Probability measures how likely an event is, as a number from 0 to 1. Class 10 moves from experimental probability, found by repeating trials, to theoretical probability, found by counting outcomes when they are equally likely. The chapter covers P(E) = favourable outcomes / all outcomes, elementary events, sure and impossible events, and complementary events. Board questions test careful counting with coins, dice, a deck of 52 cards, balls in a bag and defective items, including two dice thrown together and a second draw after one item is removed.
Key Concepts
1. Experiments, Outcomes and Equally Likely Outcomes
An experiment here means an action whose result depends on chance: tossing a coin, throwing a die, drawing a card or picking a ball from a bag. Each possible result is an outcome. An event is a collection of one or more outcomes that we are interested in, such as “getting an even number” when a die is thrown.
NCERT starts with a coin that is fair or unbiased: it is symmetrical, so there is no reason for it to fall more often on one side than the other. A random toss means the coin falls freely, with no bias or interference. The coin lands head up or tail up (landing on its edge is ignored), and each of these is as likely as the other. We say the outcomes head and tail are equally likely.
In this chapter a die always means a fair die. When it is thrown once, the outcomes are 1, 2, 3, 4, 5 and 6, and each number has the same chance of showing up.
Outcomes that are not equally likely
A bag has 4 red balls and 1 blue ball. If you draw one ball without looking, “red” and “blue” are the two colours you can get, but red is more likely because there are four red balls. So the outcomes “a red ball” and “a blue ball” are not equally likely. Each individual ball, though, is equally likely to be drawn. The trick in many problems is to count the individual balls, cards or coins as the outcomes, because those are the equally likely ones.
Some everyday “two-way” results are not equally likely either. A car starts or does not start: this depends on the car’s condition, so the two results are not equally likely. A player shoots a basketball and scores or misses: this depends on the player’s skill. From here on, NCERT assumes that all experiments in the chapter have equally likely outcomes.
Experimental probability versus theoretical probability
In Class 9 you met the experimental (empirical) probability:
P(E) = (number of trials in which the event happened) / (total number of trials)
This needs the experiment to be repeated many times. That works for coins and dice, but it is expensive or impossible for some situations, such as repeatedly launching a satellite to see how often it fails, or waiting for repeated earthquakes to see how often a tall building is destroyed. When we can make certain assumptions, such as equally likely outcomes, we can calculate the exact theoretical probability without repeating anything.
Experimental probability is based on what actually happened; theoretical probability predicts what will happen from assumptions. NCERT’s note to the reader links them: as the number of trials increases, we may expect the experimental and theoretical probabilities to be nearly the same.
2. Theoretical (Classical) Probability
The theoretical probability (also called classical probability) of an event E, written P(E), is
P(E) = (number of outcomes favourable to E) / (number of all possible outcomes of the experiment)
where the outcomes of the experiment are assumed to be equally likely. In this chapter “probability” means theoretical probability. This definition was given by Pierre Simon Laplace in 1795.
Probability theory began in the 16th century with J. Cardan’s The Book on Games of Chance, the first book on the subject.
A three-step method for every question
- List or count all possible outcomes. Make sure they are equally likely. For a bag of balls, count every ball.
- Count the outcomes favourable to the event. Write them out when the list is short, so the examiner can see them.
- Divide and simplify. Give the answer as a fraction in lowest terms (or a decimal if the question uses decimals).
Illustration: a die is thrown once. Find the probability of a number greater than 4. All outcomes: 1, 2, 3, 4, 5, 6, so 6 outcomes. Favourable: 5 and 6, so 2 outcomes. P = 2/6 = 1/3.
Elementary events
An event that has only one outcome of the experiment is an elementary event. When a coin is tossed, “head” and “tail” are elementary events. When a ball is drawn from a bag holding one red, one blue and one yellow ball, the three events “red”, “blue” and “yellow” are elementary, each with probability 1/3.
“A number greater than 4” on a die is not an elementary event, because it contains two outcomes (5 and 6).
Key fact: the sum of the probabilities of all the elementary events of an experiment is 1. For one coin, 1/2 + 1/2 = 1. For one die, six elementary events of 1/6 each add to 1. This gives a quick check on your working: if a bag has 3 blue, 2 white and 4 red marbles, then P(white) + P(blue) + P(red) = 2/9 + 3/9 + 4/9 = 1.
3. Sure Events, Impossible Events and the Range 0 to 1
Impossible event: an event that cannot happen. No outcome is favourable, so its probability is 0. Getting 8 on a single throw of a die is impossible, and P(getting 8) = 0/6 = 0.
Sure (certain) event: an event that is certain to happen. Every outcome is favourable, so its probability is 1. Every face of a die shows a number less than 7, so P(number less than 7) = 6/6 = 1.
In the formula, the number of favourable outcomes can never be more than the total number of outcomes, and it can never be negative. So for every event E:
0 ≤ P(E) ≤ 1
| Value | Can it be a probability? | Reason |
|---|---|---|
| 2/3 | Yes | Lies between 0 and 1 |
| −1.5 | No | Negative |
| 15% | Yes | 15% = 0.15, which lies between 0 and 1 |
| 0.7 | Yes | Lies between 0 and 1 |
| 5/4 | No | Greater than 1 |
NCERT Exercise 14.1 Q4 asks exactly this with the options 2/3, −1.5, 15% and 0.7: the answer is −1.5.
Illustration: a bag contains lemon flavoured candies only, and Malini takes one out without looking. P(orange flavoured) = 0, an impossible event. P(lemon flavoured) = 1, a sure event.
