Probability: Class 10 Maths · Chapter 14

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
3 simulations
33 practice questions
25 MCQs

What you will learn in Probability

Decide whether the outcomes of an experiment are equally likely, and explain why theoretical probability depends on that assumption.
Find the theoretical probability of an event as favourable outcomes divided by all outcomes, for coins, dice, cards, spinners and draws from a bag.
Recognise elementary events and use the fact that their probabilities add up to 1.
Use the complement rule to find the probability of "not E", and choose whichever of an event and its complement is easier to count.
Identify impossible and sure events and explain why every probability lies from 0 to 1.
List ordered outcomes for two coins and two dice, handle draws without replacement, and spot arguments that treat unequal results as equally likely.
Extend probability to lengths and areas when outcomes cannot be counted (beyond the board syllabus).

Original companion text aligned to NCERT Mathematics Class 10, Chapter 14, pp. 202–217. Not affiliated with or endorsed by NCERT.

14.1 Probability — A Theoretical Approach

Before a gully-cricket match, one captain flips a coin and the other calls. Nobody can say which face will land up, yet both captains accept the toss as fair. Why? Because each face has the same chance. In Class 9 you estimated chances by repeating an experiment many times and counting the results. In this chapter you will calculate chances exactly, by reasoning about the possible outcomes, as long as one condition holds: the outcomes must be equally likely.

Equally likely outcomes

A coin shows heads (H) or tails (T). When do these two results have exactly the same chance? Two things must be true. The coin must be fair: evenly made, with neither side heavier or shaped to land up more often. And the toss must be made at random (यादृच्छया): the coin spins and drops freely, and nobody steers it. Then H and T are equally likely (समप्रायिक) outcomes.

A die behaves in the same way. If its six faces differ only in their dots, none of the numbers 1 to 6 is favoured. Unless a question says otherwise, every coin and every die in this chapter is a fair one.

Definition

Equally likely outcomes

Outcomes of an experiment are equally likely when each of them has the same chance of happening, because nothing in the set-up favours any one of them.

ExampleThe six faces of a fair die; the 52 cards of a well-shuffled deck.

Be careful, though. The results we care about are not always equally likely, even when the objects are. At a school mela, a bowl holds six identical folded chits: five say Try again and one says Prize. Is winning as likely as not winning? Follow Fig. 14.1.

3/3Try again: $\frac{5}{6} .$ Prize: $\frac{1}{6} .$ The two results are not equally likely.

Fig. 14.1The six chits are equally likely; the two results are not.NCERT p. 203

Six identical folded paper chits numbered 1 to 6 stand in a row. In the first step they are closed, and a line below says that any one chit has probability one-sixth. In the second step the chits are opened: chits 1 to 5 show a cross, meaning try again, and chit 6 shows a star, meaning prize; braces group the five try-again chits and the single prize chit. In the third step two bars compare the two results: the try-again bar is five times as long as the prize bar, and they are labelled five-sixths and one-sixth.

So the safe way to list outcomes is chit by chit, not result by result. Prize and try again are not equally likely, but each of the six chits is. The same trap appears whenever objects come in unequal numbers, as in a bag holding more balls of one colour than of another. Every problem in the rest of this chapter takes one thing for granted: the outcomes we list are equally likely.

SimulationTwo colours, unequal sectors

A spinner whose red sector covers three-quarters of the wheel and whose blue sector covers the remaining quarter. You can spin it 1, 10, 100 or 1000 times at a press; bars show how often each colour has come up, next to the theoretical shares of three-quarters and one-quarter.

Fig. 14.2A wheel that is three-quarters red. Are red and blue equally likely?

Try this

Before you spin: is the chance of red $\frac{1}{2}$ ? Spin 100 times and check.
Imagine the red part cut into three equal sectors. How many equal sectors does the wheel have then, and how many of them are red?

Theoretical probability

In Class 9 you met experimental probability: repeat an experiment many times, count how often an event $E$ happens, and divide by the number of trials.

Experimental probability

P (E) = \frac{f}{N}

(14.1)

where

$f$: the number of trials in which $E$ happened
$N$: the total number of trials

Could engineers judge the chance that a new dam fails by waiting for a hundred floods? Could an airline crash a hundred new aircraft to see how often a wing breaks? Clearly not. Repeating trials is cheap for coins and dice, but for many real questions it is far too costly, dangerous or slow. We need a way to find a probability without repeating anything.

The way out is an assumption. If we can argue that the outcomes are equally likely, we need no data at all: we simply count. Among all the possible outcomes, the ones that make the event happen are called favourable outcomes (अनुकूल परिणाम).

Definition

Theoretical probability

When all the outcomes of an experiment are equally likely, the theoretical (or classical) probability of an event $E$ is the number of outcomes favourable to $E$ divided by the number of all possible outcomes.

ExampleOne chit from the mela bowl: $P (prize) = \frac{1}{6} .$

Theoretical probability

P (E) = \frac{m}{n}

(14.2)

where

$m$: the number of outcomes favourable to $E$
$n$: the number of all possible outcomes, all equally likely

Whenever this chapter says just probability, it means this theoretical kind.

Quick check

1 of 3

In which experiment are the listed outcomes equally likely?

