Probability and Random Events Practice Questions for XAT: 93+ Solved Questions with Step-by-Step Solutions

Solve 93+ Probability and Random Events practice questions for XAT with answers and detailed solutions. Free sample questions below.

Random Experiments and Sample Space

Random Experiment

An action with well-defined, uncertain outcomes.

Sample Space ( $S$ )

The set of all possible outcomes.

Event ( $E$ )

A subset of the sample space.

Example: Tossing two coins

$S = {H H, H T, T H, T T}$

Event $E$ (getting exactly one head) $= {H T, T H}$

Classical Definition of Probability

P (E) = \frac{Number of favorable outcomes n ( E )}{Total number of outcomes n ( S )}

Condition: All outcomes in the sample space must be equally likely.

Example: Rolling a fair six-sided die. Find P(prime number).

Total outcomes

n (S) = 6

Favorable outcomes (primes: 2, 3, 5)

n (E) = 3

P (E) = \frac{3}{6} = \frac{1}{2}

Probability and Random Events: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Let $S$ be the set of all permutations of the digits ${1, 2, 3, 4, 5, 6}$ . A permutation is chosen uniformly at random from $S$ .

Let $A$ be the event that the digit 1 appears before the digit 2.

Let $B$ be the event that the digit 3 appears before the digit 4.

Let $C$ be the event that the digit 5 appears before the digit 6.

What is the probability that exactly two of these events occur?

A.
1/4
B.
3/8
C.
1/2
D.
5/8

Correct Answer:

Step-by-Step Solution

Key idea: Symmetry and Independence of relative orderings.

Step 1: Analyze individual probabilities.

In any random permutation, for any distinct pair of digits (e.g., 1 and 2), the probability that one appears before the other is $1/2$ .

So, $P (A) = 1/2$ , $P (B) = 1/2$ , $P (C) = 1/2$ .

Step 2: Analyze independence.

The relative ordering of disjoint pairs (1,2), (3,4), and (5,6) are mutually independent events.

Why? The positions of 1 and 2 are symmetric with respect to 3 and 4. There is no bias introduced by the relative order of 3 and 4 on the relative order of 1 and 2.

Thus, $A, B, C$ are independent events.

Step 3: Define the target event.

We want exactly two events to occur.

Possible scenarios:

$A \cap B \cap C^{c}$
$A \cap B^{c} \cap C$
$A^{c} \cap B \cap C$

Step 4: Calculate probability for one scenario.

$P (A \cap B \cap C^{c}) = P (A) P (B) P (C^{c})$ (due to independence)

$= (1/2) \times (1/2) \times (1/2) = 1/8$ .

Step 5: Sum the probabilities.

Since the scenarios are mutually exclusive:

$P (Exactly 2) = 1/8 + 1/8 + 1/8 = 3/8$ .

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

A box contains 4 Red, 4 Green, and 4 Blue balls. Three balls are drawn simultaneously at random.

Let $X$ be the number of different colors among the drawn balls.

What is the expected value of $X$ ?

Correct Answer:

2.2

Step-by-Step Solution

Key idea: Linearity of Expectation with Indicator Variables.

Step 1: Define Indicator Variables.

Let $I_{R}$ be 1 if at least one Red ball is drawn, 0 otherwise.

Let $I_{G}$ be 1 if at least one Green ball is drawn, 0 otherwise.

Let $I_{B}$ be 1 if at least one Blue ball is drawn, 0 otherwise.

Then $X = I_{R} + I_{G} + I_{B}$ .

By Linearity of Expectation:

$E [X] = E [I_{R}] + E [I_{G}] + E [I_{B}]$ .

Step 2: Calculate $E [I_{R}]$ .

$E [I_{R}] = P (I_{R} = 1) = P (At least one Red)$ .

$P (At least one Red) = 1 - P (No Red)$ .

Total balls = 12. Draw 3.

Total ways = $(3 12) = \frac{12 \times 11 \times 10}{6} = 220$ .

Ways to draw No Red (only from 4G, 4B = 8 balls):

$(3 8) = \frac{8 \times 7 \times 6}{6} = 56$ .

$P (No Red) = \frac{56}{220}$ .

$P (At least one Red) = 1 - \frac{56}{220} = \frac{164}{220}$ .

