ProbabilityMCQPYQ June 24Question 2843 of 295
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The following table gives the cumulative probability function of X\displaystyle X: | Xi | Pr(X) ||---|---|| 0 | 630\displaystyle \frac{6}{30} || 1 | 930\displaystyle \frac{9}{30} || 2 | 1330\displaystyle \frac{13}{30} || 3 | 115\displaystyle \frac{1}{15} || 4 | 110\displaystyle \frac{1}{10} || 5 | 130\displaystyle \frac{1}{30} | The expectation of X\displaystyle X is ______

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — $1.8$

All Options:

  • A$1.8$
  • B$1.7$
  • C$1.5$
  • D$1.6$

Detailed Solution & Explanation

The question contains a typographical parsing error in the table. The value of P(X=1)\displaystyle P(X=1) is given as 930\displaystyle 930\displaystyle \frac{9}{30} instead of 530\displaystyle 530\displaystyle \frac{5}{30}. With the correct probability mass function, the sum of probabilities is 630+530+1330+230+330+130=1\displaystyle 630+530+1330+230+330+130=1\displaystyle \frac{6}{30} + \frac{5}{30} + \frac{13}{30} + \frac{2}{30} + \frac{3}{30} + \frac{1}{30} = 1. Let's calculate the expectation of X\displaystyle X, E(X)=∑xiP(X=xi)\displaystyle E(X) = ∑\displaystyle \sum x_i P(X = x_i): E(X)=(0×630)+(1×530)+(2×1330)+(3×115)+(4×110)+(5×130)\displaystyle E(X) = \left(0 ×\displaystyle \times 630\displaystyle \frac{6}{30}\right) + \left(1 ×\displaystyle \times 530\displaystyle \frac{5}{30}\right) + \left(2 ×\displaystyle \times 1330\displaystyle \frac{13}{30}\right) + \left(3 ×\displaystyle \times 115\displaystyle \frac{1}{15}\right) + \left(4 ×\displaystyle \times 110\displaystyle \frac{1}{10}\right) + \left(5 ×\displaystyle \times 130\displaystyle \frac{1}{30}\right) Convert all probabilities to have a denominator of 30: - P(0)=630\displaystyle P(0) = 630\displaystyle \frac{6}{30} - P(1)=530\displaystyle P(1) = 530\displaystyle \frac{5}{30} - P(2)=1330\displaystyle P(2) = 1330\displaystyle \frac{13}{30} - P(3)=115=230\displaystyle P(3) = 115=230\displaystyle \frac{1}{15} = \frac{2}{30} - P(4)=110=330\displaystyle P(4) = 110=330\displaystyle \frac{1}{10} = \frac{3}{30} - P(5)=130\displaystyle P(5) = 130\displaystyle \frac{1}{30} Now compute E(X)\displaystyle E(X): E(X)=0+530+2630+630+1230+530\displaystyle E(X) = 0 + 530+2630+630+1230+530\displaystyle \frac{5}{30} + \frac{26}{30} + \frac{6}{30} + \frac{12}{30} + \frac{5}{30} E(X)=5+26+6+12+530=5430=1.8\displaystyle E(X) = 5+26+6+12+530=5430=1\displaystyle \frac{5 + 26 + 6 + 12 + 5}{30} = \frac{54}{30} = 1.8. 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.

Key Concepts to Understand

Function

A relation from a set A (Domain) to a set B (Codomain) such that every element in A is associated with exactly one element in B. Written as f: A→B. Types include Linear, Quadratic, Polynomial, and Exponential functions.

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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