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CBSE · Class 10 · Mathematics

Probability 14.1: A Theoretical Approach

5 concepts · 40 practice questions with answers · Free to read

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Basic terms and the probability formula

From daily life

Before a cricket match the toss is a fair way to choose, because heads and tails are equally likely. Probability puts a number between 0 and 1 on such chances.

Simple definition

An experiment gives outcomes; all outcomes together form the sample space; an event is a set of outcomes we care about. When all outcomes are equally likely, P(E) = number of favourable outcomes ÷ total number of outcomes. P(E) always lies from 0 to 1; a sure event has P = 1 and an impossible event has P = 0.

Formula: P(E) = favourable ÷ total

0½1impossibleeven chancesureP(E) = favourable outcomes ÷ all outcomes0 ≤ P(E) ≤ 1

In the figure: P(E) lies between 0 and 1: 0 is impossible, ½ is an even chance, 1 is sure.

Rule 1: Count favourable and total — Both counts must be of equally likely outcomes.

  • Easy: Cards 1 to 10, event “even”: favourable 5 (2, 4, 6, 8, 10), total 10.
  • Medium: P(even) = 5/10 = 1/2.
  • Exam-type: P(number greater than 10) = 0/10 = 0, an impossible event.

Rule 2: Valid and invalid probabilities — A probability can never be negative or more than 1.

  • Easy: 3/5 can be a probability.
  • Medium: 8/5 cannot, since it is more than 1, and −1/2 cannot, since it is negative.
  • Exam-type: P = 1 means certain: the sum of the numbers 1 to 10 is greater than 0.
Remember

Count outcomes one by one before dividing; a missed outcome changes the answer.

Common mistake

Using favourable ÷ unfavourable instead of favourable ÷ total.

Memory clue

"Wanted over all, never above one."

Practice questions

Easy
  1. 1In a class activity, a die is thrown once. How many outcomes are favourable to getting a multiple of 3?
  2. 2A die has how many equally likely outcomes when thrown once?
  3. 3In a class activity, a bag contains 7 yellow, 12 black and 3 white balls. One ball is drawn at random. How many equally likely outcomes are possible?
Medium
  1. 4For a lucky-draw game, a die is thrown once. How many possible outcomes are there?
  2. 5In a board game, a coin is tossed once. How many possible outcomes are there?
  3. 6A bag has only 10 blue balls. A ball is drawn at random. What is the probability that it is red?
Challenging
  1. 7In an experiment with 18 equally likely outcomes, 9 outcomes favour an event E. Find P(E) in simplest form.
  2. 8In an experiment with 39 equally likely outcomes, 25 outcomes favour an event E. Find P(E) in simplest form.
Show answers (try them yourself first)
  1. 12
  2. 26
  3. 322
  4. 46
  5. 52
  6. 60
  7. 71/2
  8. 825/39

Coins and dice

From daily life

Ludo and Snakes and Ladders both depend on a die; each face is equally likely, which is why the game is fair.

Simple definition

One coin: 2 outcomes. Two coins: HH, HT, TH, TT (4). One die: 6 outcomes. Two dice: 6 × 6 = 36 outcomes. List the sample space first, then count the favourable pairs.

Formula: outcomes: 2, 4, 6, 36

112345672234567833456789445678910556789101166789101112two dice: 36 outcomessum 7: 6 cellsP = 6/36 = 1/6

In the figure: Two dice: sum 7 occurs in 6 of 36 cells, so P = 1/6.

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Coins and dice

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Rule 1: Coins — Write all outcomes in order: H or T for each coin.

  • Easy: Two coins: HH, HT, TH, TT.
  • Medium: P(at least one head) = 3/4 (HH, HT, TH).
  • Exam-type: P(exactly two heads) = 1/4.

Rule 2: One die and two dice — For two dice use the 6 × 6 grid of pairs.

  • Easy: One die: P(prime) = 3/6 = 1/2 (2, 3, 5); P(more than 4) = 2/6 = 1/3.
  • Medium: Two dice: P(sum 8) = 5/36; P(sum 7) = 6/36 = 1/6.
  • Exam-type: P(doublet, same number on both) = 6/36 = 1/6.
Remember

In two-dice problems, (2, 5) and (5, 2) are different outcomes.

