ProbabilityMCQMTP Dec 22 - Series IQuestion 3357 of 187
All Questions

Thirty balls are serially numbered and placed in bag. Find chance that the first ball drawn is a multiple of 3\displaystyle 3 or 5\displaystyle 5

Options

A8/15\displaystyle 8/15
B7/15\displaystyle 7/15
C1/2\displaystyle 1/2
D2/15\displaystyle 2/15
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 7/15\displaystyle 7/15

All Options:

  • A8/15\displaystyle 8/15
  • B7/15\displaystyle 7/15
  • C1/2\displaystyle 1/2
  • D2/15\displaystyle 2/15

Ad

Detailed Solution & Explanation

**Probability of Selecting a Multiple of 3 or 5** Total balls = 30 (numbered 1 to 30) Let A\displaystyle A be the event of selecting a multiple of 3. A={3,6,9,12,15,18,21,24,27,30}⇒n(A)=10A = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30\} \Rightarrow n(A) = 10 Let B\displaystyle B be the event of selecting a multiple of 5. B={5,10,15,20,25,30}⇒n(B)=6B = \{5, 10, 15, 20, 25, 30\} \Rightarrow n(B) = 6 Let A∩B\displaystyle A \cap B be the event of selecting a multiple of both 3 and 5 (i.e., a multiple of 15). A∩B={15,30}⇒n(A∩B)=2A \cap B = \{15, 30\} \Rightarrow n(A \cap B) = 2 By the Addition Theorem of Probability: n(A∪B)=n(A)+n(B)−n(A∩B)=10+6−2=14n(A \cup B) = n(A) + n(B) - n(A \cap B) = 10 + 6 - 2 = 14 Thus, the probability of selecting a multiple of 3 or 5 is: P(A∪B)=1430=715P(A \cup B) = \frac{14}{30} = \frac{7}{15} This matches Option B. (Note: The official answer key incorrectly lists Option D as correct, but mathematically the probability is 715\displaystyle \frac{7}{15}). 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.

More Questions from Probability

If P(A)=12\displaystyle P(A) = \frac{1}{2}, P(B)=13\displaystyle P(B) = \frac{1}{3}, P(A∩B)=14\displaystyle P(A \cap B) = \frac{1}{4}, then the value of P(A′∪B′)\displaystyle P(A' \cup B') is

Two dice are thrown simultaneously. Find the probability that the sum of digits on the two dice would be 8 or more.

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

A father had three sons namely, Kailash, Harish and Prakash. All are above 65 years in age. Prakash happens to be the eldest while Kailash as youngest. As per the health history, it is estimated that the probability that Kailash survives another 5 years is 45\displaystyle \frac{4}{5}, Harish survives another 5 years is 35\displaystyle \frac{3}{5} and Prakash survives another 5 years is 12\displaystyle \frac{1}{2}. The probabilities that Kailash and Harish survive another 5 years is 0.46, Harish and Prakash survive another 5 years is 0.32 and Kailash and Prakash survive another 5 years is 0.48. The probability that all three sons survive another 5 years is 0.26. What shall be the probability that at least one of them survives another 5 years?

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is

Two cards are drawn from a pack of 52 cards. The probability that one is a spade and one is a heart; is

Ready to Master Probability?

Practice all 187 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free