A batter faces one ball: out or not outTomorrow in Shillong: rain or no rainA well-mixed bowl of 10 identical chits numbered 1 to 10: any one numberA bag of 7 green and 3 white marbles: green or white

Show answer

Answer (C) A well-mixed bowl of 10 identical chits numbered 1 to 10: any one number

Identical, well-mixed chits give every number the same chance. A batter's skill, the weather and the 7-to-3 split of marbles all favour one result, so those pairs are not equally likely.

Elementary events

Let us put the definition to work, starting with events made of a single outcome.

Example 14.1NCERT Example 1 · p. 204

Calling the toss

Before a match, the visiting captain calls heads. The coin is fair. What is the probability that the call is right? What is the probability that it is wrong?

Try it yourself first. Then reveal the solution one step at a time.

Example 14.2NCERT Example 2 · p. 204

Four pens in a pencil box

A pencil box holds four pens that look and feel the same, one each with red, green, blue and black ink. Aditi closes her eyes and pulls out one pen. Find the probability that it writes in (a) green, (b) black, (c) red, (d) blue.

Try it yourself first. Then reveal the solution one step at a time.

Definition

Elementary event

An event that consists of exactly one outcome of an experiment.

ExampleGetting a 5 on one roll of a die; Aditi drawing the green pen.

In Example 14.1 the events right and wrong are elementary, and so are the four colours in Example 14.2. Now add up their probabilities: $\frac{1}{2} + \frac{1}{2} = 1$ and $\frac{1}{4} \times 4 = 1.$ This is no accident.

PropertyElementary events add up to 1

For any experiment, the probabilities of all its elementary events add up to 1.

Most events we care about contain several outcomes. We then count all of them.

Example 14.3NCERT Example 3 · p. 205

Two events for one draw

A bowl holds 10 identical chits numbered 1 to 10, and one chit is drawn at random. Find the probability that its number is (a) less than 4, (b) 4 or more.

Try it yourself first. Then reveal the solution one step at a time.

Neither event in Example 14.3 is elementary: one has three outcomes and the other has seven. Yet their probabilities, $\frac{3}{10}$ and $\frac{7}{10},$ again add up to 1. The next section explains why.

Quick check

1 of 3

A die is rolled once. Which of these is an elementary event?

getting an even numbergetting a number below 3getting a prime numbergetting a 4

Show answer

Answer (D) getting a 4

Only getting a 4 has exactly one outcome. Below 3 has two outcomes (1 and 2), and even and prime have three each.

Complementary events

In Example 14.3, less than 4 and 4 or more share out the ten chits: every chit belongs to exactly one of the two events. Whenever one of them happens, the other does not. A pair like this gets its own name.

Definition

Complement of an event

For an event $E,$ the event not $E,$ written $\overset{ˉ}{E},$ happens exactly when $E$ does not. $E$ and $\overset{ˉ}{E}$ are called complementary events.

ExampleIf $E$ is a number less than 4 on a chit, then $\overset{ˉ}{E}$ is a number that is 4 or more.

Why must $P (E)$ and $P (\overset{ˉ}{E})$ add up to 1? Suppose an experiment has $n$ equally likely outcomes and $m$ of them are favourable to $E .$ Every other outcome is favourable to $\overset{ˉ}{E},$ so $\overset{ˉ}{E}$ has $n - m$ favourable outcomes and $P (\overset{ˉ}{E}) = \frac{n - m}{n} .$ Adding, $P (E) + P (\overset{ˉ}{E}) = \frac{n}{n} = 1.$ Rearranged, this is the complement rule.

Complement rule

P (\overset{ˉ}{E}) = 1 - P (E)

(14.3)

Fig. 14.3 shows the idea with a box of 12 bangles, 5 made of glass and 7 of metal, from which one is picked at random.

4/4 $\frac{5}{12} + \frac{7}{12} = 1 :$ the two events fill the bar.

Fig. 14.3An event and its complement share out all the outcomes, so together they fill the bar from 0 to 1.NCERT p. 206

Twelve bangles lie in two rows: five thin, double-ringed glass bangles and seven thick metal bangles, with a key below. In the first step the panel says to pick one of the 12 bangles at random, so n equals 12. In the second step a dashed outline surrounds the five glass bangles, event E, the panel gives P of E as five-twelfths, and the left part of a bar from 0 to 1 fills in. In the third step the outline moves to the seven metal bangles, not E, with probability seven-twelfths, and the hatched right part of the bar appears. In the fourth step both parts fill the bar exactly from 0 to 1, showing that five-twelfths plus seven-twelfths equals 1.

Quick check

1 of 3

The probability that a train arrives on time is 0.64. What is the probability that it does not arrive on time?

Show answer

Answer 0.36

On time and not on time are complementary:

1 - 0.64 = 0.36.

Impossible and sure events

Two extreme cases complete the picture. The spinner in Fig. 14.4 has five equal sectors carrying the numbers 2, 4, 6, 8 and 10.

What is the probability that it stops on an odd number? No sector is odd, so there are no favourable outcomes and $P (odd) = \frac{0}{5} = 0.$ An event that can never happen in an experiment is an impossible event (असंभव घटना); its probability is 0.

Now ask for an even number. Every sector is even, so all 5 outcomes are favourable and $P (even) = \frac{5}{5} = 1.$ An event that must happen, whatever the outcome, is a sure event, also called a certain event (निश्चित घटना); its probability is 1.