Common mistake

Taking the total for two dice as 12 instead of 36.

Memory clue

"Two dice, thirty-six pairs."

Practice questions

Easy
  1. 1At a school fair stall, 3 coins are tossed together. Find the probability of getting at least one head.
  2. 2In a probability experiment, 2 coins are tossed together. Find the probability of getting at most one head.
  3. 3In a class activity, 3 coins are tossed together. Find the probability of getting three heads.
Medium
  1. 4Two dice are thrown together. Find the probability that the sum of the numbers is 12.
  2. 5For a lucky-draw game, three coins are tossed together. Find the probability of getting a tail on the first coin.
  3. 6During a quiz round, three coins are tossed together. Find the probability of getting all three alike.
Challenging
  1. 7Two dice are thrown together. Find the probability that the sum of the numbers is 7.
  2. 8For a lucky-draw game, two dice are thrown together. Find the probability of getting a sum that is a prime number.
Show answers (try them yourself first)
  1. 17/8
  2. 23/4
  3. 31/8
  4. 41/36
  5. 51/2
  6. 61/4
  7. 71/6
  8. 85/12

Bags, numbered cards and lotteries

From daily life

Putting your hand in a bag without looking is a random draw: every ball has the same chance, just like every lottery ticket.

Simple definition

In a random draw each item is equally likely, so the total is the number of items and the favourable count is the number of items with the wanted property. For a lottery, P(winning) = tickets you hold ÷ total tickets.

Formula: P = wanted items ÷ all items

5 red3 blue2 greenbag of 10 ballsP(red) = 5/10 = 1/2P(not green) = 8/10 = 4/5

In the figure: Bag: 5 red, 3 blue, 2 green: P(red) = 1/2, P(not green) = 4/5.

Rule 1: Numbered cards — List the numbers that fit before counting.

  • Easy: Cards 1 to 20, multiples of 3: 3, 6, 9, 12, 15, 18, so P = 6/20 = 3/10.
  • Medium: Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, so P = 8/20 = 2/5.
  • Exam-type: The number 1 is not prime.

Rule 2: Bags and lotteries — Read the colours or tickets as the counts.

  • Easy: 5 red, 3 blue, 2 green balls: P(red) = 5/10 = 1/2.
  • Medium: P(not green) = 8/10 = 4/5.
  • Exam-type: Lottery of 1000 tickets, you hold 5: P(win) = 5/1000 = 1/200.
Remember

Check that the favourable count never exceeds the total.

Common mistake

Forgetting to count the drawn item itself in the total.

Memory clue

"Wanted items over all items."

Practice questions

Easy
  1. 1A bag contains 5 red, 8 black and 3 yellow balls. One ball is drawn at random from the bag. Find the probability that it is black.
  2. 2A bag contains 2 green, 3 yellow and 8 red balls. One ball is drawn at random from the bag. Find the probability that it is red.
  3. 3In a board game, a card is drawn at random from 60 cards numbered 1 to 60. Find the probability that the number on the card is an even number.
Medium
  1. 4A bag has 4 red and 9 blue balls. 2 more blue balls are put in. One ball is drawn at random. Find the probability that it is red.
  2. 5A card is drawn at random from 40 cards numbered 1 to 40. Find the probability that the number on the card is a number divisible by 3.
  3. 6In a board game, a card is drawn at random from 60 cards numbered 1 to 60. Find the probability that the number on the card is a multiple of 5.
Challenging
  1. 7A bag has 6 red balls and x blue balls. If the probability of drawing a red ball is 3/4, find x.
  2. 8A bag has 9 red balls and x blue balls. If the probability of drawing a red ball is 3/8, find x.
Show answers (try them yourself first)
  1. 11/2
  2. 28/13
  3. 31/2
  4. 44/15
  5. 513/40
  6. 61/5
  7. 72
  8. 815

Complementary events

From daily life

If the chance of rain tomorrow is 0.3, the chance of no rain is whatever is left to make up 1, so 0.7.

Simple definition

The event “not E” is the complementary event of E. P(E) + P(not E) = 1, so P(not E) = 1 − P(E). It is the fast route for “at least one” questions.

Formula: P(not E) = 1 − P(E)

0.30.7P(E)P(not E)total = 1P(E) + P(not E) = 1

In the figure: P(E) = 0.3 means P(not E) = 0.7, and together they make 1.