4. Complementary Events
The event “not E” is written Ē (E with a bar on top). It happens exactly when E does not happen. E and not E are called complementary events, and Ē is the complement of E.
Every outcome belongs to exactly one of E and not E, so their favourable counts add up to the total. That gives:
P(E) + P(not E) = 1, so P(not E) = 1 − P(E)
Checks from NCERT examples:
- Coin: P(head) + P(tail) = 1/2 + 1/2 = 1. Tail is “not head”.
- Die: P(number > 4) + P(number ≤ 4) = 1/3 + 2/3 = 1. “Not greater than 4” means “less than or equal to 4”.
- Cards: P(ace) = 4/52 = 1/13, so P(not an ace) = 1 − 1/13 = 12/13.
When to use the complement
Use 1 − P(E) when the event you want is long to count and its opposite is short. The classic signal is the phrase “at least one”.
Illustration: two coins are tossed. P(at least one head) = 1 − P(no head) = 1 − P(TT) = 1 − 1/4 = 3/4. Counting directly gives the same result: HH, HT, TH are 3 of the 4 outcomes.
Illustration with a given probability: if the probability of Sangeeta winning a tennis match against Reshma is 0.62, then Reshma’s probability of winning is 1 − 0.62 = 0.38, because one of the two must win and the events are complementary.
5. Playing Cards and Drawing at Random
The deck of 52 cards
A deck has 52 cards in 4 suits of 13 cards each: spades, hearts, diamonds and clubs. Clubs and spades are black; hearts and diamonds are red. Each suit has ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are face cards. A well-shuffled deck makes all 52 outcomes equally likely.
| Type of card | Number in the deck | Probability of drawing one |
|---|---|---|
| Any one suit (e.g. spades) | 13 | 13/52 = 1/4 |
| Red cards (or black cards) | 26 | 26/52 = 1/2 |
| Aces (or kings, or queens, or jacks) | 4 | 4/52 = 1/13 |
| Face cards (K, Q, J of all suits) | 12 | 12/52 = 3/13 |
| Red kings | 2 | 2/52 = 1/26 |
Aces are not face cards in the NCERT convention. Miscounting the face cards is a frequent slip: there are 12.
Balls, marbles, coins and defective items
“Drawn at random” is a short way of saying every item is equally likely to be drawn. The total number of outcomes is the total number of items.
In quality problems, read the acceptance rule carefully. In NCERT Example 12, a carton has 100 shirts: 88 good, 8 with minor defects and 4 with major defects. Jimmy accepts only good shirts, so his favourable count is 88. Sujatha rejects only shirts with major defects, so she accepts 88 + 8 = 96.
A second draw after one item is removed
When an item is drawn and not replaced, the next draw has one fewer item in the total, and the favourable count changes if the removed item was of that kind.
Illustration (NCERT Exercise 14.1 Q17): a lot of 20 bulbs has 4 defective. The first bulb drawn is not defective and is kept aside. Now 19 bulbs remain, of which 16 − 1 = 15 are good. P(second bulb not defective) = 15/19.
Illustration (NCERT Exercise 14.1 Q15): the ten, jack, queen, king and ace of diamonds are shuffled face down. P(queen) = 1/5. If the queen is drawn and put aside, 4 cards remain: P(ace) = 1/4 and P(queen) = 0/4 = 0.
6. Two Coins, Two Dice and Repeated Tosses
When two things happen together, list the outcomes as ordered pairs so that each listed outcome is equally likely.
Two coins
Tossing two different coins (say a ₹1 and a ₹2 coin) gives 4 equally likely outcomes: (H, H), (H, T), (T, H), (T, T). (H, T) and (T, H) are different outcomes: head on the first coin and tail on the second is different from tail on the first and head on the second.
| Event | Favourable outcomes | Probability |
|---|---|---|
| Two heads | HH | 1/4 |
| Exactly one head | HT, TH | 2/4 = 1/2 |
| At least one head | HH, HT, TH | 3/4 |
| No head | TT | 1/4 |
A common wrong argument says there are three outcomes (two heads, two tails, one of each), each with probability 1/3. These three are not equally likely, because “one of each” happens in two ways. NCERT Exercise 14.1 Q25(i) asks you to explain exactly this.
Three tosses of a coin
Tossing one coin 3 times gives 2 × 2 × 2 = 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Exactly one head: HTT, THT, TTH, probability 3/8. All the same: HHH, TTT, probability 2/8 = 1/4.
Two dice
When two dice (say one blue and one grey) are thrown together, each of the 6 numbers on the first die pairs with each of the 6 numbers on the second. So there are 6 × 6 = 36 equally likely outcomes, from (1, 1) to (6, 6). (1, 4) and (4, 1) are different outcomes. NCERT treats throwing one die twice as the same experiment as throwing two dice at once.
The six outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) are the doublets, so P(same number on both dice) = 6/36 = 1/6.