3/4Three of the five sectors fit: $P = \frac{3}{5} .$

Fig. 14.4As more sectors fit the event, its probability climbs from 0 towards 1.NCERT p. 207

On the left, a spinner with a pointer at the top has five equal sectors numbered 2, 4, 6, 8 and 10. On the right, a scale runs from 0, labelled impossible, to 1, labelled sure. In the first step the panel notes five equal sectors, all with even labels. In the second step the event is an odd number: no sector is ticked, the probability is 0 over 5, which is 0, and a marker sits at 0. In the third step the event is a number more than 5: the sectors 6, 8 and 10 are shaded and ticked, the probability is three-fifths, and the marker moves to 0.6. In the fourth step the event is an even number: every sector is shaded and ticked, the probability is 5 over 5, which is 1, and the marker reaches 1.

Every other event lies between these two extremes. For any event, the number of favourable outcomes $m$ is at least 0 and at most $n .$ Dividing $0 \leq m \leq n$ by $n$ gives the range rule.

Range of a probability

0 \leq P (E) \leq 1

(14.4)

Compare

The two extremes

Impossible event

no favourable outcome
$P = 0$
a 0 on a die

Sure event

every outcome favourable
$P = 1$
a number from 1 to 6 on a die

A deck of playing cards

Card games such as rummy and teen patti use a standard deck, and board questions often do too. Before you can count, you need to know exactly what the deck holds.

The deck is split into four suits: clubs (♣), diamonds (♦), hearts (♥) and spades (♠). Every suit has 13 cards, so the deck has $4 \times 13 = 52.$
Clubs and spades are printed in black, diamonds and hearts in red: 26 cards of each colour.
Every suit has an ace (A), the number cards 2 to 10, a jack (J), a queen (Q) and a king (K). So there are 4 cards of each rank.
Jacks, queens and kings are the face cards: 3 per suit, 12 in all.

When a deck is well shuffled, each of the 52 cards is equally likely to be the one drawn. Fig. 14.5 lays the deck out as a grid, one row per suit, so you can count the cards for an event at a glance.

2/4J, Q and K in every suit: $\frac{12}{52} = \frac{3}{13} .$

Fig. 14.5Count the bright cards and divide by 52.NCERT p. 207

The 52 cards of a deck are laid out in four rows, one for each suit: spades, hearts, diamonds and clubs, marked by their symbols, with hearts and diamonds printed in red. Each row runs ace, 2 to 10, jack, queen, king. In the first step all the cards are shown and the line below reads 4 suits times 13 cards equals 52 cards. In the second step only the jack, queen and king columns stay bright: 12 face cards, probability 12 over 52, or three-thirteenths. In the third step only the ace column stays bright: 4 aces, probability one-thirteenth. In the fourth step only the hearts row stays bright: 13 hearts, probability one-quarter.

Example 14.4NCERT Example 4 · p. 207

A heart, or not a heart

Raghav shuffles a deck of 52 cards thoroughly and turns over the top card. Find the probability that this card is (a) a heart, (b) not a heart. Do part (b) in two ways.

Try it yourself first. Then reveal the solution one step at a time.

SimulationCard-drawing lab

A virtual, well-shuffled deck of 52 cards. Each press draws 1, 10, 100 or 1000 cards, with every card going back into the deck before the next draw. Bars show how often each suit has come up, and a counter compares how often a face card (a jack, queen or king) has appeared with its theoretical probability of three-thirteenths, about 0.23.

Fig. 14.6Draw from a shuffled deck again and again, and track how often a face card turns up.

Try this

Predict first: in 130 draws, about how many face cards should you expect? Then draw 10 at a time, 13 times, and compare.
Draw 1000 at a time, a few times. Does the face-card share settle near $\frac{3}{13} \approx 0.23$ ?

Quick check

1 of 3

You pick a card at random from a full, shuffled deck. What is the probability that it is a 7?

\frac{1}{52}

\frac{1}{13}

\frac{7}{52}

\frac{4}{13}

Show answer

Answer (B) $\frac{1}{13}$

There is one 7 in each suit, so 4 sevens:

\frac{4}{52} = \frac{1}{13} .

The value

\frac{1}{52}

is for one particular 7, such as the 7 of clubs.

Using the complement

The complement rule is often the quickest route to an answer, and it is always a handy check. Watch for it in the next four examples.

Example 14.5NCERT Example 5 · p. 208

A chess final

Arjun and Meera play a chess final in which a draw is not allowed: if a game is drawn, tie-break games continue until someone wins. The probability that Arjun wins the final is 0.47. What is the probability that Meera wins it?

Try it yourself first. Then reveal the solution one step at a time.

Example 14.6NCERT Example 6 · p. 208

Two friends pick numbers

Kavya and Imran each silently choose a whole number from 1 to 20, every number being equally likely, and neither knows the other's choice. Find the probability that they choose (a) different numbers, (b) the same number.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.7NCERT Example 7 · p. 209

Who plants the first sapling?

A school eco-club has 36 members: 21 from Class 9 and 15 from Class 10. To decide who plants the first sapling on Van Mahotsav, the secretary drops 36 identical name slips into a steel dabba, shakes it hard and pulls out one slip. Find the probability that the name is of (a) a Class 10 student, (b) a Class 9 student.

Try it yourself first. Then reveal the solution one step at a time.

