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Class 9 Mathematics Notes – Chapter 13: Probability

Punjab Board | 2026

EXERCISE 13.1

Question 1

Arshad rolls a die with sides labelled L, M, N, O, P, U. Find the probability that the die lands on a consonant.

Sample Space:

\[ S=\{L,M,N,O,P,U\},\qquad n(S)=6 \]

Consonants are L, M, N, P, so \(n(E)=4\).

Using the probability formula:

\[ P(E)=\frac{n(E)}{n(S)} =\frac{4}{6}=\boxed{\frac{2}{3}} \]

Question 2

Shazia throws a pair of fair dice. Find the probability of the following events.

Total possible outcomes:

\[ n(S)=6\times6=36 \]

(i) The sum of the dots is at least 4

The outcomes with a sum less than 4 are:

\[ (1,1),(1,2),(2,1) \]

There are 3 such outcomes. Therefore, favourable outcomes are \(36-3=33\).

\[ P(\text{sum}\geq4)=\frac{33}{36} =\boxed{\frac{11}{12}} \]

(ii) The product of both dots is between 5 and 10, inclusive

Favourable outcomes are:

Product 5: \((1,5),(5,1)\) — 2 outcomes
Product 6: \((1,6),(2,3),(3,2),(6,1)\) — 4 outcomes
Product 8: \((2,4),(4,2)\) — 2 outcomes
Product 9: \((3,3)\) — 1 outcome
Product 10: \((2,5),(5,2)\) — 2 outcomes

Total favourable outcomes:

\[ n(E)=2+4+2+1+2=11 \]

\[ P(E)=\boxed{\frac{11}{36}} \]

(iii) The difference between both dots is equal to 4

Favourable outcomes:

\[ (1,5),(5,1),(2,6),(6,2) \]

\[ P(E)=\frac{4}{36}=\boxed{\frac{1}{9}} \]

(iv) At least 5 on the first die and at least 4 on the second die

First die: \(\{5,6\}\), giving 2 choices.

Second die: \(\{4,5,6\}\), giving 3 choices.

\[ n(E)=2\times3=6 \]

\[ P(E)=\frac{6}{36}=\boxed{\frac{1}{6}} \]

Question 3

One letter is selected at random from the word MATHEMATICS. Find the probability of getting the following.

The word contains 11 letters:

\[ M,A,T,H,E,M,A,T,I,C,S \]

(i) A vowel

The vowels are A, A, E, I. There are 4 vowels.

\[ P(\text{vowel})=\boxed{\frac{4}{11}} \]

(ii) A consonant

There are \(11-4=7\) consonants.

\[ P(\text{consonant})=\boxed{\frac{7}{11}} \]

(iii) The letter E

The letter E occurs once.

\[ P(E)=\boxed{\frac{1}{11}} \]

(iv) The letter A

The letter A occurs twice.

\[ P(A)=\boxed{\frac{2}{11}} \]

(v) Not M

The letter M occurs twice. The number of letters other than M is \(11-2=9\).

\[ P(\text{not }M)=\boxed{\frac{9}{11}} \]

(vi) Not T

The letter T occurs twice. The number of letters other than T is \(11-2=9\).

\[ P(\text{not }T)=\boxed{\frac{9}{11}} \]

Question 5

Aslam rolls a die. Find the probability of getting 3 or 4 and the probability of not getting 3 or 4.

Sample space:

\[ S=\{1,2,3,4,5,6\},\qquad n(S)=6 \]

Probability of getting 3 or 4

\[ P(3\text{ or }4)=\frac{2}{6}=\boxed{\frac{1}{3}} \]

Probability of not getting 3 or 4

\[ P(\text{not }(3\text{ or }4)) =1-\frac{1}{3}=\boxed{\frac{2}{3}} \]

Question 6

Abdul Hadi labels cards from 1 to 30. One card is selected at random.

