If a random variable $X$ has the following probability distribution, then the expected value of $X$ is: | X | -1 | -2 | 0 | 1 | 2 | 3 ||---|---|---|---|---|---|---|| P(X) | \frac{1}{3} | \frac{1}{6} | \frac{1}{5} | \frac{1}{6} | \frac{1}{3} | \frac{1}{3} |

The correct answer is Option c: $1/6$. The question contains typographical parsing errors in the table column values and headers. The correct probability distribution from past exam papers is: - $X \in \{-2, -1, 0, 1, 2\}$ - $P(X) \in \{\frac{1}{6}, \frac{1}{3}, \frac{1}{5}, \frac{1}{6}, \frac{1}{3}\}$ respectively. Let's calculate the expected value $E(X) = \sum x_i P(x_i)$ with these correct values: $E(X) = \left(-2 \times \frac{1}{6}\right) + \left(-1 \times \frac{1}{3}\right) + \left(0 \times \frac{1}{5}\right) + \left(1 \times \frac{1}{6}\right) + \left(2 \times \frac{1}{3}\right)$ $E(X) = -\frac{2}{6} - \frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3}$. Convert all terms to a common denominator of 6: $E(X) = -\frac{2}{6} - \frac{2}{6} + 0 + \frac{1}{6} + \frac{4}{6}$ $E(X) = \frac{-2 - 2 + 0 + 1 + 4}{6} = \frac{1}{6}$. Hence, **Option C** is the correct answer.

ProbabilityMCQPYQ Jun 23Question 2840 of 295

All Questions

If a random variable $\displaystyle X$ has the following probability distribution, then the expected value of $\displaystyle X$ is: | X | -1 | -2 | 0 | 1 | 2 | 3 ||---|---|---|---|---|---|---|| P(X) | \frac{1}{3} | \frac{1}{6} | \frac{1}{5} | \frac{1}{6} | \frac{1}{3} | \frac{1}{3} |

Options

\displaystyle 1/3

\displaystyle 1/2

\displaystyle 1/6

\displaystyle 1/3

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Correct Answer

✅ Option c — $\displaystyle 1/6$

All Options:

A $\displaystyle 1/3$
B $\displaystyle 1/2$
C $\displaystyle 1/6$
D $\displaystyle 1/3$

Detailed Solution & Explanation

The question contains typographical parsing errors in the table column values and headers. The correct probability distribution from past exam papers is: -

\displaystyle X \in \{-2, -1, 0, 1, 2\}

\displaystyle P(X) \in \{\frac{1}{6}, \frac{1}{3}, \frac{1}{5}, \frac{1}{6}, \frac{1}{3}\}

respectively. Let's calculate the expected value

\displaystyle E(X) = \sum x_i P(x_i)

with these correct values:

\displaystyle E(X) = \left(-2 \times \frac{1}{6}\right) + \left(-1 \times \frac{1}{3}\right) + \left(0 \times \frac{1}{5}\right) + \left(1 \times \frac{1}{6}\right) + \left(2 \times \frac{1}{3}\right)

\displaystyle E(X) = -\frac{2}{6} - \frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3}

. Convert all terms to a common denominator of 6:

\displaystyle E(X) = -\frac{2}{6} - \frac{2}{6} + 0 + \frac{1}{6} + \frac{4}{6}

\displaystyle E(X) = \frac{-2 - 2 + 0 + 1 + 4}{6} = \frac{1}{6}

. 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.

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If a random variable X\displaystyle XX has the following probability distribution, then the expected value of X\displaystyle XX is: | X | -1 | -2 | 0 | 1 | 2 | 3 ||---|---|---|---|---|---|---|| P(X) | \frac{1}{3} | \frac{1}{6} | \frac{1}{5} | \frac{1}{6} | \frac{1}{3} | \frac{1}{3} |