NCERT Exemplar Solutions for Class 10 Maths Chapter 13 Statistics and Probability Exercise 13.3

Chapter 13 Statistics and Probability NCERT Exemplar Solutions Exercise 13.3 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 13 Statistics and Probability Exercise 13.3
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 13.1 Exercise 13.2 Exercise 13.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 13.3 Solutions

1. Find the mean of the distribution 

Class	1 – 3	3 – 5	5 – 7	7 – 10
Frequency	9	22	27	17

Solution

We first, find the class mark xi, of each class and then proceed as follows.

Class	Class marks (xi)	Frequency (fi)	fi xi
1 – 3	2	9	18
3 – 5	4	22	88
5 – 7	6	27	162
7 – 10	8.5	17	144.5
		Σfi = 75	Σfi xi = 412.5

Therefore, mean (x) = (Σf_i x_i)/(Σf_i) = 412.5/75 = 5.5 
Hence, mean of the given distribution is  5.5.

 

2. Calculate the mean of the scores of 20 students in a mathematics test

Marks	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Number of students	2	4	7	6	1

Solution

We first, find the class mark of each class and then proceed as follows 

Marks	Class marks (xi)	Frequency (fi)	fi xi
10 – 20	15	2	30
20 – 30	25	4	100
30 – 40	35	7	245
40 – 50	45	6	270
50 – 60	55	1	55
		Σfi = 20	Σfi xi = 700

Therefore, mean (x) = (Σf_i x_i)/(Σf_i) = 700/20 = 35
Hence, the mean of scores of  20 students in mathematics test is  35. 

3. Calculate the mean of the following data 

Class	4 – 7	8 – 11	12 – 15	16 – 19
Frequency	5	4	9	10

Solution

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in

the upper limit of each class.

Now, we first find the class mark xi, of each class and then proceed as follows

Class	Class marks (xi)	Frequency (fi)	fi xi
3.5 - 7.5	5.5	5	27.5
7.5 – 11.5	9.5	4	38
11.5 – 15.5	13.5	9	121.5
15.5 – 19.5	17.5	10	175
		Σfi = 28	Σfi xi = 362

Therefore, mean (x) = (Σf_i x_i)/(Σf_i)= 362/28 = 12.93
Hence, mean of the given data is 12.93.

4. The following table gives the number of pages written by Sarika for completing her own book for 30 days.

Number of pages written per day	16 – 18	19 – 21	22 – 24	25 – 27	28 – 30
Number of days	1	3	4	9	13

Find the mean number of pages written per day.

Solution
Since, 

Class - marks	Mid - value (xi)	Number of days (fi)	fi xi
15.5 – 18.5	17	1	17
18.5 – 21.5	20	3	60
21.5 – 24.5	23	4	92
24.5 – 27.5	26	9	234
27.5 – 30.5	29	13	377
Total		Σfi = 30	Σfi xi = 780

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class. 

∴  mean (x) = (Σf_i x_i)/(Σf_i) = 780/30 = 26

Hence, the mean of pages written per day is 26.

5. The daily income of a sample of 50 employees are tabulated as follows.

Income (in ₹)	1 – 200	201 – 400	400 – 600	601 – 800
Number of employees	14	15	14	7

Find the mean daily income employees.

Solution

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in

the upper limit of each class.

Now we first, find the class mark xi, of each class and then proceed as follows

∴ Assumed mean,  a = 300.5 
Class width, h = 200
and total observations, N = 50 
By step deviation method, 

= 300.5 + 200 × (1/50) × 14 
= 300.5 + 56 = 356.5 

6. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table.

Number of seats	100 – 104	104 – 108	108 – 112	112 – 116	116 – 120
Freqency	15	20	32	18	15

Determine the mean number of seats occupied over the flights. 

Solution

We first, find the class mark xi, of each class and then proceed as follows.

∴ Assumed mean, a = 110, 
Class width, h = 4 
and total observation, N = 100 
By assumed mean method, 
Mean ( x ) = a +  (Σf_i d_i )/(Σf_i )  
= 110 + (-8/100) = 110 - 0.08 = 109.92 

7. The weights (in kg) of 50 wrestlers are recorded in the following table. Find the mean weight of the wrestlers.

Weight (in kg)	100 – 110	110 – 120	120 – 130	130 – 140	140 – 150
Number of wrestlers	4	14	21	8	3

Solution

We first the class mark of each class and then proceed as follows : 

∴ Assumed mean (a) = 125 
Class width (h) = 10 and total observation (N) = 50 
By assumed mean method, 

Mean (x) = a +  (Σf_i d_i)/(Σf_i)  
= 125 + (-80)/50 
= 125 - 16 = 123.4 kg

8. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and

details are tabulated as given below

Mileage(kmL-1 )	10 – 12	12 – 14	14 – 16	16 – 18
Number of cars	7	12	18	13

 Find the mean mileage.

