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SSC CHSL Probability

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This page covers SSC CHSL Probability with complete concept notes, 6 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Probability— Rules & Concept

Core ConceptRead this first — the foundation of the topic

Probability is the mathematical way to measure how likely an event is to happen. Think of it as a fraction that tells you the chances of something occurring. In SSC CGL, probability questions appear in almost every paper and can fetch you easy marks if you master the basic concepts and shortcuts. Core Concept: Probability = Favorable Outcomes / Total Possible Outcomes. The value always lies between 0 and 1. If probability is 0, the event will never happen. If it's 1, the event will definitely happen.

Key RulesCore rules you must know cold
Addition Rule

For mutually exclusive events A and B, P(A or B) = P(A) + P(B) 2

Multiplication Rule

For independent events A and B, P(A and B) = P(A) × P(B) 3

Complementary Events

P(A) + P(not A) = 1, so P(not A) = 1 - P(A) 4

Conditional Probability

P(A given B) = P(A and B) / P(B)

Formula BlockMemorise — at least one formula appears in every paper
• Basic Probability = Favorable outcomes / Total outcomes
• Probability of at least one = 1 - Probability of none
• Odds in favor = Favorable : Unfavorable
• Expected value = Sum of (Probability × Value)
Exam PatternsWhat examiners ask — read before attempting PYQs

SSC CGL typically asks about card problems, dice problems, ball drawing from bags, and coin tossing. Questions often involve finding probability of getting specific combinations or calculating odds.

ShortcutsUse these to save 30–60 seconds per question

#1 - For 'At Least One' Problems: Instead of calculating all possible combinations, use the complement rule. Find the probability that the event doesn't happen at all, then subtract from 1.

Worked ExampleSolve this step-by-step before moving on
1
Step 1

Count total balls = 5 + 3 + 2 = 10

2
Step 2

Count favorable outcomes (red balls) = 5

3
Step 3

Apply formula = 5/10 = 1/2 = 0.5 Answer: 1/2 or 50% Shortcut Trick #2 - Card Problems Quick Reference: Remember that a deck has 52 cards with 13 cards of each suit (hearts, diamonds, clubs, spades). Face cards are 12 in total (4 each of Jack, Queen, King). Worked Example 2: Two dice are thrown. What is the probability of getting a sum of 7?

1
Step 1

Total possible outcomes when throwing two dice = 6 × 6 = 36

2
Step 2

Find favorable outcomes for sum = 7 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes

3
Step 3

Apply formula = 6/36 = 1/6 Answer: 1/6 Shortcut Trick #3 - Dice Sum Patterns: For two dice, sum of 7 has maximum probability (6/36), while sums of 2 and 12 have minimum probability (1/36 each).

Exam TrapsCommon mistakes students make — avoid these

#1: Students often forget to consider all possible arrangements. For example, when drawing 2 balls from a bag, they count (Red, Blue) but forget (Blue, Red) as a separate case when order matters. Always check whether the problem requires considering order or not. Another frequent error is mixing up 'with replacement' and 'without replacement' scenarios.

In 'with replacement', the total number of items remains the same for each draw. In 'without replacement', the total decreases after each draw, affecting subsequent probabilities.

Key Points to Remember

  • Probability always lies between 0 and 1 inclusive
  • Basic formula: Probability = Favorable outcomes / Total outcomes
  • For 'at least one' problems, use complement: P(at least one) = 1 - P(none)
  • Standard deck has 52 cards, 13 per suit, 12 face cards total
  • Two dice have 36 total outcomes, sum of 7 has highest probability
  • Mutually exclusive events: P(A or B) = P(A) + P(B)
  • Independent events: P(A and B) = P(A) × P(B)
  • Complementary events: P(A) + P(not A) = 1
  • Always check if order matters in the problem statement
  • With replacement vs without replacement changes denominators

Exam-Specific Tips

  • A standard deck contains exactly 52 cards with 4 suits of 13 cards each
  • Number of face cards in a deck is 12 (4 Jacks, 4 Queens, 4 Kings)
  • Two dice thrown together give 36 total possible outcomes
  • Sum of 7 with two dice can occur in exactly 6 ways
  • Probability of getting heads in a fair coin toss is exactly 1/2
  • In a standard deck, probability of drawing an ace is 4/52 = 1/13
  • Maximum probability value is 1, minimum probability value is 0
  • Number of ways to arrange n distinct objects is n! factorial
Practice MCQs

Probability — Practice Questions

6graded MCQs · easy to hard · full solution & trap analysis

All MCQs →
Practice 1easy

A coin is tossed 3 times. What is the probability of getting exactly 2 heads?

Practice 2easy

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random, what is the probability that it is either red or green?

Practice 3medium

In a class of 40 students, 15 play cricket, 20 play football, and 8 play both. If a student is selected at random, what is the probability that the student plays at least one sport?

Practice 4medium

A bag contains 5 red balls, 7 blue balls, and 8 green balls. If two balls are drawn simultaneously at random, what is the probability that both balls are of the same colour?

Practice 5hard

A bag contains 5 red balls, 7 blue balls, and 8 green balls. Two balls are drawn simultaneously without replacement. What is the probability that both balls are of different colours?

Practice 6hard

In a lottery, 6 tickets are drawn from 100 tickets numbered 1 to 100. What is the probability that the sum of the numbers on the drawn tickets is even?

60-Second Revision — Probability

  • Formula: Probability = Favorable outcomes / Total outcomes
  • Remember: Use 1 - P(none) for 'at least one' questions
  • Trap: Don't confuse with replacement and without replacement scenarios
  • Quick fact: Standard deck = 52 cards, Face cards = 12
  • Pattern: Two dice sum of 7 has maximum probability of 1/6
  • Rule: Always check if order matters in arrangement problems
  • Shortcut: P(not A) = 1 - P(A) for complement problems
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