Sums on two dice: count the pairs for each sum.
| Sum | Favourable pairs | Number | Probability |
|---|---|---|---|
| 2 | (1, 1) | 1 | 1/36 |
| 3 | (1, 2), (2, 1) | 2 | 2/36 = 1/18 |
| 4 | (1, 3), (2, 2), (3, 1) | 3 | 3/36 = 1/12 |
| 5 | (1, 4), (2, 3), (3, 2), (4, 1) | 4 | 4/36 = 1/9 |
| 6 | (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) | 5 | 5/36 |
| 7 | (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) | 6 | 6/36 = 1/6 |
| 8 | (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) | 5 | 5/36 |
| 9 | (3, 6), (4, 5), (5, 4), (6, 3) | 4 | 4/36 = 1/9 |
| 10 | (4, 6), (5, 5), (6, 4) | 3 | 3/36 = 1/12 |
| 11 | (5, 6), (6, 5) | 2 | 2/36 = 1/18 |
| 12 | (6, 6) | 1 | 1/36 |
The counts 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36, so the probabilities add up to 1. The most likely sum is 7. A sum of 13 is impossible (probability 0) and a sum of at most 12 is sure (probability 1).
The 11 possible sums (2 to 12) are not equally likely, so it is wrong to give each sum probability 1/11. NCERT Exercise 14.1 Q22(ii) asks you to justify this.
7. Probability with Lengths and Areas (for understanding)
All the experiments so far had a finite number of outcomes. Some experiments have infinitely many outcomes, such as any number between 0 and 2, or any point inside a region. Outcomes cannot be counted there, so NCERT uses a ratio of lengths or areas instead. NCERT marks these examples with a star as not from the examination point of view, so read this section for understanding only.
Length: music may stop at any time within 2 minutes. P(it stops within the first half-minute) = (favourable length) / (total length) = (1/2) / 2 = 1/4.
Area: a missing helicopter is equally likely to have crashed anywhere in a rectangular region of 4.5 km by 9 km, which is 40.5 km2. A lake inside it measures 2.5 km by 3 km, which is 7.5 km2. P(crashed in the lake) = 7.5/40.5 = 75/405 = 5/27.
Formula and Theorem Sheet
| Result | Statement | Remember |
|---|---|---|
| Theoretical probability | P(E) = (outcomes favourable to E) / (all possible outcomes) | Only when outcomes are equally likely |
| Range | 0 ≤ P(E) ≤ 1 | No negative value, nothing above 1 |
| Sure event | P = 1 | Number less than 7 on a die |
| Impossible event | P = 0 | Number 8 on a die, sum 13 on two dice |
| Elementary events | Sum of their probabilities = 1 | Each has exactly one outcome |
| Complement | P(E) + P(not E) = 1; P(not E) = 1 − P(E) | Use for “not” and “at least one” |
| Coins | n coins (or n tosses) give 2n outcomes | 1 coin: 2, 2 coins: 4, 3 coins: 8 |
| Dice | One die: 6 outcomes; two dice: 36 | One die thrown twice = two dice |
| Cards | 52 cards, 26 red, 26 black, 13 per suit, 12 face cards, 4 of each rank | Aces are not face cards |
| Without replacement | Second draw: total and favourable count both change | 20 bulbs → 19 left |
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Important Definitions
- Experiment: an action with results that depend on chance, such as tossing a coin or throwing a die.
- Outcome: one possible result of an experiment.
- Event: a collection of one or more outcomes of an experiment.
- Fair (unbiased) coin: a coin that is symmetrical, so there is no reason for it to fall more often on one side than the other.
- Random toss or random draw: the coin falls freely, or the item is picked without looking, so there is no bias or interference and every outcome is equally likely.
- Equally likely outcomes: outcomes that each have the same chance of occurring.
- Experimental (empirical) probability: the number of trials in which the event happened divided by the total number of trials.
- Theoretical (classical) probability: P(E) = number of outcomes favourable to E divided by the number of all possible outcomes, assuming the outcomes are equally likely.
- Favourable outcomes: the outcomes of an experiment for which the event E happens.
- Elementary event: an event having only one outcome of the experiment.
- Sure (certain) event: an event that is certain to happen; its probability is 1.
- Impossible event: an event that cannot happen; its probability is 0.
- Complementary event: the event “not E”, written Ē, which happens exactly when E does not; P(Ē) = 1 − P(E).
- Face cards: the kings, queens and jacks of a deck of playing cards (12 in all).
Solved Examples (NCERT-Based)
Example 1: Complete the statements (NCERT Exercise 14.1 Q1)
Solution:
- Probability of an event E + Probability of the event “not E” = 1.
- The probability of an event that cannot happen is 0. Such an event is called an impossible event.
- The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
- The sum of the probabilities of all the elementary events of an experiment is 1.
- The probability of an event is greater than or equal to 0 and less than or equal to 1.
Example 2: Which experiments have equally likely outcomes? (NCERT Exercise 14.1 Q2)
Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. (iii) A trial is made to answer a true-false question. The answer is right or wrong. (iv) A baby is born. It is a boy or a girl.
Solution:
- Not equally likely. Whether the car starts depends on its condition, such as the battery and fuel.
- Not equally likely. Whether the shot goes in depends on the player’s skill.
- Equally likely. With two choices and a guess, right and wrong each have the same chance.
- Equally likely. The baby is either a boy or a girl, and the chapter takes each of the two outcomes to have the same chance.
Q3 of the same exercise: tossing a coin is a fair way to decide which team gets the ball first because the outcomes head and tail are equally likely, so each team has the same probability, 1/2, of winning the toss.
Example 3: A die is thrown once (NCERT Exercise 14.1 Q13)
A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Solution: Possible outcomes: 1, 2, 3, 4, 5, 6, so 6 outcomes.
- Primes: 2, 3, 5 (1 is not prime). 3 favourable. P = 3/6 = 1/2.
- Between 2 and 6: 3, 4, 5 (2 and 6 are excluded). 3 favourable. P = 3/6 = 1/2.
