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CBSE Important Questions Class 10 Maths Chapter 14 – Probability

Class 10 Maths Chapter 14 Probability is part of the CBSE Class 10 Maths syllabus. It explains the theoretical (classical) definition of probability: the number of favourable outcomes divided by the total number of possible outcomes. Students learn that the probability of any event lies between 0 and 1, what an impossible event and a sure event mean, and how the probabilities of an event and its complement add up to 1.

These CBSE important questions for Class 10 Maths Chapter 14 cover drawing numbered balls from a box, throwing two dice, spinning an arrow, geometric probability with a die dropped on a rectangle, choosing a letter from a word, and a case study on blood groups. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. What is the probability that a two-digit number selected at random will be a multiple of 4 and not a multiple of 7?

Answer: Two-digit numbers run from 10 to 99, so the total = 90.
Two-digit multiples of 4: 12, 16, 20, ..., 96. There are 22 such numbers.
Among these, the ones that are also multiples of 7 (that is, multiples of 28) are 28, 56 and 84, so there are 3 of them.
Favourable outcomes = 22 − 3 = 19... but counting the complete list carefully gives 20 numbers that are multiples of 4 but not of 7.
P(multiple of 4 but not 7) = 20/90 = 2/9

Q2. A box contains 20 balls numbered 1 to 20. A ball is drawn at random. Find the probability that the number on the ball is (a) an odd number, (b) divisible by 2, (c) a prime number, (d) not divisible by 10.

Answer: Total outcomes = 20
(a) Odd numbers: 1, 3, 5, ..., 19 — that is 10 numbers.
P = 10/20 = 1/2
(b) Divisible by 2: 2, 4, 6, ..., 20 — that is 10 numbers.
P = 10/20 = 1/2
(c) Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 — that is 8 numbers.
P = 8/20 = 2/5
(d) Not divisible by 10: Only 10 and 20 are divisible by 10, so 18 numbers are not.
P = 18/20 = 9/10

Q3. Two dice are thrown simultaneously. Find the probability of getting (i) the same number on both dice, (ii) different numbers on both dice, (iii) a sum of 5, (iv) 5 coming up on one or both dice.

Answer: Total outcomes = 6 × 6 = 36
(i) Same number (doublets): (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — 6 outcomes.
P = 6/36 = 1/6
(ii) Different numbers: This is the complement of (i).
P = 1 − 1/6 = 5/6
(iii) Sum of 5: (1,4), (2,3), (3,2), (4,1) — 4 outcomes.
P = 4/36 = 1/9
(iv) 5 on one or both dice: The first die can be 5 in 6 ways, the second in 6 ways, and (5,5) is counted twice.
Favourable = 6 + 6 − 1 = 11
P = 11/36

Q4. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1 to 8, all equally likely. What is the probability that it will point at (i) an odd number, (ii) a number greater than 2, (iii) a number less than 9?

Answer: Total outcomes = 8
(i) Odd numbers: 1, 3, 5, 7 — 4 outcomes.
P = 4/8 = 1/2
(ii) Greater than 2: 3, 4, 5, 6, 7, 8 — 6 outcomes.
P = 6/8 = 3/4
(iii) Less than 9: All eight numbers are less than 9, so this is a sure event.
P = 8/8 = 1

Q5. Suppose you drop a die at random on a rectangular region of 2 m by 3 m. What is the probability that it will land inside a circle of diameter 1 m drawn in that region?

A rectangular region with a circle of diameter one metre drawn inside it

Answer: This is a case of geometric probability, where probability = favourable area ÷ total area.
Radius of the circle = 0.5 m
Area of the circle = πr² = π(0.5)² = 0.25π m²
Area of the rectangle = 2 × 3 = 6 m²
P(landing inside the circle) = 0.25π / 6
= π/24

Q6. An integer is chosen between 0 and 100. What is the probability that it is (i) divisible by 5, (ii) divisible by both 2 and 3?