Example 14.8NCERT Example 8 · p. 209

A box of modaks

For Ganesh Chaturthi, a box holds 15 modaks of the same size and shape: 6 chocolate, 5 coconut and 4 kesar. One modak is picked at random. Find the probability that it is (a) coconut, (b) chocolate, (c) kesar.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

The probability that a lucky-draw ticket wins nothing is 0.92. What is the probability that it wins something?

Show answer

Answer 0.08

Wins something is the complement of wins nothing:

1 - 0.92 = 0.08.

Tossing two coins

When two coins are tossed together, make them different, say a ₹5 coin and a ₹10 coin, and write each outcome as an ordered pair: the first letter for the ₹5 coin and the second for the ₹10 coin. Then (H, T) and (T, H) are different outcomes. In one, the ₹5 coin shows heads; in the other, the ₹10 coin does.

So there are four equally likely outcomes: (H, H), (H, T), (T, H) and (T, T). Fig. 14.7 sets them out in a table.

3/3Only (H, H) has no tail: $P = 1 - \frac{1}{4} = \frac{3}{4} .$

Fig. 14.7Ordered pairs keep (H, T) and (T, H) apart.NCERT p. 210

A table has the ₹10 coin's faces, H and T, across the top and the ₹5 coin's faces, H and T, down the side, giving four cells: (H, H), (H, T), (T, H) and (T, T). In the first step the panel says there are 4 equally likely outcomes and that (H, T) is not the same as (T, H). In the second step the cells (H, H) and (T, T) are shaded for the event that both coins show the same face: 2 of the 4 cells, probability one-half. In the third step the three cells with at least one tail are shaded and (H, H) is struck through as the only cell with no tail, so the probability is 1 minus one-quarter, which is three-quarters.

Example 14.9NCERT Example 9 · p. 210

Two coins, two events

Ritu tosses a ₹5 coin and a ₹10 coin together. Find the probability that (a) both coins show the same face, (b) she gets at least one tail.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Two coins are tossed together. Which statement is true?

Two heads, two tails and one of each are equally likelyTwo tails is more likely than two headsAt least one head has probability

\frac{1}{2}

One head and one tail is twice as likely as two heads

Show answer

Answer (D) One head and one tail is twice as likely as two heads

One of each covers (H, T) and (T, H), so its probability is

\frac{2}{4},

twice the

\frac{1}{4}

of (H, H). The three kinds of result are not equally likely, and at least one head is

\frac{3}{4} .

Beyond counting: lengths and areas

In every example so far, the experiment had a finite number of outcomes that we could count. Some experiments are different. The moment you reach a bus stop can be any instant in an interval, and a raindrop can land at any point of a field. Such experiments have infinitely many outcomes, so we cannot count favourable ones. Instead we measure: if every point is equally likely, the probability is the favourable length (or area) divided by the total length (or area).

4/4Favourable area $120 m^{2} :$ $P = \frac{120}{1500} = \frac{2}{25} .$

Fig. 14.8With infinitely many outcomes, compare lengths or areas instead of counts.NCERT Fig. 14.2 · p. 211

At the top, a number line marks arrival times from 0 to 15 minutes in steps of 3. In the first step only the line is shown. In the second step the stretch from 9 to 15 minutes turns green and is labelled wait at most 6 minutes, with probability 6 over 15, or two-fifths. In the third step a rectangular field 50 metres by 30 metres appears below, with a few crosses where seed packets have landed and a pond 12 metres by 10 metres inside it; the panel gives the area of the field as 1500 square metres. In the fourth step the outline of the pond thickens and the panel adds its area, 120 square metres, and the probability 120 over 1500, or two twenty-fifths.

Example 14.10NCERT Example 10 · p. 210

Waiting for the ferry

A ferry leaves the ghat every 15 minutes. You reach the ghat at a random moment, without checking the timetable. What is the probability that you wait at most 6 minutes?

Try it yourself first. Then reveal the solution one step at a time.

Example 14.11NCERT Example 11 · p. 211

A seed packet and a pond

A drone drops a seed packet that is equally likely to land anywhere on a rectangular field 50 m long and 30 m wide. Inside the field is a rectangular pond 12 m by 10 m. What is the probability that the packet lands in the pond?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

A school bell rings at a random moment between 10:00 and 10:20. What is the probability that it rings between 10:15 and 10:20?

Show answer

Answer 0.25

Favourable length 5 minutes out of 20:

\frac{5}{20} = \frac{1}{4} .

Two buyers, two dice

The same random draw can be judged by different rules. In the next example a crate of mangoes is checked by two buyers who want different things; Fig. 14.9 shows what each one accepts.

3/3Everything except rotten: $\frac{188}{200} = 0.94.$

Fig. 14.9The same random mango, judged by two different rules.NCERT p. 212

A long bar stands for the 200 mangoes in a crate. Its plain left part is 160 perfect mangoes, a striped middle part is 28 bruised mangoes and a cross-hatched right end is 12 rotten mangoes, with a key below. In the first step only the bar is shown. In the second step a bracket over the perfect part is labelled fruit-seller 160, with probability 160 over 200, which is 0.8. In the third step a longer bracket over the perfect and bruised parts is labelled juice stall 188, with probability 188 over 200, which is 0.94.

Example 14.12NCERT Example 12 · p. 212

Mangoes for two buyers

A crate holds 200 mangoes: 160 are perfect, 28 are slightly bruised and 12 are rotten. A fruit-seller will take only perfect mangoes. A juice stall will take any mango that is not rotten. One mango is picked at random from the crate. Find the probability that it is acceptable to (a) the fruit-seller, (b) the juice stall.

