ProbabilityMCQPYQ Dec. 21Question 2837 of 295

For a probability distribution, probability is given by, $\displaystyle P(X_j) = \frac{k}{x_j^2}$ , $\displaystyle x_j = 1, 2, \dots, 9$ . The value of $\displaystyle k$ is

Options

\displaystyle 55

\displaystyle 9

\displaystyle 45

\displaystyle 81

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — $\displaystyle 45$

All Options:

A $\displaystyle 55$
B $\displaystyle 9$
C $\displaystyle 45$
D $\displaystyle 81$

Detailed Solution & Explanation

The question contains a typographical parsing error in the expression for the probability distribution. The correct expression should be

\displaystyle P(X_j) = \frac{x_j}{k}

for

\displaystyle x_j = 1, 2, \dots, 9

. Using the fundamental property of probability distributions that the sum of all probabilities must equal 1:

\displaystyle \sum_{j=1}^{9} P(X_j) = 1

\displaystyle \sum_{j=1}^{9} \frac{x_j}{k} = 1 \implies \frac{1}{k} \sum_{j=1}^{9} j = 1

. The sum of the first 9 natural numbers is:

\displaystyle \sum_{j=1}^{9} j = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45

. Therefore:

\displaystyle \frac{45}{k} = 1 \implies k = 45

. Hence, **Option C** 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.

Ready to Master Probability?

Practice all 295 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

For a probability distribution, probability is given by, P(Xj)=kxj2\displaystyle P(X_j) = \frac{k}{x_j^2}P(Xj​)=xj2​k​, xj=1,2,…,9\displaystyle x_j = 1, 2, \dots, 9xj​=1,2,…,9. The value of k\displaystyle kk is