Rule 1: Find P(not E) — Subtract from 1.

  • Easy: P(E) = 4/5 gives P(not E) = 1/5.
  • Medium: Die: P(6) = 1/6, so P(not 6) = 5/6.
  • Exam-type: P(E) = 0.3 gives P(not E) = 0.7.

Rule 2: “At least one” by complement — Count the one bad case instead of many good cases.

  • Easy: Three coins give 8 outcomes; only TTT has no head.
  • Medium: P(no head) = 1/8, so P(at least one head) = 1 − 1/8 = 7/8.
  • Exam-type: Two dice: P(sum not 7) = 1 − 1/6 = 5/6.
Remember

When the question says “at least one” or “not”, try the complement first.

Common mistake

Writing P(not E) = P(E) − 1.

Memory clue

"Chance of not equals one minus chance of."

Practice questions

Easy
  1. 1The probabilities of the elementary events of an experiment are 0.4, 0.35 and p. Find p.
  2. 2The probabilities of the elementary events of an experiment are 0.35, 0.1 and p. Find p.
  3. 3The probabilities of the elementary events of an experiment are 0.05, 0.35 and p. Find p.
Medium
  1. 4The probabilities of the elementary events of an experiment are 0.2, 0.4, 0.2 and p. Find p.
  2. 5The probability that Anil wins a game is 0.35. Find the probability that Anil loses.
  3. 6The probability that Sheela wins a game is 0.8. Find the probability that Sheela loses.
Challenging
  1. 7The probability that Reshma wins a game is 0.65. Find the probability that Reshma loses.
  2. 8The probability that Anil wins a game is 0.15. Find the probability that Anil loses.
Show answers (try them yourself first)
  1. 10.25
  2. 20.55
  3. 30.6
  4. 40.2
  5. 50.65
  6. 60.2
  7. 70.35
  8. 80.85

Playing cards and applications

From daily life

A deck of cards is the classic probability kit: 52 cards, four suits, each with 13.

Simple definition

A deck has 52 cards in four suits: spades and clubs (black), hearts and diamonds (red), 13 in each. The face cards are the king, queen and jack of each suit: 12 in all. There are 4 aces. Count with these numbers, then divide by 52.

Formula: total = 52

♠♣♥♦52 cardsface cards: 12red: 26aces: 4P(face) = 12/52= 3/13

In the figure: Face cards (K, Q, J) are the last three of each row: 12 of 52, so P = 3/13.

Rule 1: Common card events — Count from the structure of the deck.

  • Easy: P(ace) = 4/52 = 1/13; P(red card) = 26/52 = 1/2.
  • Medium: P(face card) = 12/52 = 3/13.
  • Exam-type: P(king of hearts) = 1/52; P(red king) = 2/52 = 1/26.

Rule 2: Use the complement here too — Not-a-face-card is the rest of the deck.

  • Easy: P(not a face card) = 1 − 3/13 = 10/13.
  • Medium: Check by counting: 52 − 12 = 40 and 40/52 = 10/13.
  • Exam-type: Face cards and aces are different: aces are not face cards.
Remember

Learn the deck once: 52, four suits of 13, 26 red, 12 faces, 4 aces.

Common mistake

Counting the ace as a face card.

Memory clue

"Fifty-two, twenty-six red, twelve faces."

Practice questions

Easy
  1. 1During a quiz round, a card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting a black card.
  2. 2During a quiz round, a card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting an ace.
  3. 3In a board game, one day is chosen at random. Find the probability that a day of the week starts with the letter S.
Medium
  1. 4In a class activity, a card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting a face card (jack, queen or king).
  2. 5One month is chosen at random. Find the probability that a month of the year begins with the letter J (English names).
  3. 6For a lucky-draw game, a card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability of getting a black ace.
Challenging
  1. 7The probability of selecting a faulty bulb at random from a box of 25 bulbs is 0.12. How many faulty bulbs are in the box?
  2. 8The probability of selecting a faulty bulb at random from a box of 60 bulbs is 0.15. How many faulty bulbs are in the box?
Show answers (try them yourself first)
  1. 11/2
  2. 21/13
  3. 32/7
  4. 43/13
  5. 51/4
  6. 61/26
  7. 73
  8. 89