- Odd: 1, 3, 5. 3 favourable. P = 3/6 = 1/2.
Example 4: One card from a deck of 52 (NCERT Exercise 14.1 Q14)
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.
Solution: A well-shuffled deck has 52 equally likely outcomes.
- Red kings: king of hearts and king of diamonds, so 2. P = 2/52 = 1/26.
- Face cards: 3 in each suit × 4 suits = 12. P = 12/52 = 3/13.
- Red face cards: 3 in hearts + 3 in diamonds = 6. P = 6/52 = 3/26.
- Jack of hearts: 1 card. P = 1/52.
- Spades: 13. P = 13/52 = 1/4.
- Queen of diamonds: 1 card. P = 1/52.
Example 5: Marbles in a box (NCERT Exercise 14.1 Q9)
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Solution: Total marbles = 5 + 8 + 4 = 17, so 17 equally likely outcomes.
- P(red) = 5/17.
- P(white) = 8/17.
- P(green) = 4/17, so P(not green) = 1 − 4/17 = 13/17. Directly: red + white = 13 marbles, 13/17.
Example 6: Probability of “not E” and the birthday question (NCERT Exercise 14.1 Q5 and Q7)
(a) If P(E) = 0.05, what is the probability of “not E”?
P(not E) = 1 − P(E) = 1 − 0.05 = 0.95.
(b) It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
“Same birthday” and “not the same birthday” are complementary. P(same birthday) = 1 − 0.992 = 0.008.
Example 7: The spinning arrow (NCERT Exercise 14.1 Q12)
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Solution: 8 equally likely outcomes.
- Only one 8. P = 1/8.
- Odd numbers: 1, 3, 5, 7, so 4. P = 4/8 = 1/2.
- Greater than 2: 3, 4, 5, 6, 7, 8, so 6. P = 6/8 = 3/4.
- Less than 9: all 8 numbers. P = 8/8 = 1, a sure event.
Example 8: Defective pens and the shop (NCERT Exercise 14.1 Q16 and Q21)
(a) 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Total pens = 12 + 132 = 144. P(good) = 132/144 = 11/12.
(b) A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?
Good pens = 144 − 20 = 124.
- P(she buys it) = 124/144 = 31/36.
- P(she does not buy it) = 20/144 = 5/36. Check: 31/36 + 5/36 = 1.
Example 9: Bulbs without replacement (NCERT Exercise 14.1 Q17)
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
- P(defective) = 4/20 = 1/5.
- Good bulbs at the start = 20 − 4 = 16. One good bulb is removed, so 19 bulbs remain, of which 15 are good. P(not defective) = 15/19.
Example 10: Discs numbered 1 to 90 (NCERT Exercise 14.1 Q18)
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Solution: 90 equally likely outcomes.
- One-digit numbers are 1 to 9, which is 9 numbers. Two-digit numbers = 90 − 9 = 81. P = 81/90 = 9/10.
- Perfect squares up to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81, which is 9 numbers (102 = 100 is too big). P = 9/90 = 1/10.
- Multiples of 5 up to 90: 5, 10, …, 90, which is 90 ÷ 5 = 18 numbers. P = 18/90 = 1/5.
Example 11: The lettered die (NCERT Exercise 14.1 Q19)
A child has a die whose six faces show the letters A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution: The die has 6 faces, so 6 equally likely outcomes. The letter A appears on 2 faces.
- P(A) = 2/6 = 1/3.
- P(D) = 1/6.
Example 12: Sum on two dice (NCERT Example 13)
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12?
Solution: Each die shows 1 to 6, so the number of outcomes = 6 × 6 = 36: all ordered pairs from (1, 1) to (6, 6), with the blue die first; each pair appears once in the sums table of Key Concept 6.
- Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), so 5 outcomes. P = 5/36.
- The largest possible sum is 6 + 6 = 12, so no outcome gives 13. P = 0/36 = 0.
- Every outcome has a sum of at most 12. P = 36/36 = 1.
Example 13: A die thrown twice (NCERT Exercise 14.1 Q24)
A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
Solution: Throwing a die twice is the same experiment as throwing two dice, so there are 36 outcomes.
- 5 does not come up either time: each throw has 5 choices (1, 2, 3, 4, 6), so 5 × 5 = 25 outcomes. P = 25/36.
- “At least once” is the complement of “not either time”. P = 1 − 25/36 = 11/36. Direct count: (5, 1) to (5, 6) gives 6, (1, 5) to (6, 5) gives 6, and (5, 5) was counted twice, so 6 + 6 − 1 = 11.
Example 14: Hanif’s coin game (NCERT Exercise 14.1 Q23)
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution: Outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, so 8 in all.
Hanif wins on HHH and TTT: P(win) = 2/8 = 1/4.
P(lose) = 1 − 1/4 = 3/4. The 6 losing outcomes are HHT, HTH, THH, HTT, THT, TTH, and 6/8 = 3/4 agrees.
Competency-Based Questions (with answers)
1. Case-based: The class representative
A Class X section has 40 students: 25 girls and 15 boys. The teacher writes each name on an identical card, stirs the cards in a bag and draws one to choose the class representative.
(a) What is the probability that a girl is chosen? P(girl) = 25/40 = 5/8.
(b) What is the probability that a boy is chosen? P(boy) = 15/40 = 3/8, or 1 − 5/8 = 3/8.
(c) Before the draw, 5 girls are absent and their cards are removed. Find the new probability that a girl is chosen. Cards left = 35, girls = 20. P(girl) = 20/35 = 4/7.