Try it yourself first. Then reveal the solution one step at a time.

Two dice give a bigger set of outcomes. Use dice of different colours, say red and green, and write each outcome as an ordered pair (red, green). The red die can show any of 6 numbers, and for each of these the green die can show any of 6, so there are $6 \times 6 = 36$ equally likely outcomes. The pair (2, 5) means red 2 and green 5; it is a different outcome from (5, 2).

4/4Sums near 7 have more cells, so they are more likely.

Fig. 14.10Each of the 36 cells is equally likely, but the sums are not: 7 can happen in 6 ways, 2 and 12 in only 1.NCERT Fig. 14.3 · p. 213

A grid has the green die's numbers 1 to 6 across the top and the red die's numbers 1 to 6 down the side; each of the 36 cells shows the sum of its two numbers. In the first step the panel says there are 36 outcomes and each cell shows the sum. In the second step the four cells with sum 9 are shaded and ringed, running diagonally, and the panel gives probability 4 over 36, or one-ninth. In the third step the six doubles along the main diagonal are shaded and ringed, giving probability 6 over 36, or one-sixth. In the fourth step a bar chart beside the grid shows how many cells give each sum: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 and 1 for the sums 2 to 12, with the bar for 7 the tallest.

Example 14.13NCERT Example 13 · p. 212

Sums on two dice

A red die and a green die are rolled together. Find the chance that the two numbers add up to (a) 9, (b) 1, (c) more than 1. Also find (d) the chance of a double, that is, the same number on both dice.

Try it yourself first. Then reveal the solution one step at a time.

SimulationTwo-dice lab

Two virtual dice rolled together 1, 10, 100 or 1000 times at a press. A bar chart shows how often each sum from 2 to 12 has come up, beside the theoretical values, and a counter compares how often a total of 9 has appeared with its theoretical probability of one-ninth.

Fig. 14.11Roll two dice in batches and compare how often each sum appears with the theory.

Try this

Roll 10 at a time, three times. How far is the share of 9s from $\frac{1}{9} \approx 0.11$ ?
Now roll 1000 at a time. Which sum is the most common, and does the shape of the bars match the bar chart in the figure above?

Quick check

1 of 3

Two dice are rolled. What is the probability that the sum is 5?

Show answer

Answer 0.1111111111111111

(1, 4), (2, 3), (3, 2) and (4, 1):

\frac{4}{36} = \frac{1}{9} \approx 0.111.

