ProbabilityMTP Dec 22 - Series 1Question 2808 of 295

Let P be a probability function on $\displaystyle S = \{X1, X2, X3\}$ If $\displaystyle P(X1)=\frac{1}{4}$ and $\displaystyle P(X3) = \frac{1}{3}$ then $\displaystyle P(X2)$ is

Options

\displaystyle \frac{5}{12}

\displaystyle \frac{7}{12}

\displaystyle \frac{3}{4}

DNone of these

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Correct Answer

✅ Option a — $\displaystyle \frac{5}{12}$

All Options:

A $\displaystyle \frac{5}{12}$
B $\displaystyle \frac{7}{12}$
C $\displaystyle \frac{3}{4}$
DNone of these

Detailed Solution & Explanation

Given that

\displaystyle P

is a probability function defined on the sample space

\displaystyle S = \{X_1, X_2, X_3\}

containing mutually exclusive and exhaustive simple events: 1. The sum of the probabilities of all outcomes in a sample space must equal 1:

P(X_1) + P(X_2) + P(X_3) = 1

2. Substitute the given values

\displaystyle P(X_1) = \frac{1}{4}

and

\displaystyle P(X_3) = \frac{1}{3}

\frac{1}{4} + P(X_2) + \frac{1}{3} = 1

3. Combine the fractions:

\frac{3}{12} + P(X_2) + \frac{4}{12} = 1

\frac{7}{12} + P(X_2) = 1

4. Solve for

\displaystyle P(X_2)

P(X_2) = 1 - \frac{7}{12} = \frac{5}{12}

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.

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Let P be a probability function on S={X1,X2,X3}\displaystyle S = \{X1, X2, X3\}S={X1,X2,X3} If P(X1)=14\displaystyle P(X1)=\frac{1}{4}P(X1)=41​ and P(X3)=13\displaystyle P(X3) = \frac{1}{3}P(X3)=31​ then P(X2)\displaystyle P(X2)P(X2) is