2. Case-based: The school fair game
At a fair stall, a player throws two dice together. The player wins a prize if the sum is 7 or 11, and wins a small sticker if both dice show the same number.
(a) How many outcomes are possible? 6 × 6 = 36.
(b) Find P(prize). Sum 7: 6 pairs. Sum 11: (5, 6), (6, 5), 2 pairs. Total 8. P = 8/36 = 2/9.
(c) Find P(sticker). Doublets: (1, 1) to (6, 6), 6 outcomes. P = 6/36 = 1/6.
3. Case-based: Shirts in a carton (NCERT Example 12)
A carton has 100 shirts: 88 good, 8 with minor defects and 4 with major defects. Jimmy accepts only good shirts. Sujatha rejects only shirts with major defects. One shirt is drawn at random.
(a) P(acceptable to Jimmy)? 88/100 = 0.88.
(b) P(acceptable to Sujatha)? (88 + 8)/100 = 96/100 = 0.96.
(c) P(rejected by both)? Only a shirt with a major defect is rejected by Sujatha, and Jimmy also rejects it. P = 4/100 = 0.04.
4. Source-based: Laplace’s definition
The chapter states: “The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as P(E) = Number of outcomes favourable to E / Number of all possible outcomes of the experiment, where we assume that the outcomes of the experiment are equally likely.” It also says the definition was given by Pierre Simon Laplace in 1795.
(a) State the assumption on which the definition depends. All outcomes of the experiment are equally likely.
(b) Explain why this definition cannot give the probability of the colour of a ball from a bag of 4 red and 1 blue ball by treating “red” and “blue” as the only two outcomes. The two colours are not equally likely. Treating the 5 balls as the outcomes gives P(red) = 4/5 and P(blue) = 1/5.
(c) Name the earlier type of probability that needs repeated trials. Experimental (empirical) probability.
5. Assertion-Reason
Assertion (A): The probability of getting a number less than 7 in a single throw of a die is 1.
Reason (R): The probability of a sure event is 1.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). Every face shows a number less than 7, so the event is sure and its probability is 6/6 = 1. R explains A.
6. Assertion-Reason
Assertion (A): When two coins are tossed, the probability of getting one head and one tail is 1/3.
Reason (R): When two coins are tossed, the possible results are two heads, two tails, and one of each.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (d). R correctly names the three kinds of result, but they are not equally likely. The equally likely outcomes are HH, HT, TH, TT, so P(one head and one tail) = 2/4 = 1/2. A is false.
7. Assertion-Reason
Assertion (A): If P(E) = 0.05, then P(not E) = 0.95.
Reason (R): For any event E, P(E) + P(not E) = 1.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). R is the complement rule, which is true for every event. It gives P(not E) = 1 − P(E) = 1 − 0.05 = 0.95, so A is true and R explains it.
8. Error analysis: The eleven sums
A student argues: “When two dice are thrown, there are 11 possible sums, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11.” (NCERT Exercise 14.1 Q22(ii)) Do you agree? Justify.
Answer: No. The 11 sums are not equally likely. The 36 ordered pairs are the equally likely outcomes. Sum 2 arises only from (1, 1), so P(2) = 1/36, while sum 7 arises from 6 pairs, so P(7) = 6/36 = 1/6. The probabilities of the sums are 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36.
9. Competency MCQ: The second card
From the five cards ten, jack, queen, king and ace of diamonds, the king is drawn and put aside. A second card is drawn at random. The probability that it is the king is:
(a) 1/5 (b) 1/4 (c) 0 (d) 1
Answer: (c). The only king has been removed, so the event is impossible and its probability is 0/4 = 0.
Important Questions for Board Exams
1-Mark Questions
- Which of the following cannot be the probability of an event? (A) 2/3 (B) −1.5 (C) 15% (D) 0.7 (B) −1.5, because a probability cannot be negative.
- If P(E) = 0.05, what is the probability of “not E”? 1 − 0.05 = 0.95.
- A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy? 0, an impossible event.
- One card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is an ace. 4/52 = 1/13.
- Two coins are tossed simultaneously. Find the probability of getting no head. Only TT is favourable, so 1/4.
2-Mark Questions
- A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red? Total = 8. (i) 3/8. (ii) 1 − 3/8 = 5/8.
- Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish? Total = 13. P(male) = 5/13.
- Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). Hamida’s birthday can be any of 365 equally likely days. (i) Different: 364 favourable days, P = 364/365. (ii) Same: 1 − 364/365 = 1/365.
- Harpreet tosses two different coins simultaneously. What is the probability that she gets at least one head? Outcomes HH, HT, TH, TT. At least one head: HH, HT, TH. P = 3/4.
3-Mark Questions
- A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a ₹5 coin? Total = 100 + 50 + 20 + 10 = 180. (i) 100/180 = 5/9. (ii) (180 − 10)/180 = 170/180 = 17/18.
- Five cards, the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen? (i) 1/5. (ii) 4 cards remain: (a) 1/4; (b) 0, because no queen is left.
- A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red? Total = 9. (i) 2/9. (ii) 3/9 = 1/3. (iii) 4/9. Check: 2/9 + 3/9 + 4/9 = 1.
