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Probability – Class 10 Maths Chapter 14 Notes & NCERT PDF

Class 10MathematicsChapter 14NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 14 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 14: Probability

Introduction

"The theory of probabilities and the theory of errors now constitute a formidable body of great mathematical interest and of great practical importance." — R.S. Woodward

In Class IX you met experimental (empirical) probability: the number of trials in which an event happened divided by the total number of trials. But repeating an experiment is not always possible — you cannot launch satellites again and again, or repeat an earthquake, to measure a probability. When the outcomes are equally likely, we can calculate the probability directly. This is theoretical (classical) probability, the subject of this chapter.

Key Concepts

1. Equally likely outcomes

A fair coin gives head or tail with equal chance; a fair die gives 1 to 6 with equal chance. Not every experiment is like this: a bag with 4 red balls and 1 blue ball does not give "red" and "blue" equally often. In this chapter all experiments are assumed to have equally likely outcomes.

2. Theoretical probability

P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes of the experiment

This definition was given by Pierre Simon Laplace in 1795. Probability began in the 16th century with the Italian J. Cardan's The Book on Games of Chance; James Bernoulli, A. de Moivre and Laplace (whose Théorie Analytique des Probabilités, 1812, is considered the greatest single contribution) developed it further.

3. Important facts

  • An elementary event has only one outcome. The sum of the probabilities of all the elementary events of an experiment is 1.
  • Complementary events: "not E" (written Ē) — P(Ē) = 1 − P(E), i.e. P(E) + P(Ē) = 1.
  • Impossible event: probability 0 (getting 8 on a die).
  • Sure (certain) event: probability 1 (getting a number less than 7 on a die).
  • For every event, 0 ≤ P(E) ≤ 1.
  • 4. A deck of cards

    52 cards in 4 suits of 13 — spades ♠ and clubs ♣ (black), hearts ♥ and diamonds ♦ (red). Each suit has ace, king, queen, jack, 10, 9, …, 2. Kings, queens and jacks are face cards (12 in all).

    5. Beyond counting (not for examination)

    When outcomes are all points on a line or in a region, probability can be found as a ratio of lengths or areas. The music stopping in the first half-minute of 2 minutes: P = (1/2) ÷ 2 = 1/4. A helicopter crashing in a 2.5 km × 3 km lake inside a 4.5 km × 9 km region: P = 7.5 ÷ 40.5 = 5/27.

    Methods Step by Step

  • List (or count) all possible outcomes and check they are equally likely.
  • Count the outcomes favourable to the event.
  • Divide and simplify.
  • If "not E" or "at least one" is easier to count the other way, use P(E) = 1 − P(not E).
  • For two dice or two coins, write the full list of ordered pairs — (1, 4) and (4, 1) are different outcomes.
  • Worked Examples (from the chapter)

  • Die (Example 3): P(number > 4) = 2/6 = 1/3; P(number ≤ 4) = 4/6 = 2/3.
  • Cards (Example 4): P(ace) = 4/52 = 1/13; P(not an ace) = 48/52 = 12/13 = 1 − 1/13.
  • Tennis (Example 5): P(Sangeeta wins) = 0.62, so P(Reshma wins) = 1 − 0.62 = 0.38.
  • Birthdays (Example 6): P(different birthdays) = 364/365; P(same birthday) = 1 − 364/365 = 1/365.
  • Class representative (Example 7): 25 girls, 15 boys → P(girl) = 25/40 = 5/8, P(boy) = 3/8.
  • Marbles (Example 8): 3 blue, 2 white, 4 red → P(white) = 2/9, P(blue) = 1/3, P(red) = 4/9 (they add to 1).
  • Two coins (Example 9): outcomes HH, HT, TH, TT → P(at least one head) = 3/4 = 1 − P(no head).
  • Shirts (Example 12): 88 good, 8 minor defects, 4 major → P(acceptable to Jimmy) = 0.88; P(acceptable to Sujatha) = 96/100 = 0.96.
  • Two dice (Example 13): 36 outcomes. P(sum 8) = 5/36; P(sum 13) = 0; P(sum ≤ 12) = 1.
  • Board Exam Focus

  • Cards, dice, coins, balls in a bag, defective items and numbered discs appear almost every year.
  • "At least one" questions are usually quickest with 1 − P(none).
  • Two dice: remember 36 outcomes, and P(5 does not come up either time) = 25/36, so P(5 comes at least once) = 11/36.
  • A probability can never be negative or more than 1 — so −1.5 cannot be a probability.
  • Formula Sheet

  • P(E) = favourable outcomes ÷ total outcomes
  • P(E) + P(Ē) = 1, so P(Ē) = 1 − P(E)
  • P(impossible event) = 0; P(sure event) = 1; 0 ≤ P(E) ≤ 1
  • Sum of probabilities of all elementary events = 1
  • Common Mistakes

  • Counting unequal outcomes as equal — e.g. saying two coins give 3 outcomes (two heads, two tails, one of each) each with probability 1/3. Correctly, "one of each" has probability 2/4 = 1/2.
  • Saying the 11 possible sums of two dice (2 to 12) each have probability 1/11 — they are not equally likely.
  • Treating (1, 4) and (4, 1) as the same outcome.
  • Forgetting that face cards are only kings, queens and jacks (12 cards, not 16).
  • Giving an answer greater than 1 or less than 0.
  • 💡 Key Learning Points

    • ✓Theoretical probability P(E) = favourable outcomes ÷ all possible outcomes, when outcomes are equally likely.
    • ✓Complementary events: P(E) + P(not E) = 1, so P(not E) = 1 − P(E).
    • ✓An impossible event has probability 0, a sure event has probability 1, and 0 ≤ P(E) ≤ 1.
    • ✓The sum of the probabilities of all the elementary events of an experiment is 1.
    • ✓List all outcomes carefully for coins, dice and cards — ordered pairs like (1, 4) and (4, 1) are different.

    👨‍🏫 Teaching Tips

    • →Start with the book's question: why can't we find the probability of a satellite failure by repeating the launch?
    • →Use real coins, dice and a pack of cards; let students list outcomes before calculating.
    • →Draw the 6 × 6 grid of two-dice outcomes on the board and shade favourable cells for different sums.
    • →Discuss the common wrong arguments from Exercise 14.1 (Q22, Q25) as a class debate.
    • →Teach '1 − P(not E)' as a shortcut and let students check it both ways.
    • →Use a spinner divided into 8 equal parts to show equally likely outcomes.
    • →Connect to history: Cardan, Bernoulli, de Moivre and Laplace, as mentioned in the chapter.

    📋 Assessment Questions

    1. Can the student list all possible outcomes for one die, two dice and two coins?
    2. Can the student find probabilities of card events such as an ace, a face card or a red king?
    3. Can the student use P(not E) = 1 − P(E) correctly?
    4. Can the student identify impossible and sure events and explain why 0 ≤ P(E) ≤ 1?
    5. Can the student spot the error in arguments that treat unequal outcomes as equally likely?

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