The manufacturer claimed that the mileage of the model was 16 kmL -1.
Do you agree with this claim?

Solution

Here, Σf_i = 50 
and Σf_i x_i = 724 

∴ Mean (x) = (Σf_i x_i)/Σf_i 
= 724/50 = 14.48

Hence, mean mileage is 14.48 kmL^-1 .

No, the manufacturer is claiming mileage 1.52 kmh^-1 more than average mileage.

9. The following is the distribution of weights (in kg) of 40 persons. 

Weight (in kg)	40 – 45	45 – 50	50 – 55	55 – 60	60 – 65	65 – 70	70 – 75	75 – 80
Number of persons	4	4	13	5	6	5	2	1

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Solution

The cumulative distribution (less than type) table is shown below 

Weight (in kg)	Cumulative frequency
Less than 45	4
Less than 50	4 + 4 = 8
Less than 55	8 + 13 = 21
Less than 60	21 +5 = 26
Less than 65	26 + 6 = 32
Less than 70	32 + 5 = 37
Less than 75	37 + 2 = 39
Less than 80	39 + 1 = 40

10. The following table shows the cumulative frequency distribution of marks of 800 students in an examination.

Marks	Number of students
Below 10	10
Below 20	50
Below 30	130
Below 40	270
Below 50	440
Below 60	570
Below 70	670
Below 80	740
Below 90	780
Below 100	800

Construct a frequency distribution table for the data above.

Solution

Here, we observe that 10 students have scored marks below 10 i.e., it lies between class interval 0-10. Similarly, 50 students have scored marks below 20. So, 50 -10 = 40 students lies in the interval 10-20 and so on. The table of a frequency distribution for the given data is

11. From the frequency distribution table from the following data.

Marks (Out of 90)	Number of candidates
More than or equal to 80	4
More than or equal to 70	6
More than or equal to 60	11
More than or equal to 50	17
More than or equal to 40	23
More than or equal to 30	27
More than or equal to 20	30
More than or equal to 10	32
More than or equal to 0	34

Solution

Here, we observe that, all 34 students have scored marks more than or equal to 0. Since, 32 students have scored marks more than or equal to 10. So, 34- 32 =2 students lies in the interval 0-10 and so on.

Now, we construct the frequency distribution table.

Class interval	Number of candidates
0 – 10	34 – 32 = 2
10 – 20	32 – 30 = 2
20 – 30	30 – 27 = 3
30 – 40	27 – 23 = 4
40 – 50	23 – 17 = 6
50 – 60	17 – 11 = 6
60 – 70	11 – 6 = 5
70 – 80	6 – 4 = 2
80 – 90	4

12. Find the unknown entries o, b, c, d, e and f in the following distribution of heights of students in a class

Height (in cm)	Frequency	Cumulative frequency
150 – 155	12	A
155 – 160	b	25
160 – 165	10	C
165 – 170	d	43
170 – 175	e	48
175 – 180	2	f
Total	50

Solution

On comparing last two tables, we get 

a = 12 
∴ 12 + b = 25 
⇒ b = 25 - 12 = 13 
22 + b = c
⇒ c = 22 + 13 = 35 
22 + b + d = 43 
⇒ 22 + 13 + d = 43 
⇒ d = 43 - 35 = 8 
and 22 + b + d + e = 48 
⇒ 22 + 13 + 8 + e = 48 
⇒ e = 48 - 43 = 5
and 24 + b + d + e = f 
⇒ 24 + 13 + 8 + 5 = f 
∴  f = 50

13. The following are the ages of 300 patients getting medical treatment in a hospital on a particular day

Age (in year)	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Number of patients	60	42	55	70	53	20

Form

(i) less than type cumulative frequency distribution.

(ii) More than type cumulative frequency distribution.

Solution

(i) We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is O’. Similarly, less than 20 include the number of patients which take medical treatment from 0-10 as well as the number of patients which take medical treatment from 10-20.

So, the total number of patients less than 20 is 0 + 60 = 60, we say that the cumulative frequency of the class 10-20 is 60. Similarly, for other class.