- Which of the following arguments are correct and which are not correct? Give reasons. (i) If two coins are tossed simultaneously there are three possible outcomes, two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3. (ii) If a die is thrown, there are two possible outcomes, an odd number or an even number. Therefore, the probability of getting an odd number is 1/2. (i) Incorrect: the equally likely outcomes are HH, HT, TH, TT, so P(two heads) = 1/4, P(two tails) = 1/4 and P(one of each) = 2/4 = 1/2. (ii) Correct: odd (1, 3, 5) and even (2, 4, 6) each have 3 of the 6 outcomes, so P(odd) = 1/2.
5-Mark Questions
- Two dice are thrown at the same time. Find the probability that (i) the sum is 8 (ii) the sum is a prime number (iii) the same number appears on both dice (iv) the product is 6 (v) the sum is at least 10. Total outcomes = 36.
(i) (2, 6), (3, 5), (4, 4), (5, 3), (6, 2): 5/36.
(ii) Prime sums 2, 3, 5, 7, 11 have 1 + 2 + 4 + 6 + 2 = 15 pairs: 15/36 = 5/12.
(iii) Doublets: 6/36 = 1/6.
(iv) Product 6: (1, 6), (6, 1), (2, 3), (3, 2): 4/36 = 1/9.
(v) Sum 10, 11 or 12: 3 + 2 + 1 = 6 pairs: 6/36 = 1/6. - A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that it is (i) a red king (ii) a face card (iii) a red face card (iv) a black card that is not a face card (v) neither a heart nor a king.
(i) 2/52 = 1/26.
(ii) 12/52 = 3/13.
(iii) 6/52 = 3/26.
(iv) Black cards 26, black face cards 6, so 20. P = 20/52 = 5/13.
(v) Hearts 13, kings not in hearts 3, so “heart or king” covers 16 cards. Neither: 52 − 16 = 36. P = 36/52 = 9/13.
Common Mistakes and Examiner Tips
- Treating unequal results as equally likely. Two coins do not give “three outcomes each 1/3”, and two dice do not give “eleven sums each 1/11”. Fix: always list the basic ordered outcomes (HH, HT, TH, TT; the 36 pairs) and count from them.
- Writing (1, 4) and (4, 1) as the same outcome. Fix: with two dice or two coins, the first entry is always the first die or coin. (1, 4) and (4, 1) are different outcomes.
- Counting 16 face cards. Fix: face cards are kings, queens and jacks only, 3 per suit, 12 in all. Aces are not face cards.
- Using the number of colours as the total. A bag with 3 red and 5 black balls has 8 outcomes, not 2. Fix: the total is the number of items.
- Forgetting to reduce both counts after removing an item. Fix: for a second draw without replacement, reduce the total by 1, and reduce the favourable count by 1 if the removed item was of that kind (20 bulbs, 16 good → 19 bulbs, 15 good).
- Including the end numbers in “between”. “Between 2 and 6” on a die means 3, 4, 5. Fix: read “between”, “less than”, “at most” and “at least” carefully and write the favourable outcomes out.
- Counting 1 as a prime. Primes on a die are 2, 3 and 5 only. Fix: 1 is neither prime nor composite.
- Giving an answer above 1 or below 0. Fix: if a probability comes out as 7/5 or negative, the counts are swapped or wrong. Check that 0 ≤ P(E) ≤ 1 before moving on.
- Counting “at least one” the long way and missing cases. Fix: use P(at least one) = 1 − P(none). For a die thrown twice, P(5 at least once) = 1 − 25/36 = 11/36.
- Counting a double outcome twice. When counting “5 on the first throw or the second”, (5, 5) appears in both lists. Fix: subtract it once, giving 11.
Quick Revision Points
- Theoretical probability: P(E) = favourable outcomes / all possible outcomes, with equally likely outcomes.
- Experimental probability: trials in which E happened / total trials.
- Laplace gave the classical definition in 1795; J. Cardan wrote the first book on probability.
- In a bag of 4 red and 1 blue ball, “red” and “blue” are not equally likely; each ball is.
- Sure event: probability 1. Impossible event: probability 0.
- 0 ≤ P(E) ≤ 1 for every event. Negative values and values above 1 are never probabilities.
- An elementary event has exactly one outcome.
- Probabilities of all elementary events of an experiment add up to 1.
- P(E) + P(not E) = 1, so P(not E) = 1 − P(E).
- “At least one” = 1 − P(none).
- One coin: 2 outcomes. Two coins: 4. Three tosses: 8.
- One die: 6 outcomes. Two dice, or one die thrown twice: 36.
- Most likely sum on two dice is 7 (probability 1/6); sums 2 and 12 each have probability 1/36.
- Doublets on two dice: 6 of 36, probability 1/6.
- Deck: 52 cards, 4 suits of 13; hearts and diamonds red, spades and clubs black.
- Face cards: kings, queens, jacks, so 12; red face cards 6; aces 4.
- Without replacement: the total drops by 1 for the next draw.
- Probability by lengths or areas is for understanding; NCERT marks those examples as not from the examination point of view.
Weightage in Board Exams
Check the current CBSE Class 10 Maths course structure for the marks given to this chapter. The way it is assessed is predictable.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Which value cannot be a probability; P(not E) from P(E); sure and impossible events; the two-coin or eleven-sums fallacy |
| Short answers | Balls, marbles, coins or pens in a bag; one card from a deck; one die with primes, odd numbers or “between” |
| Longer answers | Two dice with several parts (sums, doublets, products); cards with several parts; numbered discs or tickets |
| Case-based | A real setting such as choosing a representative, a fair game or quality checking, split into two or three parts |
The skills that decide the marks are writing the total number of outcomes correctly, listing the favourable ones, and simplifying the fraction. Work through every question of NCERT Exercise 14.1 (Q20 is starred as not from the examination point of view), then practise two-dice and card questions with several parts.