Total outcomes:

\[ n(S)=30 \]

(i) Getting the number 25

\[ P(25)=\boxed{\frac{1}{30}} \]

(ii) Getting a number between 17 and 22

The numbers are 18, 19, 20, and 21, giving 4 favourable outcomes.

\[ P(E)=\frac{4}{30}=\boxed{\frac{2}{15}} \]

(iii) Getting a number at least 20

The numbers are 20 through 30, giving 11 outcomes.

\[ P(E)=\boxed{\frac{11}{30}} \]

(iv) Getting a number that is neither 27 nor 29

There are \(30-2=28\) favourable outcomes.

\[ P(E)=\frac{28}{30}=\boxed{\frac{14}{15}} \]

(v) Getting a number not between 12 and 15

The excluded numbers are 12, 13, 14, and 15. Therefore, there are \(30-4=26\) favourable outcomes.

\[ P(E)=\frac{26}{30}=\boxed{\frac{13}{15}} \]

Question 7

The probability that Ayesha will pass an examination is 0.85. Find the probability that she will not pass.

\[ P(\text{not pass})=1-P(\text{pass}) \]

\[ =1-0.85=\boxed{0.15} \]

Question 8

Taabish tosses a fair coin and rolls a fair die once.

Total outcomes:

\[ n(S)=2\times6=12 \]

(i) A tail on the coin and at least 4 on the die

Favourable outcomes: \((T,4),(T,5),(T,6)\).

\[ P(E)=\frac{3}{12}=\boxed{\frac{1}{4}} \]

(ii) A head on the coin and 2 or 3 on the die

Favourable outcomes: \((H,2),(H,3)\).

\[ P(E)=\frac{2}{12}=\boxed{\frac{1}{6}} \]

(iii) Both head and tail on the coin and 6 on the die

This is an impossible event.

\[ \boxed{P(E)=0} \]

(iv) Not tail on the coin and 5 on the die

Not tail means head. The favourable outcome is \((H,5)\).

\[ P(E)=\boxed{\frac{1}{12}} \]

(v) Not head on the coin and 5 or 2 on the die

Not head means tail. The favourable outcomes are \((T,5)\) and \((T,2)\).

\[ P(E)=\frac{2}{12}=\boxed{\frac{1}{6}} \]

Question 10

A card is selected at random from a well-shuffled pack of 52 playing cards.

(i) Selecting a queen

There are 4 queens in a pack.

\[ P(\text{queen})=\frac{4}{52}=\boxed{\frac{1}{13}} \]

(ii) Selecting neither a queen nor a jack

There are 4 queens and 4 jacks, giving 8 excluded cards.

Remaining cards:

\[ 52-8=44 \]

\[ P(E)=\frac{44}{52}=\boxed{\frac{11}{13}} \]

Question 11

A card is selected at random from a pack of 52 playing cards.

(i) Getting a jack

There are 4 jacks in a pack.

\[ P(\text{jack})=\frac{4}{52}=\boxed{\frac{1}{13}} \]

(ii) Getting no diamond

There are 13 diamonds and \(52-13=39\) non-diamond cards.

\[ P(\text{no diamond})=\frac{39}{52}=\boxed{\frac{3}{4}} \]

EXERCISE 13.2

Question 1

Research data on deaths from horse-kicks:

Number of Deaths	Frequency (\(f\))	Relative Frequency
0	60	\(60/294\approx0.204\)
1	50	\(50/294\approx0.170\)
2	87	\(87/294\approx0.296\)
3	40	\(40/294\approx0.136\)
4	32	\(32/294\approx0.109\)
5	15	\(15/294\approx0.051\)
6	10	\(10/294\approx0.034\)
Total	294	—

The relative frequency is calculated using:

\[ \text{Relative Frequency}=\frac{f}{\sum f} \]

Question 2

Frequency of defective products in 750 samples:

Frequency	Relative Frequency
120	\(120/750=0.160\)
140	\(140/750\approx0.187\)
94	\(94/750\approx0.125\)
85	\(85/750\approx0.113\)
105	\(105/750=0.140\)
50	\(50/750\approx0.067\)
40	\(40/750\approx0.053\)
66	\(66/750=0.088\)
50	\(50/750\approx0.067\)
Total: 750	—

Formula:

\[ \text{Relative Frequency}=\frac{\text{Frequency}}{750} \]

Question 3

Quiz competition correct answers:

Frequencies: 10, 23, 15, 25, 18, 9.