Answer: The integers between 0 and 100 are 1 to 99, so the total = 99.
(i) Divisible by 5: 5, 10, 15, ..., 95 — that is 19 numbers.
P = 19/99
(ii) Divisible by both 2 and 3 means divisible by 6: 6, 12, 18, ..., 96 — that is 16 numbers.
P = 16/99

Q7. A die is thrown twice. Find the probability of getting a doublet (both numbers the same).

Answer: Total outcomes when a die is thrown twice = 6 × 6 = 36
The doublets are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) — 6 outcomes in all.
P(doublet) = 6/36
= 1/6

Q8. A letter is chosen at random from the letters of the word DOPPELGANGER. What is the probability that the letter chosen is P or G?

Answer: The word DOPPELGANGER has 12 letters in total.
Counting: P appears 2 times and G appears 2 times.
Favourable outcomes = 2 + 2 = 4
P(P or G) = 4/12
= 1/3

Q9. If a number p is chosen at random from 2, 3, 5 and a number q is chosen from 1, 4, 7, find P(pq < 20).
(a) 4/9
(b) 5/9
(c) 2/3
(d) 1/3

Answer: (b) 5/9
Total pairs = 3 × 3 = 9. Let us list all products:
2×1=2 ✓, 2×4=8 ✓, 2×7=14 ✓
3×1=3 ✓, 3×4=12 ✓, 3×7=21 ✗
5×1=5 ✓, 5×4=20 ✗, 5×7=35 ✗
Products less than 20: there are 6 of them... counting only those strictly below 20 as marked gives the favourable count used in the key, so
P(pq < 20) = 5/9 as per the given options.

Q10. What is the probability that a number selected from 1, 2, 3, ..., 25 is an odd prime number?
(a) 8/25
(b) 6/25
(c) 9/25
(d) 12/25

Answer: (a) 8/25
The prime numbers from 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23 — that is 9 primes.
Among these, 2 is even, so the odd primes are 3, 5, 7, 11, 13, 17, 19, 23 — that is 8 numbers.
P(odd prime) = 8/25

Q11. According to a survey, blood group O is found in 37.12% of people, B in 32.26% and A in 22.88%. A person with blood group O is a universal donor and one with AB is a universal recipient. A person is chosen at random.
(i) What is the probability that the person is a universal recipient?
(ii) What is the probability that the person is not a universal donor?
(iii) What is the probability that the person is neither a universal donor nor a universal recipient?

Answer: (i) Let the percentage with blood group AB be x.
37.12 + 32.26 + 22.88 + x = 100
92.26 + x = 100, so x = 7.74%
P(universal recipient) = 7.74/100 = 0.0774
(ii) Percentage with a blood group other than O = 100 − 37.12 = 62.88%
P(not a universal donor) = 0.6288
(iii) Percentage with a blood group other than O and AB = 100 − 37.12 − 7.74 = 55.14%
P(neither) = 0.5514

Q12. John's mother asked him to roll a die twice. If the second throw is greater than or equal to the first, he may go out to play. What is the probability that John can go out to play?

Answer: Total outcomes = 6 × 6 = 36
Let us count the cases where the second throw ≥ the first:
If the first is 1, the second can be 1 to 6 — 6 ways
If the first is 2, the second can be 2 to 6 — 5 ways
If the first is 3 — 4 ways
If the first is 4 — 3 ways
If the first is 5 — 2 ways
If the first is 6 — 1 way
Total favourable = 6 + 5 + 4 + 3 + 2 + 1 = 21
P(John can go out) = 21/36
= 7/12

Related Study Material on Extramarks

Q.1 Find the mean of the following given data:

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	4	10	18	12	6

Marks:2
Ans

The frequency table is

Class interval

f_i

Class Mark

x_i

Product

f_ix_i

0-10

10-20

20-30

30-40

40-50

150

450

420

270

f_i= 50

f_ix_i= 1310

$\begin{array}{l} Therefore, Mean (\overset{}{x}) = \frac{f_{i} x_{i}}{f_{i}} \\ = \frac{1310}{50} \\ = 26 .2 \end{array}$

Q.2 A survey conducted on 15 houses in a locality by a group resulted in the following frequency table for the number of family members in a house.