Class 10 Maths Β· Chapter 14 β swipe through all 10 cards to understand the whole chapter.
Equally Likely Outcomes
Outcomes that each have the same chance of happening, like head and tail on a fair coin.
A bag of 4 red and 1 blue ball: the colours are not equally likely, but each ball is.
- A fair die gives 1, 2, 3, 4, 5, 6, each equally likely.
- A car starting or not starting is not equally likely.
- Count individual items as outcomes, never colours or kinds.
Theoretical Probability
Count the favourable outcomes and divide by all possible outcomes.
Valid only when all outcomes are equally likely. Given by Laplace in 1795.
- Die, number greater than 4: 2/6 = 1/3.
- Bag of 3 red and 5 black, red: 3/8.
- Always simplify the fraction to lowest terms.
Experimental Probability
Found from actual trials rather than by counting outcomes.
Repeating some experiments, like satellite launches, is costly or impossible.
- Based on what has actually happened.
- Theoretical probability predicts using assumptions.
- With many trials, the two are expected to come close.
Sure and Impossible Events
Every probability lies between 0 and 1, both included.
Negative values and values above 1 are never probabilities.
- Impossible: 8 on a die, P = 0.
- Sure: a number less than 7 on a die, P = 1.
- β1.5 cannot be a probability; 15% = 0.15 can.
Elementary Events
An event having only one outcome of the experiment.
‘An even number on a die’ is not elementary: it has 3 outcomes.
- Head and tail on one coin: 1/2 + 1/2 = 1.
- 3 blue, 2 white, 4 red marbles: 3/9 + 2/9 + 4/9 = 1.
- Use the sum as a check on your answers.
Complementary Events
‘Not E’ happens exactly when E does not.
Use it for every ‘at least one’ question: 1 β P(none).
- P(E) = 0.05 gives P(not E) = 0.95.
- P(ace) = 1/13, so P(not an ace) = 12/13.
- Same birthday = 1 β 364/365 = 1/365.
The Deck of 52 Cards
4 suits of 13: spades and clubs black, hearts and diamonds red.
Face cards are kings, queens and jacks only. Aces are not face cards.
- Red king: 2/52 = 1/26.
- Red face card: 6/52 = 3/26.
- A spade: 13/52 = 1/4; jack of hearts: 1/52.
Two Coins and Three Tosses
List ordered outcomes so that each one is equally likely.
HT and TH are different outcomes.
- Two coins: HH, HT, TH, TT.
- At least one head: 3/4; exactly one head: 1/2.
- Three tosses: 8 outcomes; all the same = 2/8 = 1/4.
Two Dice
Each die shows 1 to 6, so there are 36 ordered pairs.
One die thrown twice is the same experiment as two dice thrown together.
- Sum 8: 5/36; sum 7: 6/36 = 1/6; sum 13: 0.
- Doublets (same number): 6/36 = 1/6.
- The 11 sums are not equally likely, so none is 1/11.
Drawing Without Replacement
After an item is removed, the next draw has one fewer item.
Reduce the favourable count too if the removed item was of that kind.
- 20 bulbs, 4 defective: P(defective) = 1/5.
- Good bulb removed: P(next is good) = 15/19.
- Queen removed from 5 cards: P(queen) = 0.
π Practice Probability - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why C: A year has 52 full weeks plus extra days. A leap year has 366 = 52 Γ 7 + 2 days, and the 2 extra days are one of 7 equally likely pairs (Sun-Mon, Mon-Tue, …, Sat-Sun); Monday is in 2 of them, so P = 2/7 and A is true. A non-leap year has 365 = 52 Γ 7 + 1 day, the extra day is one of 7 days, so P(53 Mondays) = 1/7, not 5/7; R is false.
Why not A: R is false, so it cannot be true or explain A.
Why not B: R is false, since one extra day gives 1/7.
Why not D: A is true; the 2 extra days of a leap year give Monday 2 chances out of 7 pairs.
Remember: leap year 2 extra days, 2/7; ordinary year 1 extra day, 1/7.
Why B: Since P(E) β€ 1, 0.2p β€ 1 gives p β€ 5, so A is true. R, the complement rule, is also true. But A follows from the fact that a probability cannot exceed 1, not from the complement rule, so R does not explain A.
Why not A: R is true but is not the reason; the bound comes from P(E) β€ 1.
Why not C: R is true; P(Δ) = 1 – P(E) holds for every event.
Why not D: A is true; if p were more than 5, 0.2p would be above 1, which no probability can be.
Remember: to bound an unknown probability use 0 β€ P(E) β€ 1.
Why C: Step 1 (the event ‘3’): P(3) = 1/6. Step 2 (complement rule): P(other than 3) = 1 – 1/6 = 5/6, which matches the five faces 1, 2, 4, 5, 6.
Why not A: 1/6 is the probability of getting 3 itself.
Why not B: 3/6 confuses the number 3 with the count of favourable outcomes.
Why not D: 1 would make ‘other than 3’ a sure event, but 3 can still turn up.
Remember: other than one face of a die = 5/6.
Why B: Step 1 (count): there are 4 tens and 4 kings, and no card is both, so 8 favourable cards. Step 2 (classical probability): P = 8/52 = 2/13.
Why not A: 1/26 = 2/52 counts only one ten and one king, not one of each suit.
Why not C: 1/13 = 4/52 counts only the tens or only the kings.