Total frequency:

\[ \sum f=10+23+15+25+18+9=100 \]

Frequency	Relative Frequency
10	0.10
23	0.23
15	0.15
25	0.25
18	0.18
9	0.09
Total	1.00

Question 4

A survey of 50 students records their favourite foods:

Food	Number of Responses
Biryani	40
Fresh Juice	7
Chicken	21
Bar-B-Q	15
Sweets	25

The total number of selections is:

\[ 40+7+21+15+25=108 \]

(i) Percentage who like biryani

If the responses are treated as 108 total selections:

\[ \frac{40}{108}\times100\approx\boxed{37.04\%} \]

If each of the 50 students selected exactly one food, the data would need to be checked because the responses total 108.

(ii) Percentage who like chicken

Using 50 students as the denominator, as in the supplied text:

\[ \frac{21}{50}\times100=\boxed{42\%} \]

(iii) Least-liked food

Answer: Fresh Juice, with 7 responses.

(iv) Most-preferred food

Answer: Biryani, with 40 responses.

Question 5

In 500 trials of throwing two dice, find the expected frequency that the sum is greater than 8.

Possible sums greater than 8 are 9, 10, 11, and 12.

Sum 9: 4 outcomes
Sum 10: 3 outcomes
Sum 11: 2 outcomes
Sum 12: 1 outcome

Total favourable outcomes:

\[ 4+3+2+1=10 \]

Probability:

\[ P(E)=\frac{10}{36}=\frac{5}{18} \]

Expected frequency:

\[ 500\times\frac{5}{18}\approx\boxed{139} \]

Question 6

A person receives Rs. 120 if at least 2 heads occur when three coins are tossed. Find the expected amount.

Total possible outcomes:

\[ n(S)=2^3=8 \]

Outcomes with at least 2 heads:

\[ \{HHH,HHT,HTH,THH\} \]

There are 4 favourable outcomes.

\[ P(E)=\frac{4}{8}=\frac{1}{2} \]

Expected amount:

\[ 120\times\frac{1}{2}=\boxed{\text{Rs. }60} \]

Question 7

Find the expected frequencies when the experiment is repeated 200 times.

Expected frequency is calculated using:

\[ \text{Expected Frequency}=n\times P(x) \]

Outcome	\(P(x)\)	Expected Frequency
0	0.11	\(200\times0.11=22\)
1	0.21	\(200\times0.21=42\)
2	0.17	\(200\times0.17=34\)
3	0.18	\(200\times0.18=36\)
4	0.09	\(200\times0.09=18\)
5	0.17	\(200\times0.17=34\)
6	0.07	\(200\times0.07=14\)
Total	1.00	200

Question 8

The supplied text states that the probability of getting five sixes is \(2/5\), but the full experiment description is incomplete.

Note: The expected frequency cannot be determined without knowing the total number of trials and the precise event being considered.