(ii) Also, we observe that all 300 patients which take medical treatment more than or equal to 10. Since, there are 60 patients which take medical treatment in the interval 10-20, this means that there are 300 –60 = 240 patients which take medical treatment more than or equal to 20. Continuing in the same manner.

14. Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class

Marks	Below 20	Below 40	Below 60	Below 80	Below 100
Number of students	17	22	29	37	50

Form the frequency distribution table for the data.  

Solution

Here, we observe that, 17 students have scored marks below 20 i.e., it lies between class interval 0-20 and 22 students have scored marks below 40, so 22 -17 = 5 students lies in the class interval 20-40 continuing in the same manner, we get the complete frequency distribution table for given data.

Marks	Number of students
0 – 20	17
20 – 40	22 – 17 = 5
40 – 60	29 – 22 = 7
60 – 80	37 – 29 = 8
80 – 100	50 – 37 = 13

15. Weekly income of 600 families is tabulated below

Weekly Income(in ₹)	Number of families
0 – 1000	250
1000 – 2000	190
2000 – 3000	100
3000 – 4000	40
4000 – 5000	15
5000 – 6000	5
Total	600

Compute the median income

Solution

First we construct a cumulative frequency table 

It is given that, n = 600 
∴ n/2 = 600/2 = 300 
Since, cumulative frequency 440 lies in the interval 1000 - 2000. 
Here, (lower median class)l = 1000, 
f = 190, cf = 250, (class width) h = 1000
and (total observation ) n = 600 

= 1000 + 263.15 = 1263.15 

Hence, the median income is ₹ 1263.15.

16. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows

Speed (in km/h)	85 – 100	100 – 115	115 – 130	130 – 145
Number of players	11	9	8	5

Calculate the median bowling speed. 

Solution

First we construct the cumulative frequency table 

It is given that, n = 33
∴ n/2 = 33/2 = 16.5 
So, the median class is  100 - 115, 

where, lower limit (l) = 100, 
frequency (f) = 9, 
cumulative frequency (cf) = 11 
and class width (h) = 15 

= 109.17 
Hence, the median bowling speed is 109.17 km/h. 

17. The monthly income of  100 families are given as below 

Income (in ₹)	Number of families
0 – 5000	8
5000 – 10000	26
10000 – 15000	41
15000 – 20000	16
20000 – 25000	3
25000 – 30000	3
30000 – 35000	2
35000 – 40000	1

Calculate the model income.

Solution

In a given data, the highest frequency is 41, which lies in the interval 10000-15000.

Hence, the modal income is ₹ 11875. 

18. The weight of coffee in  70 packets are shown in the following table 

Weight (in g)	Number of packets
200 – 201	12
201 – 202	26
202 – 203	20
203 – 204	9
204 – 205	2
205 – 206	1

Determine the model weight.

Solution

In the given data, the highest frequency is 26, which lies in the interval 201 - 202. 

Hence, the modal weight is 201.7 g. 

19. Two dice are thrown at the same time. Find the probability of getting

(i) same number on both dice.

(ii) different number on both dice.

Solution

Two dice are thrown at the same time. [given]

So, total number of possible outcomes = 36

(i) We have, same number on both dice.

So, possible outcomes are (1,1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
∴ Number of possible outcomes = 6 
Now, required probability = 6/36 = 1/6 

(ii) We have, different number on both dice.

So, number of possible outcomes 
= 36 - Number of possible outcomes for same number on both dice
= 36 - 6 = 30 
∴ Required probability = 30-36 = 5/ 6 

20. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

(i) 7 ?
(ii) a prime number ?
(iii) 1 ?

Solution

Two dice are thrown simultaneously. [given]

So, total number of possible outcomes = 36

(i) Sum of the numbers appearing on the dice is 7.

So, the possible ways are (1, 6), (2, 5), (3, 4), (4, 3) , (5, 2) and  (6, 1 )
Number of possible ways = 6 
∴ Required probability = 6/36 = 1/6 

(ii) Sum of the numbers appearing on the dice is a prime number i.e. 2, 3, 5, 7 and 11. 
So, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5) . 

Number of possible ways = 15 
∴ Required probability = 15/36 = 5/12 

(iii) Sum of the numbers appearing on the dice is 1. 
It is not possible, so its probability is zero. 

21. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

(i) 6
(ii) 12
(iii) 7

Solution

Number of total outcomes = 36 

(i) When product of the numbers on the top of the dice is  6. 