Practice · 25 questions

Practice 14.1: questions with answers

1
Quick answerEasyNCERT Exercise 14.1, Q1
Fill in the blanks.
1. Whatever the event $E,$ the sum $P (E) + P (\overset{ˉ}{E})$ is always
  Show answer
  Answer 1
  An event and its complement share out all the outcomes.
2. An event that can never happen has probability and is called an event
  Show answer
  Answer (1) 0, (2) impossible
  No outcome is favourable, so $m = 0.$
3. An event that is certain to happen has probability and is called a event
  Show answer
  Answer (1) 1, (2) sure
  Every outcome is favourable, so $m = n .$
4. If you add the probabilities of every elementary event of an experiment, you get . No probability is less than or more than
  Show answer
  Answer (1) 1, (2) 0, (3) 1
  Elementary events together cover every outcome once, and $0 \leq m \leq n$ gives $0 \leq P (E) \leq 1.$
2
Short answerEasyNCERT Exercise 14.1, Q2
Decide whether the two outcomes in each experiment are equally likely, and give a reason.
1. A slip is drawn from a well-mixed bowl holding two identical slips, one marked Yes and one marked No. The result is Yes or No.
  equally likelynot equally likely
  Show solution
  Equally likely: the slips are identical and well mixed, so each has the same chance of being drawn.
2. A kabaddi raider enters the other half. She scores a point or she is caught.
  equally likelynot equally likely
  Show solution
  Not equally likely: the result depends on her skill and on the defenders, so there is no reason for equal chances.
3. On a given day in July, it rains in Chennai or it does not.
  equally likelynot equally likely
  Show solution
  Not equally likely: the weather depends on the season and on conditions that day; nothing makes the two results balance.
4. An ordinary die is rolled. It shows an even number or an odd number.
  equally likelynot equally likely
  Show solution
  Equally likely: each result covers three of the six equally likely faces (2, 4, 6 and 1, 3, 5), so each has probability $\frac{1}{2} .$
Short answerEasyNCERT Exercise 14.1, Q3
Before a kabaddi match, the referee tosses a coin and the captain who wins the toss chooses which team raids first. Why is this a fair way to decide?
Show solution
A fair coin tossed at random is equally likely to land heads or tails. Whichever face a captain calls, his chance of winning the toss is $\frac{1}{2},$ exactly the same as the other captain's. Neither team is favoured, and nobody can control the result.
Quick answerEasyNCERT Exercise 14.1, Q4
Which of these cannot be the probability of an event?
$\frac{7}{9}$ $108 %$ $0.001$ $0$
Show answer
Answer (B) $108 %$
Every probability lies from 0 to 1. $108 % = 1.08,$ which is more than 1. The value 0 is allowed: it is the probability of an impossible event.
NumericalEasyNCERT Exercise 14.1, Q5
If $P (E) = 0.17,$ what is the probability of not $E$ ?
Show solution
$P (\overset{ˉ}{E}) = 1 - 0.17 = 0.83.$
6
Short answerEasyNCERT Exercise 14.1, Q6
Every laddoo in Dadi's tin is a coconut laddoo. With his eyes shut, Arjun grabs one. Find the chance that he has picked
1. a besan laddoo;
  Show solution
  No laddoo in the tin is besan, so this is an impossible event: $P = 0.$
2. a coconut laddoo.
  Show solution
  Every laddoo is coconut, so this is a sure event: $P = 1.$
NumericalEasyNCERT Exercise 14.1, Q7
The probability that two people chosen at random were born on different days of the week is $\frac{6}{7} .$ What is the probability that they were born on the same day of the week?
Show solution
$P (same day) = 1 - \frac{6}{7} = \frac{1}{7} \approx 0.143.$
8
Short answerEasyNCERT Exercise 14.1, Q8
Riya's pouch has 7 green and 4 yellow pencils of the same length. She pulls one out without looking. How likely is it to be
1. green;
  Show solution
  $n = 7 + 4 = 11$ and $m = 7,$ so $P (green) = \frac{7}{11} .$
2. not green.
  Show solution
  $P (not green) = 1 - \frac{7}{11} = \frac{4}{11},$ the share of yellow pencils.
9
Short answerEasyNCERT Exercise 14.1, Q9
At a birthday party, a jar holds 20 toffees in identical wrappers: 6 mango, 9 orange and 5 lime. The first guest picks one without looking. What is the probability that it is
1. mango;
  Show solution
  $n = 20,$ so $P (mango) = \frac{6}{20} = \frac{3}{10} .$
2. orange;
  Show solution
  $P (orange) = \frac{9}{20} .$
3. not lime.
  Show solution
  $20 - 5 = 15$ toffees are not lime, so $P = \frac{15}{20} = \frac{3}{4} .$
10
Short answerMediumNCERT Exercise 14.1, Q10
Meena's clay gullak has 150 coins: 60 of ₹1, 45 of ₹2, 30 of ₹5 and 15 of ₹10. She shakes it over her bed until a single coin drops out through the slot; assume each coin is equally likely to be the one. What is the probability that the coin is
1. a ₹2 coin;
  Show solution
  $n = 150,$ and 45 coins are ₹2 coins, so the probability is $\frac{45}{150} = \frac{3}{10} .$
2. anything except a ₹10 coin?
  Show solution
  $P = 1 - \frac{15}{150} = \frac{135}{150} = \frac{9}{10} .$
NumericalEasyNCERT Exercise 14.1, Q11
A nursery tray has 7 rose saplings and 11 marigold saplings that look alike before they flower. A gardener picks one at random. What is the probability that it is a rose sapling?
Show solution
$n = 7 + 11 = 18$ and $m = 7,$ so $P (rose) = \frac{7}{18} \approx 0.39.$
12
Short answerMediumNCERT Exercise 14.1, Q12
A wheel has 10 equal sectors numbered 1 to 10, and its arrow is equally likely to stop on any sector. Find the probability that it stops on
1. 7;
  Show solution
  One sector shows 7: $P = \frac{1}{10} .$
2. a prime number;
  Show solution
  Primes: 2, 3, 5, 7, so $P = \frac{4}{10} = \frac{2}{5} .$
3. a number that is at least 4;
  Show solution
  4 to 10 is 7 numbers: $P = \frac{7}{10} .$
4. a number less than 11.
  Show solution
  All 10 sectors: $P = 1,$ a sure event.
13
Short answerEasyNCERT Exercise 14.1, Q13
A die is rolled once. Find the probability of getting
1. a factor of 6;
  Show solution
  Factors of 6 on a die: 1, 2, 3, 6, so $P = \frac{4}{6} = \frac{2}{3} .$
2. a number strictly between 1 and 5;
  Show solution
  2, 3 and 4, so $P = \frac{3}{6} = \frac{1}{2} .$
3. a perfect square.
  Show solution
  1 and 4, so $P = \frac{2}{6} = \frac{1}{3} .$
14
Short answerMediumNCERT Exercise 14.1, Q14