REVIEW EXERCISE 13

Question 1: Multiple Choice Questions (MCQs)

No.	Question	Correct Answer
(i)	Each element of the sample space is called:	(c) Sample point
(ii)	An outcome that represents how many times we expect something to happen is called:	(b) Favourable outcome
(iii)	Which tells us how often a specific event occurs relative to the total trials?	(c) Relative frequency
(iv)	Estimated probability of an event occurring is also known as:	(a) Relative frequency
(v)	The sum of all expected frequencies equals the fixed number of:	(a) Trials
(vi)	The chance of occurrence of a particular event is called:	(c) Probability
(vii)	An event that has a greater chance of occurring is called:	(b) Likely event
(viii)	Find the total number of possible outcomes when 4 dice are rolled.	(c) 1296
(ix)	When rolling a pair of dice, what is the probability of getting double 2?	(d) \(1/36\)
(x)	A card is chosen from a pack of 52 cards. Find the probability of getting neither a jack nor a king.	(b) \(11/13\)

Note for MCQ (viii): The supplied text gives 64, but for four standard six-sided dice the total number of outcomes is:

\[ 6^4=1296 \]

Question 2: Definitions

(i) Relative Frequency

The ratio of the frequency of a specific event to the total number of trials.

\[ \text{Relative Frequency}=\frac{f}{\sum f} \]

(ii) Expected Frequency

The predicted number of times an outcome should occur in a given number of trials.

\[ \text{Expected Frequency}=n\times P(E) \]

Question 3

An urn contains 10 red balls, 5 green balls, and 8 blue balls.

Total balls:

\[ n(S)=10+5+8=23 \]

(i) Getting a green ball

\[ P(\text{green})=\boxed{\frac{5}{23}} \]

(ii) Getting a red ball

\[ P(\text{red})=\boxed{\frac{10}{23}} \]

(iii) Getting a blue ball

\[ P(\text{blue})=\boxed{\frac{8}{23}} \]

(iv) Not getting a red ball

\[ P(\text{not red})=\frac{23-10}{23} =\boxed{\frac{13}{23}} \]

(v) Not getting a green ball

\[ P(\text{not green})=\frac{23-5}{23} =\boxed{\frac{18}{23}} \]

Question 4

Three coins are tossed together.

Total possible outcomes:

\[ n(S)=2^3=8 \]

(i) Exactly three heads

Favourable outcome: \(HHH\).

\[ P(E)=\boxed{\frac{1}{8}} \]

(ii) At least two tails

Favourable outcomes:

\[ \{TTT,TTH,THT,HTT\} \]

\[ P(E)=\frac{4}{8}=\boxed{\frac{1}{2}} \]

(iii) Not at least two heads

This means fewer than two heads, or 0 or 1 head.

Favourable outcomes:

\[ \{TTT,TTH,THT,HTT\} \]

\[ P(E)=\boxed{\frac{1}{2}} \]

(iv) Not exactly two heads

There are 3 outcomes with exactly two heads: \(HHT, HTH, THH\).

\[ P(E)=\frac{8-3}{8}=\boxed{\frac{5}{8}} \]

Question 5

A card is drawn from a well-shuffled pack of 52 playing cards.

(i) A king or jack of red colour

There are 2 red kings and 2 red jacks, giving 4 favourable cards.

\[ P(E)=\frac{4}{52}=\boxed{\frac{1}{13}} \]

(ii) Not the 2 of clubs or the 2 of spades

The excluded cards are the 2 of clubs and the 2 of spades.

Remaining cards:

\[ 52-2=50 \]

\[ P(E)=\frac{50}{52}=\boxed{\frac{25}{26}} \]

Question 6

Six coins are tossed 600 times.

The recorded frequencies are:

110, 90, 105, 80, 76, 123, 16.

Frequency	Relative Frequency
110	\(110/600\approx0.183\)
90	\(90/600=0.150\)
105	\(105/600=0.175\)
80	\(80/600\approx0.133\)
76	\(76/600\approx0.127\)
123	\(123/600=0.205\)
16	\(16/600\approx0.027\)
Total: 600	1.000

Question 7

A lot contains 25 items, of which 8 are defective.

Non-defective items:

\[ 25-8=17 \]

Relative frequency of non-defective items:

\[ \frac{17}{25}=\boxed{0.68} \]

Expected frequency of non-defective items in 25 items:

\[ 25\times\frac{17}{25}=\boxed{17} \]

Notes by Sir Dilawar

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