Step 3: Symmetry.

By symmetry, $E [I_{G}] = E [I_{B}] = E [I_{R}] = \frac{164}{220}$ .

Step 4: Sum.

$E [X] = 3 \times \frac{164}{220}$ .

Simplify fraction:

$164/220$ . Divide by 4: $41/55$ .

$E [X] = 3 \times \frac{41}{55} = \frac{123}{55}$ .

Calculate decimal:

$123 \div 55 \approx 2.236$ .

Wait, let's re-calculate $123/55$ .

$55 \times 2 = 110$ . Remainder 13.

$130/55 = 2.36...$

$123/55 = 2.23636...$

Is the answer a clean decimal?

Let's check the calculation of $(3 12)$ .

$12 \times 11 \times 10/6 = 2 \times 11 \times 10 = 220$ . Correct.

$(3 8) = 56$ . Correct.

$1 - 56/220 = 164/220$ . Correct.

$164/4 = 41$ . $220/4 = 55$ . Correct.

$3 \times 41 = 123$ .

$123/55$ .

The question asks for NAT. Usually 2 decimal places.

Answer: 2.24.

Let's double check if "Expected Value" implies a fraction or decimal. NAT usually accepts decimals.

Let's try to verify with direct counting.

Possible values for X: 1, 2, 3.

P(X=1): All same color.

3 Reds: $(3 4) = 4$ .

3 Greens: 4.

3 Blues: 4.

Total 12.

$P (X = 1) = 12/220$ .

P(X=3): All different colors (1R, 1G, 1B).

$4 \times 4 \times 4 = 64$ .

$P (X = 3) = 64/220$ .

P(X=2): 1 - P(X=1) - P(X=3) = $1 - 12/220 - 64/220 = 144/220$ .

Check sum: $12 + 64 + 144 = 220$ . Correct.

$E [X] = 1 \times \frac{12}{220} + 2 \times \frac{144}{220} + 3 \times \frac{64}{220}$ .

Numerator: $12 + 288 + 192 = 492$ .

$E [X] = \frac{492}{220}$ .

Divide by 4: $\frac{123}{55}$ .

Matches.

$123/55 = 2.23636...$

Rounding to 2 decimal places: 2.24.

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A frog is positioned at the center square of a $3 \times 3$ grid. At each step, the frog jumps to one of the adjacent squares (sharing an edge) with equal probability. If the frog reaches any corner square, it stops jumping.

What is the probability that the frog stops at a specific corner square (e.g., top-left) given that it started at the center?

A.
1/4
B.
1/8
C.
1/6
D.
1/12

Correct Answer:

Step-by-Step Solution

Key idea: This is a random walk problem with absorbing states. We can use symmetry and systems of linear equations for probabilities.

Step 1: Define States.

Let $P_{C}$ be the probability of stopping at the specific target corner (say, Top-Left, TL) starting from the Center.

Let $P_{E}$ be the probability of stopping at TL starting from an Edge center (e.g., Top-Middle).

Let $P_{O}$ be the probability of stopping at TL starting from an Other corner (non-target).

Note: By symmetry, all 4 edge centers are equivalent with respect to the target TL? No.

The grid has rotational symmetry, but fixing a target breaks it.

Let's classify positions relative to the Target Corner (TL):

Target (TL): Absorbing. Prob = 1.
Adjacent Corners (TR, BL): Absorbing. Prob = 0 (since we stop there, and it's not TL).
Opposite Corner (BR): Absorbing. Prob = 0.
Center (C): Start position.
Edges:

$E_{1}$ : Top-Middle (adjacent to TL).
$E_{2}$ : Left-Middle (adjacent to TL).
$E_{3}$ : Right-Middle (adjacent to TR, BR).
$E_{4}$ : Bottom-Middle (adjacent to BL, BR).

Due to symmetry across the main diagonal passing through TL and BR:

$P (E_{1}) = P (E_{2})$ . Let's call this $x$ .

$P (E_{3}) = P (E_{4})$ . Let's call this $y$ .

Let $P (C) = z$ .

Step 2: Set up Equations.

From Center (C), the frog jumps to any of the 4 edges with prob 1/4.