A card is pulled out at random from a thoroughly shuffled pack of 52. What is the probability that it is
1. a black queen;
  Show solution
  Queens of spades and clubs: $\frac{2}{52} = \frac{1}{26} .$
2. not a face card;
  Show solution
  $52 - 12 = 40$ cards: $\frac{40}{52} = \frac{10}{13} .$
3. a black face card;
  Show solution
  J, Q, K of spades and of clubs: $\frac{6}{52} = \frac{3}{26} .$
4. the 7 of clubs.
  Show solution
  Exactly one card: $\frac{1}{52} .$
15
Short answerMediumNCERT Exercise 14.1, Q15
Kabir takes the six clubs from the ace to the 6, mixes them and lays them in a row, face down. He flips one card at random.
1. What is the probability that it is the 4?
  Show solution
  6 equally likely cards, one of them the 4: $P = \frac{1}{6} .$
2. It is the 4, and it is kept aside. Then one of the remaining cards is turned over at random. Find the chance that it is the ace.
  Show solution
  5 cards remain and one is the ace: $P = \frac{1}{5} .$
3. In part (b), what is the probability that the second card is the 4?
  Show solution
  The 4 is no longer among the cards: an impossible event, $P = 0.$
NumericalEasyNCERT Exercise 14.1, Q16
A batch of 300 phone chargers contains 15 that do not work; the faulty ones look exactly like the rest. One charger is taken at random. What is the probability that it works?
Show solution
$300 - 15 = 285$ chargers work, so $P = \frac{285}{300} = \frac{19}{20} = 0.95.$
17
Short answerMediumNCERT Exercise 14.1, Q17
A basket has 32 mangoes. 8 of them are spoilt inside, although they look fine.
1. One mango is taken at random. What is the probability that it is spoilt?
  Show solution
  $P = \frac{8}{32} = \frac{1}{4} .$
2. The mango taken in (a) turns out to be good and is eaten. Then a second mango is chosen at random from those left in the basket. Find the chance that it is good.
  Show solution
  31 mangoes remain, of which $24 - 1 = 23$ are good, so $P = \frac{23}{31} .$
18
Short answerMediumNCERT Exercise 14.1, Q18
At a tambola evening, the caller's bag holds tokens numbered 1 to 60. The first token is pulled out at random. Find the probability that its number is
1. a single-digit number;
  Show solution
  1 to 9 is 9 numbers: $P = \frac{9}{60} = \frac{3}{20} .$
2. a perfect cube;
  Show solution
  1, 8 and 27 (the next cube, 64, is too big): $P = \frac{3}{60} = \frac{1}{20} .$
3. a multiple of 6.
  Show solution
  6, 12, …, 60 is 10 numbers: $P = \frac{10}{60} = \frac{1}{6} .$
19
Short answerEasyNCERT Exercise 14.1, Q19
The six faces of a cube are lettered B, A, N, A, N and A. The cube is rolled once. Find the probability that the top face shows
1. A;
  Show solution
  Three of the six faces show A: $P = \frac{3}{6} = \frac{1}{2} .$
2. B.
  Show solution
  One face shows B: $P = \frac{1}{6} .$
NumericalHardNCERT Exercise 14.1, Q20
(Not for the board exam.) A rectangular garden is 14 m long and 10 m wide, and a circular fountain of radius 3.5 m stands inside it. A seed blown by the wind lands at a random point of the garden. What is the probability that it lands in the fountain? Use $π = \frac{22}{7} .$
Show solution
Garden: $14 \times 10 = 140 m^{2} .$ Fountain: $\frac{22}{7} \times {3.5}^{2} = 38.5 m^{2} .$
$P = \frac{38.5}{140} = \frac{11}{40} = 0.275.$
21
Short answerEasyNCERT Exercise 14.1, Q21
Of 200 biscuit packets on a shelf, 16 are past their expiry date, which is printed inside the flap. Meera's rule is simple: she keeps a packet if it is in date and returns it otherwise. Her brother brings her one packet from the shelf at random. Find the probability that Meera
1. keeps it;
  Show solution
  $200 - 16 = 184$ packets are in date: $P = \frac{184}{200} = \frac{23}{25} .$
2. returns it.
  Show solution
  $P = 1 - \frac{23}{25} = \frac{2}{25} .$
22
Long answerHardNCERT Exercise 14.1, Q22
Two dice are rolled together, and we note the difference between the larger and the smaller number (0 if they are equal).
1. Make a table of the probabilities of the differences 0, 1, 2, 3, 4 and 5. What is the probability that the difference is 1?
  Show solution
  Counting the pairs for each difference:
  Table 14.1Difference of two dice: the number of ordered pairs and the probability $P$
  Difference Pairs $P$
  0 6 $\frac{1}{6}$
  1 10 $\frac{5}{18}$
  2 8 $\frac{2}{9}$
  3 6 $\frac{1}{6}$
  4 4 $\frac{1}{9}$
  5 2 $\frac{1}{18}$
  The counts add up to 36. The difference is 1 for (1, 2), (2, 1), (2, 3), (3, 2), …, (5, 6), (6, 5): 10 pairs, so $P = \frac{10}{36} = \frac{5}{18} .$
2. A classmate says: "The difference can only be 0, 1, 2, 3, 4 or 5, so each has probability $\frac{1}{6} .$ " Do you agree? Explain.
  Show solution
  No. The six differences are not equally likely; only the 36 ordered pairs are. From the table, a difference of 1 has probability $\frac{5}{18}$ but a difference of 5 only $\frac{1}{18} .$
23
Short answerMediumNCERT Exercise 14.1, Q23
Priya tosses a coin three times and notes the result each time. She wins a prize if she gets exactly two heads. Find the probability that she
1. wins;
  Show solution
  Exactly two heads: HHT, HTH, THH, so 3 of the 8 equally likely results: $P = \frac{3}{8} .$
2. does not win.
  Show solution
  $P = 1 - \frac{3}{8} = \frac{5}{8} .$
24
Short answerMediumNCERT Exercise 14.1, Q24
A die is rolled twice. Find the probability that
1. both rolls show even numbers;
  Show solution
  3 even faces for the first roll and 3 for the second: $3 \times 3 = 9$ pairs, so $P = \frac{9}{36} = \frac{1}{4} .$
2. at least one roll shows an odd number.
  Show solution
  This is the complement of (a): $P = 1 - \frac{1}{4} = \frac{3}{4} .$
25
Short answerMediumNCERT Exercise 14.1, Q25
Say whether each argument is correct, and give a reason.
1. When a die is rolled, either a 6 comes up or it does not. So the probability of a 6 is $\frac{1}{2} .$
  correctnot correct
  Show solution
  Not correct. A 6 and not a 6 are not equally likely: not a 6 covers five of the six equally likely faces. In fact $P (6) = \frac{1}{6} .$
2. A card drawn from a well-shuffled deck is either red or black. So the probability that it is red is $\frac{1}{2} .$
  correctnot correct
  Show solution
  Correct, but only because the colours have equal numbers of cards: 26 red out of 52, so $P (red) = \frac{26}{52} = \frac{1}{2} .$

