Probability and Random Events Previous Year Questions (PYQs) for XAT: 3+ Solved Questions with Step-by-Step Solutions

Solve 3+ Probability and Random Events previous year 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 (3 Problems)

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

Three categories of candidates appear for an admission test: diligent(10%), lazy(30%) and confused (60%). A diligent candidate is 10 times more likely to clear the admission test compared to a lazy candidate.

If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?

A.
13/37
B.
13/90
C.
2/3
D.
6/7
E.
37/100

Correct Answer:

Step-by-Step Solution

Key idea: This is a Bayes' theorem problem with optimization, recognisable because we're given a posterior probability and asked to maximise a prior-to-posterior likelihood.

Step 1: Define variables.

Let $P (Pass ∣ L) = p$ . Then $P (Pass ∣ D) = 10 p$ . Let $P (Pass ∣ C) = q$ .

Priors: $P (D) = 0.1$ , $P (L) = 0.3$ , $P (C) = 0.6$ .

Step 2: Use the given posterior.

$P (C ∣ Pass) = 0.4$ .

By Bayes' theorem:

$P (C ∣ Pass) = \frac{P ( Pass ∣ C ) \cdot P ( C )}{P ( Pass )}$

Where:

$P (Pass) = 0.1 (10 p) + 0.3 (p) + 0.6 (q) = 1.3 p + 0.6 q$

Step 3: Set up the equation.

$0.4 = \frac{0.6 q}{1.3 p + 0.6 q}$

$0.4 (1.3 p + 0.6 q) = 0.6 q$

$0.52 p + 0.24 q = 0.6 q$

$0.52 p = 0.36 q$

$q = \frac{0.52}{0.36} p = \frac{13}{9} p$

Step 4: Apply probability constraints to maximise $q$ .

Since $q = \frac{13}{9} p$ , to maximise $q$ we maximise $p$ .

Constraints:

$q \leq 1 ⟹ p \leq \frac{9}{13} \approx 0.69$
$P (Pass ∣ D) = 10 p \leq 1 ⟹ p \leq 0.1$

The binding constraint is $p \leq 0.1$ .

Max $p = 0.1 ⟹$ Max $q = \frac{13}{9} \times 0.1 = \frac{13}{90}$ .

Answer: 13/90

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

Common Description: Aman has come to the market with Rs. 100. If he buys 5 kilograms of cabbage and 4 kilograms of potato, he will have Rs. 20 left; or else, if he buys 4 kilograms of cabbage and 5 kilograms of onion, he will have Rs. 7 left.
The per kilogram prices of cabbage, onion and potato are positive integers(in rupees), and any type of these vegetables can only be purchased in positive integer kilogram, or none at all. Aman decides to buy only onion and potato, both in positive integer kilogram, in such a way that the money left with him after the purchase will be insu cient to buy a full kilogram of either of the two vegetables.
If all such permissible combinations of purchases are equally likely, what is the probability that Aman buys more onion than potato?

A.
10/3
B.
6/5
C.
9/2
D.
20/7
E.
10/4

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-step problem combining Linear Diophantine Equations (to find integer prices) with probability (counting valid purchase combinations). Recognisable because we must solve for unknown integer prices first, then compute probability over valid purchases.

Step 1: Find prices of Cabbage ( $c$ ), Potato ( $p$ ), Onion ( $o$ ).

Condition 1: $5 c + 4 p = 100 - 20 = 80$

Condition 2: $4 c + 5 o = 100 - 7 = 93$

From (1): $c = \frac{80 - 4 p}{5} = 16 - \frac{4 p}{5}$ .

For $c$ to be a positive integer, $p$ must be a multiple of 5, and $16 - 4 p /5 > 0 ⟹ p < 20$ .

Test $p = 5 k$ for $k = 1, 2, 3$ :

$k = 1$ : $p = 5, c = 12$ . Then $4 (12) + 5 o = 93 ⟹ 5 o = 45 ⟹ o = 9$ . Valid!
$k = 2$ : $p = 10, c = 8$ . Then $4 (8) + 5 o = 93 ⟹ 5 o = 61$ . Not integer.
$k = 3$ : $p = 15, c = 4$ . Then $4 (4) + 5 o = 93 ⟹ 5 o = 77$ . Not integer.

Unique solution: $c = 12$ , $p = 5$ , $o = 9$ .

Step 2: Find valid purchase combinations.

Aman buys $x$ kg potato and $y$ kg onion ( $x \geq 1, y \geq 1$ ).

Cost = $5 x + 9 y$ . Left = $100 - 5 x - 9 y$ .

Condition: Left < min(9, 5) = 5, so $0 \leq 100 - 5 x - 9 y \leq 4$ .

Equivalently: $96 \leq 5 x + 9 y \leq 100$ .

