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Quantitative AptitudeBeginner6 min study

How to Solve Probability Questions: Formulas, Rules & Solved Examples

प्रायिकता (Probability) के प्रश्न कैसे हल करें: सूत्र, नियम और हल सहित उदाहरण

Probability (प्रायिकता) is the mathematical quantification of the chance that a particular event will occur, represented on a continuous scale from 0 (impossible event / असंभव घटना) to 1 (certain event / निश्चित घटना). In competitive examinations across Madhya Pradesh (MP Police Constable, MP Sub-Inspector, MP Patwari, MPPSC CSAT) and central recruitment (SSC CGL/CHSL/MTS, RRB NTPC), probability problems most frequently test four standard random experiments: tossing fair coins, rolling 6-faced dice, drawing cards from a standard 52-card pack, and drawing colored balls from an urn without replacement. The central pillar of solving any exam problem is accurately computing the total sample space n(S) and the favorable outcomes n(E), while strategically employing the complementary event shortcut P(At least one) = 1 - P(None) to save crucial exam minutes.

Key Rules & Core Formulas

Understand these foundational properties before solving questions

1Classical Probability Definition (प्रायिकता का मूल सूत्र)

P(E) = n(E) / n(S) = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)

The foundational ratio where n(E) represents the total outcomes meeting the condition of event E, and n(S) is the exhaustive sample space of equally likely elementary outcomes. The probability always satisfies 0 ≤ P(E) ≤ 1.

For an impossible event, P(E) = 0. For a sure / certain event, P(E) = 1. Probability can never be negative or exceed 1.

2Complementary Event Rule & "At Least One" Shortcut (पूरक घटना का नियम)

P(E') = 1 - P(E) ⇒ P(At least one) = 1 - P(None)

The probability that event E will NOT occur equals 1 minus the probability that E occurs. Whenever a question contains the phrase "at least one" (कम से कम एक), calculating P(None) and subtracting from 1 is exponentially faster than calculating multiple individual positive cases.

For n tosses of a fair coin: P(At least one Head) = 1 - (1/2ⁿ) = (2ⁿ - 1) / 2ⁿ.

3Addition Theorem for Events (योग प्रमेय - Keyword "OR")

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Used when finding the probability of event A OR event B occurring. If events A and B are mutually exclusive (परस्पर अपवर्जी - they cannot happen simultaneously, meaning A ∩ B = ∅), then P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B).

When drawing a card that is a "King OR a Spade", King of Spades belongs to both sets. Never forget to subtract P(A ∩ B) to avoid double counting.

4Multiplication Theorem for Independent Events (गुणन प्रमेय - Keyword "AND")

P(A ∩ B) = P(A) × P(B)

If two events A and B are independent (स्वतंत्र घटनाएं - occurrence of one does not affect the likelihood of the other, such as throwing a die and flipping a coin), the probability of both occurring together is the product of their individual probabilities.

If draws are "without replacement" (प्रतिस्थापन बिना), events become dependent: P(A and B) = P(A) × P(B | A).

5Selection Formula with Combinations (संयोजन सूत्र - ⁿCᵣ)

ⁿCᵣ = n! / [r! × (n - r)!]

Used when r items are drawn simultaneously (or successively without replacement) from an available pool of n items where order does not matter (e.g., drawing 2 balls from 10 balls, or 3 cards from 52).

Fast competitive shortcut: ⁿC₂ = [n × (n - 1)] / 2; ⁿC₁ = n; ⁿC₀ = 1.

6Standard Sample Space Quick Reference (सिक्के, पासा और ताश)

Coins: n(S) = 2ⁿ | Dice: n(S) = 6ⁿ | Cards: Total = 52

• Coins: 1 coin = 2 outcomes, 2 coins = 4 (HH, HT, TH, TT), 3 coins = 8 (2³). • Dice: 1 die = 6, 2 dice = 36 (6²). • Cards: 52 total = 26 Red (13 Hearts ♥, 13 Diamonds ♦) + 26 Black (13 Spades ♠, 13 Clubs ♣). 4 Suits with 13 cards each (Ace, 2–10, Jack, Queen, King). 12 Face cards (4 Jacks, 4 Queens, 4 Kings). 4 Aces.

