ProbabilityMCQMTP May 20Question 3352 of 187
All Questions

Find the probability that a four-digit number comprising the digits $2, 5, 6and\displaystyle and7$ would be divisible by $4$.

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 1/3\displaystyle 1/3

All Options:

  • A1/4\displaystyle 1/4
  • B1/3\displaystyle 1/3
  • C1/2\displaystyle 1/2
  • D1\displaystyle 1

Detailed Solution & Explanation

**Probability of a 4-Digit Number Being Divisible by 4** Given digits: $2, 5, 6, 7$ (no digit is repeated). **Step 1: Total possible 4-digit numbers** The number of permutations of 4 distinct digits is: Total outcomes=4!=24Total outcomes=4\displaystyle \text{Total outcomes} = 4! = 24 **Step 2: Divisibility by 4 rule** A number is divisible by 4 if the last two digits form a number divisible by 4. Let's find the favorable two-digit combinations from {2,5,6,7}\displaystyle \{2, 5, 6, 7\}: - $52\displaystyle ($52 = 4 ×\displaystyle \times 13, Yes) - $56\displaystyle ($56 = 4 ×\displaystyle \times 14, Yes) - $72\displaystyle ($72 = 4 ×\displaystyle \times 18, Yes) - $76\displaystyle ($76 = 4 ×\displaystyle \times 19, Yes) There are 4 favorable last-two-digit pairs. **Step 3: Calculating favorable outcomes** For each pair, the remaining 2 digits can be arranged in the first two positions in $2! = 2$ ways. Favorable outcomes=4×2=8Favorable outcomes=4\displaystyle \text{Favorable outcomes} = 4 ×\displaystyle \times 2 = 8 **Step 4: Probability calculation** P=824=13P = 824=13\displaystyle \frac{8}{24} = \frac{1}{3} 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

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.

More Questions from Probability

If P(A)=12\displaystyle P(A) = 12\displaystyle \frac{1}{2}, P(B)=13\displaystyle P(B) = 13\displaystyle \frac{1}{3}, P(A∩B)=14\displaystyle P(A \cap B) = 14\displaystyle \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 45\displaystyle \frac{4}{5}, Harish survives another 5 years is 35\displaystyle 35\displaystyle \frac{3}{5} and Prakash survives another 5 years is 12\displaystyle 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