Probability: Complete Master Guide for UPSC/MPSC CSAT (English)

Introduction to Probability

Probability is a branch of mathematics that quantifies the likelihood or chance of a specific event occurring. In the context of CSAT for UPSC and MPSC, it is a high-yield topic that logically extends the concepts of Permutation and Combination. By understanding how to count possibilities, you can effortlessly transition to finding the probability of those events.

Questions from Probability test an aspirant’s analytical reasoning and decision-making skills under uncertainty. Common question types involve tossing coins, rolling dice, drawing cards from a deck, or picking colored balls from a bag. A firm grip on the basics and consistent practice are the keys to scoring full marks in this section.

Core Concepts and Formulas

1. Basic Terminology

• Experiment: An operation that produces some well-defined outcomes (e.g., rolling a die).
• Sample Space (S): The set of all possible outcomes of an experiment. For a coin toss, S = {Heads, Tails}.
• Event (E): A subset of the sample space representing the outcome we are interested in.

2. Basic Formula of Probability

The probability of an event E happening is defined as:
P(E) = (Number of favorable outcomes) / (Total number of possible outcomes)
P(E) = n(E) / n(S)

Key Properties:
1. 0 ≤ P(E) ≤ 1 (Probability can never be negative or greater than 1).
2. P(Sure Event) = 1 (Event that is guaranteed to happen).
3. P(Impossible Event) = 0 (Event that cannot happen).
4. P(E’) = 1 – P(E), where P(E’) is the probability of the event NOT occurring.

3. Addition Theorem of Probability

For any two events A and B:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
– If A and B are mutually exclusive (cannot happen at the same time, like getting both a Head and a Tail on a single coin toss), then P(A ∩ B) = 0. Therefore, P(A or B) = P(A) + P(B).

4. Multiplication Theorem (Independent Events)

If two events A and B are independent (the outcome of one does not affect the other), then the probability of both occurring is:
P(A and B) = P(A) × P(B)

Tricks and Shortcuts

• “At least one” problems: Calculating the probability of “at least one” is often lengthy. Use the shortcut: P(At least one) = 1 – P(None).
• Deck of Cards: Memorize the structure of a standard 52-card deck. 4 suits (Spades, Clubs, Hearts, Diamonds) with 13 cards each. 26 Red, 26 Black. Face cards = 12 (Jacks, Queens, Kings).
• Dice Sums: When throwing two dice, the total outcomes are 36. The sum ranges from 2 to 12. The number of ways to get sums from 2 to 7 is (Sum – 1). For sums from 8 to 12, it is (13 – Sum). E.g., Ways to get sum 5 = 5 – 1 = 4 ways. Ways to get sum 10 = 13 – 10 = 3 ways.

Solved Examples

Example 1: Tossing Coins

Question: Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?

Step-by-step Solution:
1. Total possible outcomes when 3 coins are tossed = 2³ = 8.
Sample Space (S) = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}.
2. Favorable outcomes (Event E: exactly two heads) = {HHT, HTH, THH}.
n(E) = 3.
3. Probability P(E) = n(E) / n(S) = 3/8.

Example 2: Rolling Dice (Using the trick)

Question: Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is 8?

Step-by-step Solution:
1. Total possible outcomes when 2 dice are rolled = 6 × 6 = 36.
2. Favorable outcomes (Sum = 8). Using the trick: For sum > 7, ways = 13 – Sum = 13 – 8 = 5 ways.
(The actual pairs are: (2,6), (3,5), (4,4), (5,3), (6,2)).
3. Probability P(E) = 5/36.

Example 3: Drawing Cards

Question: One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either a red card or a King?

Step-by-step Solution:
1. Total outcomes n(S) = 52.
2. Let A = Event of getting a red card. P(A) = 26/52.
3. Let B = Event of getting a King. P(B) = 4/52.
4. A and B are not mutually exclusive. There are 2 Red Kings (King of Hearts and King of Diamonds). So, P(A ∩ B) = 2/52.
5. Applying Addition Theorem: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= (26/52) + (4/52) – (2/52) = 28/52 = 7/13.

Example 4: Colored Balls (Combination based)

Question: A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random. What is the probability that all of them are red?

Step-by-step Solution:
1. Total number of balls = 4 + 5 + 6 = 15 balls.
2. Total ways to draw 3 balls from 15 = ¹⁵C₃
= (15 × 14 × 13) / (3 × 2 × 1) = 455.
3. Favorable ways to draw 3 red balls from 5 red balls = ⁵C₃
= (5 × 4) / (2 × 1) = 10.
4. Probability P(E) = 10 / 455 = 2/91.

Pro-Tips to Avoid Common Mistakes

• “With Replacement” vs “Without Replacement”: Read carefully. If balls/cards are replaced, the total number remains the same for the next draw (Independent Events). If not replaced, the total number decreases (Dependent Events).
• And/Or Confusion: ‘AND’ means multiplication of probabilities. ‘OR’ means addition of probabilities.
• Exhaustive Counting: When dealing with coins or dice, writing down the sample space helps avoid missing any specific case, especially for “at least” or “at most” conditions if the sample space is small.

Practice Questions

1. A coin is tossed three times. Find the probability of getting at least two heads.
2. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other.
3. From a well-shuffled deck of 52 cards, two cards are drawn simultaneously. Find the probability that both are face cards.
4. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
5. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Answers: 1. 1/2; 2. 11/36; 3. 11/221; 4. 4/7; 5. 2/7.