Crucial Exam Tip: Aces (इक्के) are Honor cards, NOT face cards (तस्वीर वाले पत्ते). Total face cards in a pack are strictly 12 (3 in each of 4 suits).
Standard Method

Step-by-Step Method to Solve

Follow these universal steps in order to solve any Quantitative Aptitude question accurately:

  1. 1

    Step 1: Determine the Experiment & Calculate Sample Space n(S)

    Read the problem carefully to identify the physical action being conducted (e.g., tossing n coins, rolling 2 dice, or drawing r balls). Establish whether outcomes are drawn with or without replacement. Calculate total possible outcomes n(S).

    Pro-Tip: For 2 dice, n(S) = 36. For drawing 2 balls from a bag of 12 balls, n(S) = ¹²C₂ = (12 × 11) / 2 = 66.
  2. 2

    Step 2: Define the Favorable Event E & Count n(E)

    Identify all specific conditions that fulfill the question's criteria. Systematically list or use combination formulas (ⁿCᵣ) to compute the count of successful outcomes n(E).

    Pro-Tip: When counting dice sums, write pairs systematically in ascending order of the first die so you don't miss reverse pairs (e.g., (2, 5) and (5, 2)).
  3. 3

    Step 3: Evaluate the Complementary Event Shortcut

    If the question asks for 'at least one' (कम से कम एक) or 'not all', check if calculating the opposite event E' ('none' or 'zero occurrences') is faster. If so, apply P(E) = 1 - P(E').

    Pro-Tip: If a problem asks for at least 1 head in 4 tosses, calculating 1, 2, 3, and 4 heads requires 4 calculations. Calculating 0 heads (TTTT) takes 1 second: 1 - (1/16) = 15/16!
  4. 4

    Step 4: Formulate the Ratio & Reduce to Irreducible Form

    Substitute values into P(E) = n(E) / n(S). Cancel out common factors between numerator and denominator to express the answer in lowest terms as typically presented in exam option choices.

    Pro-Tip: Sanity check: Numerator must never exceed denominator. P(E) must always lie strictly within [0, 1].
  5. 5

    Step 5: Differentiate Between 'AND' (Multiply) vs 'OR' (Add)

    If an event consists of multiple sequential or simultaneous requirements (e.g., 1st Red AND 2nd Blue), multiply their individual probabilities. If the event consists of alternate acceptable cases (Red OR Green), add the probabilities.

    Pro-Tip: Keyword decoder: 'AND' / 'तथा' / 'और' → Multiply (×). 'OR' / 'या' / 'अथवा' → Add (+).

Worked Examples & Walkthroughs

Real questions solved step-by-step showing every calculation

1Example 1Method: Systematic Sample Space Pairing Method

Question

Two standard fair six-faced dice are rolled simultaneously. What is the probability of obtaining a total sum of numbers that is strictly greater than 8?

Step-by-Step Solution:

Step 1: Calculate total sample space n(S):Each standard die has 6 faces numbered 1 to 6. When two dice are thrown together, the total number of equally likely outcomes is n(S) = 6 × 6 = 36.
Step 2: Identify favorable outcomes where sum > 8:A sum strictly greater than 8 means the possible sums can be 9, 10, 11, or 12. Let us list all valid pairs (Die 1, Die 2): • Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 pairs • Sum = 10: (4, 6), (5, 5), (6, 4) → 3 pairs • Sum = 11: (5, 6), (6, 5) → 2 pairs • Sum = 12: (6, 6) → 1 pair
Step 3: Total the favorable outcomes n(E):Total favorable outcomes n(E) = 4 + 3 + 2 + 1 = 10.
Step 4: Compute probability P(Sum > 8):P(Sum > 8) = n(E) / n(S) = 10 / 36. Dividing numerator and denominator by 2 gives 5 / 18.
Final Result:
5/18
Exam Shortcut / Verification: Exam Shortcut: The sum frequencies of two dice follow a symmetric pyramid: Sums 2, 3, 4, 5, 6, 7 have frequencies 1, 2, 3, 4, 5, 6; and sums 8, 9, 10, 11, 12 decrease as 5, 4, 3, 2, 1. For sum > 8, mentally sum the counts for 9, 10, 11, and 12: 4 + 3 + 2 + 1 = 10. Result = 10/36 = 5/18 in under 5 seconds!
2Example 2Method: Addition Theorem for Non-Mutually Exclusive Events

