ProbabilityPYQ Jan 26Question 4290 of 187
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Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Options

A920\displaystyle \frac{9}{20}
B12\displaystyle \frac{1}{2}
C25\displaystyle \frac{2}{5}
D815\displaystyle \frac{8}{15}
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Correct Answer

✅ Option a — 920\displaystyle \frac{9}{20}

All Options:

  • A920\displaystyle \frac{9}{20}
  • B12\displaystyle \frac{1}{2}
  • C25\displaystyle \frac{2}{5}
  • D815\displaystyle \frac{8}{15}

Detailed Solution & Explanation

The sample space S\displaystyle S consists of tickets numbered from 1\displaystyle 1 to 20\displaystyle 20: S={1,2,3,…,20}  ⟹  n(S)=20S = \{1, 2, 3, \dots, 20\} \implies n(S) = 20
Let A\displaystyle A be the event of drawing a ticket that is a multiple of 3\displaystyle 3: A={3,6,9,12,15,18}  ⟹  n(A)=6A = \{3, 6, 9, 12, 15, 18\} \implies n(A) = 6
Let B\displaystyle B be the event of drawing a ticket that is a multiple of 5\displaystyle 5: B={5,10,15,20}  ⟹  n(B)=4B = \{5, 10, 15, 20\} \implies n(B) = 4
The event A∩B\displaystyle A \cap B represents drawing a ticket that is a multiple of both 3\displaystyle 3 and 5\displaystyle 5 (which is a multiple of 15\displaystyle 15): A∩B={15}  ⟹  n(A∩B)=1A \cap B = \{15\} \implies n(A \cap B) = 1
Using the addition theorem of probability, the number of favorable outcomes for a multiple of 3\displaystyle 3 or 5\displaystyle 5 is: n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) n(A∪B)=6+4−1=9n(A \cup B) = 6 + 4 - 1 = 9
Thus, the probability of drawing a ticket that is a multiple of 3\displaystyle 3 or 5\displaystyle 5 is: P(A∪B)=n(A∪B)n(S)=920P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{9}{20} Hence, **Option A** 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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