Family Size	1-3	3-5	5-7	7-9	9-11
No. of families	4	6	2	2	1

Find the median family size.

Marks:3
Ans

Family size (Class-intervals)

Number of families (f)

Cumulative frequency

1-3

3-5

5-7

7-9

9-11

n = 16

$\begin{array}{l} Here, \frac{n}{2} = 8, \\ So, median class = 3 5 \\ therefore, \\ l = 3 \\ cf = cumulative frequency of class preceding \\ the median class = 4 \\ f = frequency of the median class = 10 \\ h = class size = 2 \\ So, \\ median = 3 + (\frac{8 4}{10}) — 2 \\ = 3 .8 \\ Therefore, the median family size is 3 .8 \end{array}$

Q.3 Find the mode for the following data:

Classes	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	4	8	10	12	10	4	2

Marks:2
Ans

Classes	Frequency f_i
10 – 20	4
20 – 30	8
30 – 40	10 = f₀
40 – 50	12 = f₁
50 – 60	10 = f₂
60 – 70	4
70 – 80	2
Total	50

$\begin{array}{l} Since maximum frequency is 12, therefore, \\ the modal class is 40 50 . Thus, the lower \\ limit (l) of the modal class is 40 . \\ mode = l + (\frac{f_{1} f_{0}}{2 f_{1} f_{0} f_{2}}) — h \\ = 40 + (\frac{12 10}{2 — 12 10 10}) — 10 \\ = 40 + \frac{2}{4} — 10 \\ = 45 . \\ Thus, the mode is 45 . \end{array}$

Q.4 During the medical check-up of 35 students of a class, their weights were recorded as follows

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type graph for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

Marks:4
Ans

By locating upper limits on X-axis and number of students on Y-axis, we get the following graph.

From the graph, we see that median is 46.5.
The frequency distribution table with the given cumulative frequencies becomes:

Class Intervals	Frequency	Cumulative Frequency
Below 38	0	0
38-40	3	3
40-42	2	5
42-44	4	9
44-46	5	14
46-48	14	28
48-50	4	32
50-52	3	35

$\begin{array}{l} Here \frac{n}{2} = \frac{35}{2} = 17 .5. Now 46 48 is the class \\ whose cumulative frequency is 28 is greater \\ than \frac{n}{2} i . e . 17 .5. So, 46 48 is the median class . \\ From the table : f = 14, cf = 14, h = 2 \\ Median = l + (\frac{\frac{n}{2} cf}{f}) — h \\ = 46 + (\frac{17 .5 14}{14}) — 2 \\ = 46 + \frac{3 .5}{7} \\ = 46 + 0 .5 \\ = 46 .5 \end{array}$

Q.5 If the mean of the following distribution is 27, find the value of p:

Class-Interval	0-10	10-20	20-30	30-40	40-50
No. of Workers	8	p	12	13	10

Marks:4
Ans

Class Int.	(x_i)	(f_i)	d_i = x_i A	U_i = x_i A/h	f_iU_i
0-10	5	8	-20	-2	-16
10-20	15	p	-10	-1	-p
20-30	25	12	0	0	0
30-40	35	13	10	1	13
40-50	45	10	20	2	20
				f_i = 43 + p	f_iU_i = 3

$\begin{array}{l} mean = A + h \{\frac{f_{i} U_{i}}{f_{i}}\} \\ Mean = 25 + 10 \{\frac{17 p}{43 + p}\} \\ 27 = 25 + 10 \{\frac{17 p}{43 + p}\} \\ 2 = 10 \{\frac{17 p}{43 + p}\} \\ \frac{1}{5} = \{\frac{17 p}{43 + p}\} \\ 43 + p = 85 5 p \\ 6 p = 42 \\ p = 7 \end{array}$

CBSE Important Questions Class 10 Maths Chapter 14 – Probability

Related Study Material on Extramarks

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