Why not D: 8/26 has the right 8 favourable cards but divides by 26 instead of 52.
Remember: each rank has 4 cards, one in each suit.
Why C: Step 1 (each die): numbers less than 4 are 1, 2, 3, so 3 choices per die. Step 2 (pairs): 3 Γ 3 = 9 ordered pairs out of 36. Step 3 (simplify): 9/36 = 1/4.
Why not A: 2/9 = 8/36 is a miscount of the favourable pairs.
Why not B: 7/36 counts too few pairs, as if some ordered pairs like (1, 2) and (2, 1) were merged.
Why not D: 2/3 = 24/36 is far more than the 9 pairs in which both dice show 1, 2 or 3.
Remember: ‘less than 4’ means 1, 2, 3, and both dice gives 3 Γ 3.
Why A: Step 1 (classical probability): P(win) = tickets she holds / 800 = 0.08. Step 2 (solve): tickets = 0.08 Γ 800 = 64.
Why not B: 640 multiplies by 0.8 instead of 0.08.
Why not C: 100 comes from 800 / 8, dividing instead of multiplying.
Why not D: 10 comes from 800 / 80 or 0.08 read as 1/80, also a division slip.
Remember: favourable = P(E) Γ total.
Why C: 10/0.2 = 100/2 = 50, which is greater than 1. A probability must lie between 0 and 1, so 50 cannot be a probability.
Why not A: 39/100 = 0.39 lies between 0 and 1, so it is valid.
Why not B: 0.001/20 = 0.00005 is tiny but still between 0 and 1.
Why not D: 10% = 0.1, which is a valid probability.
Remember: dividing by a decimal below 1 makes a number bigger; check the result against 1.
Why C: Step 1 (check each face): 36 is divisible by 1, 2, 3, 4 and 6 but not by 5, so 5 faces are factors of 36. Step 2 (complement rule): P(not a factor) = 1 – 5/6 = 1/6; only the face 5 is favourable.
Why not A: 1/2 assumes three faces are not factors, but only 5 fails.
Why not B: 2/3 counts four faces as non-factors, which is a checking slip.
Why not D: 5/6 is the probability that the number IS a factor of 36.
Remember: test each face 1 to 6; only 5 does not divide 36.
Why D: Step 1 (deck facts): each suit has 3 face cards (jack, queen, king), and the red suits are hearts and diamonds, so red face cards = 2 Γ 3 = 6. Step 2 (classical probability): P = 6/52 = 3/26.
Why not A: 3/13 = 12/52 counts all 12 face cards, both red and black.
Why not B: 2/13 = 8/52 counts 4 face cards per red suit, wrongly taking the ace as a face card.
Why not C: 1/2 is P(red card), which ignores the face-card condition.
Remember: face cards are J, Q, K only; 12 in all, 6 red.
Why D: Step 1 (sample space): two dice give 36 equally likely ordered pairs. Step 2 (more than 10 means 11 or 12): sum 11 is (5, 6) and (6, 5); sum 12 is (6, 6); so 3 favourable pairs. Step 3 (classical probability): P = 3/36 = 1/12.
Why not A: 1/9 = 4/36 is a miscount of the favourable pairs.
Why not B: 1/6 = 6/36 is P(sum β₯ 10), because it also counts the three pairs with sum exactly 10.
Why not C: 7/12 = 21/36 is far too many; only three pairs reach 11 or 12.
Remember: ‘more than 10’ starts at 11; only (5, 6), (6, 5), (6, 6) qualify.
Chapter Navigation
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Frequently Asked Questions
Experimental probability is based on what actually happened: trials in which the event happened divided by the total number of trials. Theoretical probability predicts what will happen by counting: favourable outcomes divided by all possible outcomes, assuming the outcomes are equally likely. As the number of trials grows, the two are expected to come close to each other.
An elementary event is an event with only one outcome of the experiment. Getting a head on one coin, or getting a 4 on one die, are elementary events. Getting an even number on a die is not, because it has three outcomes. The probabilities of all the elementary events of an experiment always add up to 1.
Any negative number and any number greater than 1. For every event, 0 β€ P(E) β€ 1, because the favourable outcomes can never be negative or more than the total. So β1.5, 1.02 and 5/4 are not probabilities, while 2/3, 0.7 and 15% (which is 0.15) can be. Probability 0 means an impossible event and 1 means a sure event.
The event ‘not E’ happens exactly when E does not, so P(E) + P(not E) = 1 and P(not E) = 1 β P(E). If P(E) = 0.05, then P(not E) = 0.95. It saves time for ‘at least one’ questions: for a die thrown twice, P(5 at least once) = 1 β P(no 5) = 1 β 25/36 = 11/36.
There are 12 face cards: the king, queen and jack of each of the four suits. Aces are not face cards. Six face cards are red (hearts and diamonds) and six are black (spades and clubs). So P(face card) = 12/52 = 3/13 and P(red face card) = 6/52 = 3/26.
Each of the 6 numbers on the first die can pair with each of the 6 numbers on the second, giving 6 Γ 6 = 36 ordered pairs. (1, 4) and (4, 1) are different outcomes. Throwing one die twice is treated as the same experiment. The 11 possible sums are not equally likely: sum 7 has 6 pairs, sum 2 has only 1.
The total number of items for the next draw drops by one, and if the removed item was of the kind you want, the favourable count also drops by one. From 20 bulbs with 4 defective, if a good bulb is removed, 19 remain with 15 good ones, so the chance that the next bulb is good is 15/19.
