 24 Series IQuestion 3374 of 187

All Questions

A number is selected at random from the set $\displaystyle {1, 2, ..., 99}$ . The probability that it is divisible by $\displaystyle 9$ or $\displaystyle 11$ is

Options

\displaystyle 19/100

\displaystyle 19/99

\displaystyle 10/100

\displaystyle 10/99

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Correct Answer

✅ Option b — $\displaystyle 19/99$

All Options:

A $\displaystyle 19/100$
B $\displaystyle 19/99$
C $\displaystyle 10/100$
D $\displaystyle 10/99$

Detailed Solution & Explanation

**Probability of Divisibility by 9 or 11** Total numbers in the set

\displaystyle \{1, 2, \dots, 99\}

is:

n(S) = 99

Let

\displaystyle A

be the event that the number is divisible by 9:

n(A) = \lfloor \frac{99}{9} \rfloor = 11

Let

\displaystyle B

be the event that the number is divisible by 11:

n(B) = \lfloor \frac{99}{11} \rfloor = 9

The event

\displaystyle A \cap B

represents numbers divisible by both 9 and 11 (i.e., divisible by 99):

n(A \cap B) = \lfloor \frac{99}{99} \rfloor = 1

Using the Addition Theorem of Probability:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 11 + 9 - 1 = 19

Thus, the probability of selecting a number divisible by 9 or 11 is:

P(A \cup B) = \frac{19}{99}

Hence, **Option B** 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

Set

A well-defined collection of distinct objects (called elements or members). Sets are denoted by capital letters and elements are listed within curly braces. Key operations include Union (∪), Intersection (∩), and Complement.

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 set 1,2,...,99\displaystyle {1, 2, ..., 99}1,2,...,99. The probability that it is divisible by 9\displaystyle 99 or 11\displaystyle 1111 is