So, the possible ways are (1, 6), (2, 3), (3, 2) and (6, 1). 
Number of possible ways  = 4 
∴ Required probability = 4/36 = 1/9 

(ii) When product of the numbers on the top of the dice is 12. 

So, the possible ways are (2, 6), (3, 4), (4, 3) and (6, 2). 
Number of possible ways = 4 
∴ Required probability = 4/36 = 1/9 

(iii) Product of the numbers on the top of the dice cannot be 7. So, its probability is zero. 

22. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Solution

Number of total outcomes = 36

When product of numbers appearing on them is less than 9, then possible ways are (1,6),

(1,5) (1,4), (1,3), (1,2), (1,1), (2, 2), (2, 3), (2, 4), (3, 2), (4, 2), (4,1), (3,1), (5,1), (6,1) and (2,1).

Number of possible ways = 16

Required probability = 16/36 = 4/9 

23. Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9, separately.

Solution

Number of total outcomes = 36

24. A coin is tossed two times. Find the probability of getting at most one head.

Solution

The possible outcomes, if a coin is tossed 2 times is
S = {(HH), (TT), (HT), (TH)}
∴ n(S) = 4 
Let E = Event of getting at most one head 

= {(TT), (HT), (TH)} 
∴ n(E) = 3 
Hence, required probability = n(E)/n(S) = 3/4 

25. A coin is tossed 3 times. List the possible outcomes. Find the probability of getting

(i) all heads
(ii) at least 2 heads

Solution

The possible outcomes if a coin is tossed 3 times is

S = {(HHH), (TTT), (HTT), (THT), (TEH), (THH), (HTH), (HHT)}

26. Two dice are thrown at the same time. Determine the probability that the difference of the

numbers on the two dice is 2.

Solution

The total number of sample space in two dice, n(S) = 6 × 6 = 36 
Let,   E = Event of getting the numbers whose difference is 2 
= {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
∴ n(E) = 8 
∴ P(E)  = n(E)/n(S) = 8/36 = 2/9 

27. A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a

(i) red ball
(ii) green ball
(iii) not a blue ball

Solution

if a ball is drawn out of 22 balls (5 blue + 7 green + 10 red), then the total number of outcomes are

n(S) = 22 

28. The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now, one card is drawn at fandom from the remaining cards. Determine the probability that the card is

(i) a heart
(ii) a king

Solution

If we remove one king, one queen and one jack of clubs from 52 cards, then the remaining cards left, n(S) = 49

29. Refer to Q.28. What is the probability that the card is

(i) a club
(ii) 10 of hearts

Solution

30. All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value.

(i) 7
(ii) greater than 7
(iii) Less than 7

Solution

In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining

cards are left, n(S) = 52 – 3 × 4 = 40.

(i) Let E₁ = Event of getting a card whose value is 7 
E = Card value 7 may be of a spade, a  diamond, a club or a heart 

∴ n(E₁) = 4 
∴ P(E₁) = n(E₁)/n(S) = 4/40 = 1/10 

(ii) Let E₂ = Event of getting a card whose value is greater than 7 

= Event of getting a card whose value is 8, 9 or 10 
∴ n(E₂) = 3 × 4 = 12 
P(E₂ ) = n(E₂)/n(S) = 12/40 = 3/10 

(iii) Let E₃ = Event of getting a card whose value is less than 7 
= Event of getting a card whose value is 1, 2, 3, 4, 5 or  6 
∴   n(E₃) = 6 × 4 = 24 
∴  P(E₃) = n(E₃)/n(S) = 24/40 = 3/5 

31. An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7?
(ii) not divisible by 7?

Solution

The number of integers between 0 and  100 is 
n(S)= 99 

(i) Let E₁ = Event of choosing an integer which is divisible by 7 
= Event of choosing an integer which is multiple of 7

= {7, 14, 21, 28, 35, 42, 49, 56,63, 70, 77, 84, 91, 98}
∴  n(E₁) = 14 
∴ P(E₁) = n(E₁)/n(S) = 14/99

(ii) Let E₂ = Event of choosing an integer which is not divisible by 7 

∴ n(E₂) = n(S) - n(E₁) = 99 - 14 = 85  

P(E₂) = n(E₂)/n(S) = 85/99

32. Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has

(i) an even number
(ii) a square number

Solution

Total number of out comes with numbers  2 to 101, n(S) = 100 

(i) Let E₁ = Event of selecting a card which is an even number = {2, 4, 6, ....100}
[in an AP, l = a + (n-1)d, here l = 100, a = 2 and d = 2

