Probability PYQs for UPSC (2011-2024) | Solved Questions & PDF Download

Master Probability Concepts for UPSC with solved Previous Year Questions (PYQs) from 2011 to 2024. This comprehensive guide provides detailed solutionsshortcut formulas, and a free downloadable PDF to help you crack UPSC CSAT's probability problems.


Why Probability Matters in UPSC CSAT?

Probability questions are consistently asked in UPSC CSAT (Paper II) to test:

• Your ability to analyze uncertain situations
• Understanding of basic and conditional probability
• Skills in solving real-world scenario-based problems
This page covers all probability PYQs with smart calculation techniques.

Year-Wise Probability PYQs (2011-2024)

Q1. A round archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white. The radius of the red band is 0.20 m. The width of all the remaining bands is equal. If archers throw arrows towards the target, what is the probability that the arrows fall in the red region of the archery target? [CSAT 2016]

(a) 0.40

(b) 0.20

(c) 0.16

(d) 0.04

Solution:

Given that,

A around archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white.

The radius of the red band is 0.20 m.

The width of all the remaining bands is equal.

Now,

The area of red band = π(0.20)2 = 0.04 π

Radius of round archery target = 1/2 = 0.5m

Total area = π(0.5)2= 0.25 π

Probability = 0.04 π/0.25 π = 0.16

Hence option (c) is correct


Q2. A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning? [CSAT 2018]

(a) 1/2

(b) 4/7

(c) 5/9

(d) 12/13

Solution:

Given that,

A bag contains 15 red balls and 20 black balls.

Each ball is numbered either 1 or 2 or 3.

20% of the red balls are numbered 1 and 40% of them are numbered 3.

Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3.

A boy picks a ball at random.

He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2.

Now,

Total number of balls = 20 + 15 = 35

Red balls numbered 3 = (40/100) x 15 = 6

Red balls numbered 1 = 20% of 15 = 3

Black balls numbered 2 = 45% of 20 = 9

Black balls numbered 3 = 30% of 20 = 6

Black balls numbered 1 = 20 - (9 + 6) = 5

Black balls numbered 1 or 2 = 9 + 5 = 14

Probability of winning = (6 + 14)/35 = 4/7

Hence option (b) is correct