$z = \frac{1}{4} (x + x + y + y) = \frac{1}{2} x + \frac{1}{2} y$ .

From $E_{1}$ (Top-Middle), neighbors are TL (Target), C, TR (Other Corner).

Jumps to TL (Prob 1/3) -> Success.

Jumps to C (Prob 1/3) -> Value $z$ .

Jumps to TR (Prob 1/3) -> Fail (Value 0).

$x = \frac{1}{3} (1) + \frac{1}{3} (z) + \frac{1}{3} (0) = \frac{1}{3} + \frac{1}{3} z$ .

From $E_{3}$ (Right-Middle), neighbors are TR (Fail), C, BR (Fail).

Jumps to TR (Prob 1/3) -> 0.

Jumps to C (Prob 1/3) -> $z$ .

Jumps to BR (Prob 1/3) -> 0.

$y = \frac{1}{3} (0) + \frac{1}{3} (z) + \frac{1}{3} (0) = \frac{1}{3} z$ .

Step 3: Solve the System.

Substitute $y$ into the equation for $x$ :

$x = \frac{1}{3} + \frac{1}{3} z$ .

Substitute $x$ and $y$ into the equation for $z$ :

$z = \frac{1}{2} (\frac{1}{3} + \frac{1}{3} z) + \frac{1}{2} (\frac{1}{3} z)$ .

$z = \frac{1}{6} + \frac{1}{6} z + \frac{1}{6} z$ .

$z = \frac{1}{6} + \frac{1}{3} z$ .

$\frac{2}{3} z = \frac{1}{6}$ .

$z = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}$ .

The probability is 1/4.

Intuition Check: There are 4 corners. By symmetry, is the probability of ending at any specific corner equal?

Yes, the grid and start position are symmetric with respect to the 4 corners. Since the frog MUST eventually stop at a corner (random walk on finite connected graph with absorbing states), and the sum of probabilities for all 4 corners must be 1, each corner has probability 1/4.

Answer: 1/4.

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Out of 100 employees, 40 are in Sales, 30 are in Marketing, and 10 are in both. If an employee is chosen at random from the Sales department, what is the probability that they are also in Marketing?

A.
1/10
B.
3/10
C.
1/3
D.
1/4

Correct Answer:

Step-by-Step Solution

Key idea: This is a conditional probability question, recognisable because it restricts the sample space to a specific group ("chosen from the Sales department").

Step 1: Identify the condition. The employee is in Sales. Total in Sales = 40. This is our new sample space.

Step 2: Identify the favorable outcome within that condition. The employee is also in Marketing. Both Sales and Marketing = 10.

Step 3: Calculate the conditional probability. P(Marketing | Sales) = (Both) / (Sales) = 10 / 40 = 1/4.

Answer: 1/4

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A fair six-sided die is rolled. Using the classical definition of probability, what is the probability of rolling an even number?

A.
1/6
B.
1/2
C.
1/3
D.
2/3

Correct Answer:

Step-by-Step Solution

Key idea: This is a classical probability question, recognisable because all outcomes of a fair die are equally likely.

Step 1: Identify the total number of outcomes $n (S)$ . A standard die has 6 faces, so $n (S) = 6$ .

Step 2: Identify the favorable outcomes $n (E)$ . The even numbers on a die are 2, 4, and 6. So $n (E) = 3$ .

Step 3: Apply the classical probability formula.

$P (E) = \frac{n ( E )}{n ( S )} = \frac{3}{6} = \frac{1}{2}$ .

Answer: 1/2

Probability and Random Events Practice Questions for XAT: 93+ Solved Questions with Step-by-Step Solutions

Random Experiments and Sample Space

Random Experiments and Sample Space

Random Experiment

Sample Space ( $S$ )

Event ( $E$ )

Example: Tossing two coins

Classical Definition of Probability

Classical Definition of Probability

Probability and Random Events: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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Probability and Random Events Practice Questions for XAT: 93+ Solved Questions with Step-by-Step Solutions

Random Experiments and Sample Space

Random Experiments and Sample Space

Random Experiment

Sample Space (S)

Event (E)

Example: Tossing two coins

Classical Definition of Probability

Classical Definition of Probability

Probability and Random Events: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit

A question from this chapter

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Sample Space ( $S$ )

Event ( $E$ )