Question

A card is drawn at random from a standard, well-shuffled pack of 52 playing cards. What is the probability that the card drawn is either a King (बादशाह) or a Spade (हुकुम)?

Step-by-Step Solution:

Step 1: Determine sample space n(S):A standard deck of playing cards contains 52 cards, so n(S) = 52.
Step 2: Identify individual event counts:• Let event A be drawing a King: There are 4 Kings in a deck (Heart, Diamond, Club, Spade), so n(A) = 4, giving P(A) = 4/52. • Let event B be drawing a Spade: There are 13 Spades in a deck, so n(B) = 13, giving P(B) = 13/52.
Step 3: Identify the overlapping intersection (A ∩ B):Event (A ∩ B) is drawing the King of Spades (हुकुम का बादशाह). Since this single card belongs to both categories, n(A ∩ B) = 1, giving P(A ∩ B) = 1/52.
Step 4: Apply the Addition Theorem P(A ∪ B):P(King or Spade) = P(A) + P(B) - P(A ∩ B) = (4 / 52) + (13 / 52) - (1 / 52) = (4 + 13 - 1) / 52 = 16 / 52. Dividing numerator and denominator by 4 gives 4 / 13.
Final Result:
4/13
Exam Shortcut / Verification: Direct Counting Shortcut: Take all 13 cards of the Spade suit, then add the remaining Kings from the other 3 suits (Heart, Diamond, Club) = 13 + 3 = 16 cards. Directly calculate 16 / 52 = 4 / 13 without writing formulas!
3Example 3Method: Combination Method (ⁿCᵣ) for Mutually Exclusive Cases

Question

An urn contains 5 red balls, 4 green balls, and 3 blue balls. If 2 balls are drawn at random simultaneously, what is the probability that both balls are of the same color?

Step-by-Step Solution:

Step 1: Calculate total balls and sample space n(S):Total balls = 5 Red + 4 Green + 3 Blue = 12 balls. Total ways to select any 2 balls from 12 is: n(S) = ¹²C₂ = (12 × 11) / (2 × 1) = 66.
Step 2: Calculate favorable outcomes for each color:Both balls can be Red, OR both Green, OR both Blue (mutually exclusive): • Both Red: ⁵C₂ = (5 × 4) / 2 = 10 ways • Both Green: ⁴C₂ = (4 × 3) / 2 = 6 ways • Both Blue: ³C₂ = (3 × 2) / 2 = 3 ways
Step 3: Sum favorable outcomes n(E):Total favorable outcomes n(E) = 10 + 6 + 3 = 19 ways.
Step 4: Compute probability:P(Both same color) = n(E) / n(S) = 19 / 66.
Final Result:
19/66
Exam Shortcut / Verification: Mental Combinations Shortcut: For any ⁿC₂, just multiply n by (n - 1) and halve it. Red: 5×4/2=10; Green: 4×3/2=6; Blue: 3×2/2=3. Total = 19. Total sample space = 12×11/2=66. P = 19/66 in 15 seconds flat.
4Example 4Method: Complementary Event Shortcut Method (1 - P(None))

Question

Three unbiased coins are tossed simultaneously. What is the probability of getting at least one Head (कम से कम एक चित्त)?