Enumerate by target sum $T = 5 x + 9 y$ :

$T = 100$ : $y \equiv 0 (mod 5)$ . $y = 5 ⟹ x = 11$ ; $y = 10 ⟹ x = 2$ . (2 pairs)
$T = 99$ : $y \equiv 1 (mod 5)$ . $y = 1 ⟹ x = 18$ ; $y = 6 ⟹ x = 9$ ; $y = 11 ⟹ x = 0$ (invalid). (2 pairs)
$T = 98$ : $y \equiv 2 (mod 5)$ . $y = 2 ⟹ x = 16$ ; $y = 7 ⟹ x = 7$ . (2 pairs)
$T = 97$ : $y \equiv 3 (mod 5)$ . $y = 3 ⟹ x = 14$ ; $y = 8 ⟹ x = 5$ . (2 pairs)
$T = 96$ : $y \equiv 4 (mod 5)$ . $y = 4 ⟹ x = 12$ ; $y = 9 ⟹ x = 3$ . (2 pairs)

Total valid pairs $(x, y)$ : 10.

List: (11,5), (2,10), (18,1), (9,6), (16,2), (7,7), (14,3), (5,8), (12,4), (3,9).

Step 3: Count favorable outcomes (more onion than potato: $y > x$ ).

Check each: (2,10)✓, (5,8)✓, (3,9)✓. Total = 3.

Step 4: Probability.

$P = \frac{3}{10}$

Note: The given options are all greater than 1, which is impossible for a probability. The option 10/3 is the reciprocal of the correct answer 3/10, suggesting a question-setting error. We select 10/3 as the most likely intended answer.

Answer: 10/3 (intended; actual probability is 3/10)

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

In a computer game, each move requires pressing a button. When the button is pressed for the first time, as a move, the computer randomly chooses a cell from a 4x4 grid of sixteen cells and puts an “X” mark on that cell. When the button is pressed subsequently, the computer randomly chooses a cell from the remaining unmarked cells and puts an “X” mark on that cell. This goes on till the end of the game. The game ends when either all the cells in any one row, or all the cells in any one column, are marked with “X”.

What is the maximum possible number of times a player has to press the button to finish the game?

A.
16
B.
10
C.
6
D.
13
E.
4

Correct Answer:

Step-by-Step Solution

Key idea: This is a worst-case (Pigeonhole Principle) problem, recognisable because we need the maximum moves before a termination condition is forced.

Step 1: Understand the termination condition.

Game ends when ANY row or ANY column is fully marked with X's on a $4 \times 4$ grid.

Step 2: Think in terms of empty cells.

A row is full $⟺$ it has 0 empty cells.

To avoid ending, every row must have $\geq 1$ empty cell, and every column must have $\geq 1$ empty cell.

Step 3: Minimise empty cells to maximise X's.

We need to place empty cells such that all 4 rows and all 4 columns are "covered" (each has at least one empty cell).

Minimum empty cells needed = 4 (e.g., placed on the main diagonal: (1,1), (2,2), (3,3), (4,4)).

Step 4: Verify 12 X's don't end the game.

With 4 empties on the diagonal:

Each row has exactly 1 empty → 3 X's → not full.
Each column has exactly 1 empty → 3 X's → not full.

So 12 X's is a valid non-ending state.

Step 5: Show 13 X's must end the game.

With 13 X's, there are only 3 empty cells.

By Pigeonhole Principle: 3 empty cells can cover at most 3 rows. At least one row has 0 empty cells → that row is full → game ends.

Step 6: Conclusion.

The game can last up to 12 presses without ending. The 13th press forces a completion.

Maximum presses to finish = 13.

Answer: 13

A question from this chapter

Question 1

Three categories of candidates appear for an admission test: diligent(10%), lazy(30%) and confused (60%). A diligent candidate is 10 times more likely to clear the admission test compared to a lazy candidate. If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?

Question 2

Common Description: Aman has come to the market with Rs. 100. If he buys 5 kilograms of cabbage and 4 kilograms of potato, he will have Rs. 20 left; or else, if he buys 4 kilograms of cabbage and 5 kilograms of onion, he will have Rs. 7 left. The per kilogram prices of cabbage, onion and potato are positive integers(in rupees), and any type of these vegetables can only be purchased in positive integer kilogram, or none at all. Aman decides to buy only onion and potato, both in positive integer kilogram, in such a way that the money left with him after the purchase will be insu cient to buy a full kilogram of either of the two vegetables. If all such permissible combinations of purchases are equally likely, what is the probability that Aman buys more onion than potato?

Question 3

In a computer game, each move requires pressing a button. When the button is pressed for the first time, as a move, the computer randomly chooses a cell from a 4x4 grid of sixteen cells and puts an “X” mark on that cell. When the button is pressed subsequently, the computer randomly chooses a cell from the remaining unmarked cells and puts an “X” mark on that cell. This goes on till the end of the game. The game ends when either all the cells in any one row, or all the cells in any one column, are marked with “X”. What is the maximum possible number of times a player has to press the button to finish the game?

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Probability and Random Events Previous Year Questions (PYQs) for XAT: 3+ 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 (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit

A question from this chapter

Login to view the complete previous-year questions and solutions

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Built around you, not around a syllabus PDF

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Questions calibrated to the real exam

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Unlock the whole course

Sample Space ( $S$ )

Event ( $E$ )