Difference	Pairs	$P$
0	6	$\frac{1}{6}$
1	10	$\frac{5}{18}$
2	8	$\frac{2}{9}$
3	6	$\frac{1}{6}$
4	4	$\frac{1}{9}$
5	2	$\frac{1}{18}$

14.2 Summary

4/4Their probabilities add up to 1.

Fig. 14.12Four tools solve almost every problem in this chapter.NCERT p. 217

Four tiles appear one after another. Tile 1, Count, shows P of E equals favourable outcomes over all outcomes. Tile 2, Range, adds 0 is at most P of E, which is at most 1, with 0 marked impossible and 1 marked sure. Tile 3, Complement, adds P of E plus P of not E equals 1. Tile 4, Elementary events, adds that the probabilities of all the elementary events add up to 1.

Recap

Probability: key points and summary

Outcomes are equally likely when nothing in the set-up favours any of them; list outcomes so that this is true (Equally likely outcomes).
Theoretical probability: $P (E) = \frac{m}{n},$ the favourable outcomes over all the outcomes, for equally likely outcomes (Theoretical probability).
An elementary event has exactly one outcome, and the probabilities of all the elementary events add up to 1 (Elementary events).
For any event, $P (E) + P (\overset{ˉ}{E}) = 1,$ so $P (\overset{ˉ}{E}) = 1 - P (E)$ (Complementary events).
An impossible event has probability 0 and a sure event has probability 1; always $0 \leq P (E) \leq 1$ (Impossible and sure events).
A deck has 52 cards: 4 suits of 13, 26 red and 26 black, 4 of each rank and 12 face cards (A deck of playing cards).
Use ordered pairs for two coins (4 outcomes) and two dice (36 outcomes); the sums of two dice are not equally likely (Tossing two coins, Two buyers, two dice).
Beyond the board syllabus (Beyond counting: lengths and areas): when outcomes fill a length or an area, divide the favourable length or area by the total.
With many trials, experimental probability tends to come close to theoretical probability (Key idea).

Flash cards · 7

Key terms

1 / 7

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In your NCERT textbook	Practice question here
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NCERT Exercise 14.1, Q24	Practice Q24
NCERT Exercise 14.1, Q25	Practice Q25

Quick check: MCQs on Probability

Theoretical probability3 questions
Elementary events3 questions
Complementary events3 questions
Impossible and sure events3 questions
A deck of playing cards3 questions
Using the complement3 questions
Tossing two coins2 questions
Beyond counting: lengths and areas2 questions
Two buyers, two dice3 questions

Glossary · 11

Key terms (English–Hindi)

at randomयादृच्छयाyadrichchhaya: Chosen or thrown with no bias or control, so that every outcome keeps its natural chance.First seen in the chapter →

complementary eventsपूरक घटनाएँpoorak ghatnayen: An event $E$ and the event not $E,$ written $\overset{ˉ}{E};$ exactly one of them happens, so $P (E) + P (\overset{ˉ}{E}) = 1.$

elementary eventप्रारंभिक घटनाprarambhik ghatna: An event made of exactly one outcome of an experiment.
equally likelyसमप्रायिकsamprayik: Having the same chance of happening, because nothing in the set-up favours one outcome over another.First seen in the chapter →
eventघटनाghatna: A collection of outcomes of an experiment, such as an even number on a die.First seen in the chapter →
experimental probabilityप्रायोगिक प्रायिकताprayogik prayikta: The fraction of trials in which an event actually happened.First seen in the chapter →

face card: A jack, queen or king; a deck of 52 cards has 12 of them.First seen in the chapter →
favourable outcomeअनुकूल परिणामanukool parinam: An outcome that makes a given event happen.First seen in the chapter →

impossible eventअसंभव घटनाasambhav ghatna: An event with no favourable outcome; its probability is 0.First seen in the chapter →

sure eventनिश्चित घटनाnishchit ghatna: An event for which every outcome is favourable; its probability is 1.First seen in the chapter →

theoretical probabilityसैद्धांतिक प्रायिकताsaiddhantik prayikta: For equally likely outcomes, $P (E) = \frac{m}{n} :$ favourable outcomes divided by all possible outcomes.

Probability: Class 10 Maths · Chapter 14

What you will learn in Probability

14.1 Probability — A Theoretical Approach

Equally likely outcomes

Theoretical probability

Elementary events

Complementary events

Impossible and sure events

A deck of playing cards

Using the complement

Tossing two coins

Beyond counting: lengths and areas

Two buyers, two dice

Practice 14.1: questions with answers

14.2 Summary

Probability: key points and summary

Board exam practice

Practice questions with answers (NCERT “Board exam practice”)

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Quick check: MCQs on Probability

Key terms (English–Hindi)

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