Step-by-Step Solution:

Step 1: Calculate sample space n(S):For 3 fair coins, each toss has 2 possible outcomes (Head or Tail). Total sample space n(S) = 2³ = 8. Explicitly: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
Step 2: Identify the complement event E' (No Heads):The opposite of 'at least one Head' is 'zero Heads' (getting all Tails). The only outcome with zero heads is {TTT}. Thus, n(E') = 1.
Step 3: Calculate P(E'):P(No Heads) = n(E') / n(S) = 1 / 8.
Step 4: Subtract from 1 to find P(At least one Head):P(At least one Head) = 1 - P(No Heads) = 1 - (1 / 8) = 7 / 8.
Final Result:
7/8
Exam Shortcut / Verification: Universal Coin Shortcut: For any n coin tosses, P(At least one Head) is always (2ⁿ - 1) / 2ⁿ. For 3 coins: (8 - 1)/8 = 7/8. For 4 coins: (16 - 1)/16 = 15/16. Instant 2-second answer!

Common Mistakes to Avoid

Avoid these frequent traps to protect yourself from negative marking

1

Double counting overlapping outcomes in 'A OR B' card questions

Why it happens: When asked for the probability of a King or a Spade, students simply add 4 Kings + 13 Spades = 17 cards, inadvertently counting the King of Spades twice.
Correct approach: Always apply the subtraction rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Subtract the overlapping cards (here, 1 King of Spades) to get 16/52 = 4/13.
2

Confusing 'At Least' (कम से कम) with 'At Most' (अधिक से अधिक)

Why it happens: Students misinterpret the inequality boundaries during fast exam solving. 'At least 2 Heads' in 3 tosses means 2 or 3 Heads (≥ 2), while 'At most 2 Heads' means 0, 1, or 2 Heads (≤ 2).
Correct approach: Mentally translate immediately: 'At least x' = x or more (≥ x). 'At most x' = x or fewer (≤ x, including 0). For 'at least 1', always use the complement 1 - P(None).
3

Forgetting to decrease the denominator in 'without replacement' draws

Why it happens: When drawing two balls or cards one after another without replacement, students keep the denominator as the original total count for both draws.
Correct approach: If an item is removed and not put back, the total available pool shrinks by 1 (from n to n-1), and the count of that specific category also shrinks by 1 (from m to m-1) for subsequent draws.
4

Omitting order permutations when listing dice outcomes

Why it happens: When finding pairs that sum to 7, a student counts (1, 6), (2, 5), (3, 4) as 3 outcomes, forgetting the reverse pairs (6, 1), (5, 2), (4, 3).
Correct approach: Two dice are distinct entities (Die A and Die B). Unless both dice show the exact same number (like (4, 4)), every pair (a, b) occurs in two distinct orders: (a, b) and (b, a). There are 6 pairs for sum 7, not 3.
Self-Assessment

Try It Yourself (Practice Questions)

Test your understanding on your own first, then reveal the step-by-step solution to check your work:

Practice #1

Two fair dice are thrown simultaneously. What is the probability that the product of the two numbers obtained is an odd number?

Practice #2

From a standard well-shuffled pack of 52 playing cards, two cards are drawn at random simultaneously. What is the probability that both cards are Aces (इक्के)?

Practice #3

A box contains 6 black balls and 4 white balls. Two balls are drawn one after another without replacement. What is the probability that the first ball is black and the second ball is white?

Quick Revision Summary

  • Fundamental Probability Ratio: P(E) = n(E) / n(S) = Favorable Outcomes / Total Sample Space, strictly bounded between 0 and 1.
  • 'At Least One' Golden Shortcut: P(At least 1) = 1 - P(None). For n coins, P(At least 1 Head) = (2ⁿ - 1) / 2ⁿ.
  • Two Dice Sum Frequency: 36 total outcomes. Sum frequencies peak at 7 (count 6) and follow a triangular pattern: Count = 6 - |7 - Sum|.
  • 52-Card Pack Breakdown: 26 Red, 26 Black; 4 suits of 13 cards; 12 Face cards (Jacks, Queens, Kings) and 4 Aces. Always subtract overlaps: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
  • Draws Without Replacement: Denominators decrease with each successive pick. Use combination formula ⁿCᵣ for simultaneous selections.
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