⇒ 100 = 2 + (n - 1)2 

⇒ (n - 1) = 49 ⇒ n = 50]
∴ n(E₁ ) = 50 
∴ Required probability = n(E₁ )/n(S) = 50/100 = 1/2 

(ii) Let E₂ = Event of selecting a card which is a square number 

= {4, 9, 16, 25, 36, 49, 64, 81, 100}
= {(2)² ,(3)² , (4)² , (5)² , (6)² , (7)² , (8)² , (9)² , (10)² }
∴ n(E₂ ) = 9 
Hence, required probability = n(E₂ )/n(S) = 9/100 

33. A letter of english alphabets is chosen at random. Determine the probability that the letter is a consonant.

Solution

We know that, in english alphabets, there are (5 vowels + 21 consonants)=26 letters. So, total number of outcomes in english alphabets
are, n(S) = 26 
Let  E = Event of choosing a english alphabet, which is a consonant 

= {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
∴ n(E) = 21 

Hence, required probability = n(E)/n(S) = 21/26 

34. There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of ₹ 100 each, 100 of them contain a cash prize of ₹ 50 each and 200 of them contain a cash prize of ₹ 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

Solution

Total number of sealed envelopes in a box, n (S) = 1000

Number of envelopes containing cash prize = 10 + 100 + 200 = 310

Number of envelopes containing no cash prize,

n(E) = 1000 - 310 = 690 
∴ P(E) = n(E)/n(S) = 690/1000 = 69/100 = 0.69 

35. Box A contains 25 slips of which 19 are marked ₹ 1 and other are marked ₹ 5 each. Box B contains 50 slips of which 45 are marked ₹ 1 each and others are marked ₹ 13 each. Slips of both boxes are poured into a third box and resuffled. A slip is drawn at random. What is the probability that it is marked other than ₹ 1?

Solution

Total number of slips in a box, n(S) = 25 + 50 = 75 

From the chart it is clear that, there are 11 slips which are marked other than ₹ 1. 

∴ Required probability  = (Number of slips other than ₹ 1)/(Total number of slips) = 11/75 

36. A carton of 24 bulbs contain 6 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

Solution

∴ Total number of bulbs, n(S) = 24 

Let E₁ = Event of selecting not defective bulb = Event of selecting good bulbs 
n(E₁) = 18 
∴ P(E₁) = n(E₁)/n(S) = 18/24 = 3/4 
Suppose, the selected bulb is defective and not replaced, then total number of bulbs  remains in a carton, n(S) = 23.
In them, 18 are good bulbs and 5 are defective bulbs. 

∴ P (selecting second defective bulb) = 5/23 

37.  A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a

(i) triangle
(ii) square
(iii) square of blue colour
(iv) triangle of red

Solution

Total number of figures 

n(S) = 8 triangles + 10 squares = 18 
(i) P (lost piece is a triangle ) = 8/18 = 4/9 

(ii) P (lost piece is a square) = 10/18 = 5/9 
(iii) P (square of blue colour) = 6/18 = 1/3 
(iv) P(triangle of red colour) = 5/18 

38. In a game, the entry fee is of ₹ 5. The game consists of a tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she

(i) loses the entry fee.

(ii) gets double entry fee.

(iii) just gets her entry fee.

Solution

Total possible outcomes of tossing a coin 3 times,

S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}

∴ n (S) = 8

(i) Let E₁ = Event that Sweta losses the entry fee

= She tosses tail on three times 

n(E₁) = {(TTT)}
P(E₁) = n(E₁)/n(S) = 1/8 

(iii) Let E₃ = Event that Sweta gets her entry fee back 
= Sweta gets heads one or two times 
= (HTT), (THT), (TTH), (HHT) , (HTH), (THH)}
∴  n(E₃) = 6
∴ P(E₃) = n(E₃)/n(S) = 6/8 = 3/4 

39. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total

score is recorded.

(i) How many different scores are possible?

(ii) What is the probability of getting a total of 7 ?

Solution

Given, a die has its six faces marked {0,1,1,1,6, 6}

Total sample space, n(S) = 6² = 36

(i) The different score which are possible are 6 scores e., 0,1,2,6,7 and12.

(ii) Let E = Event of getting a sum 7 

= {(1, 6) , (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)
∴ n(E) = 12 
∴ P(E) = n(E)/n(S) = 12/36 = 1/3 

40. A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone, if it is good but the trader will only buy a mobile, if it has no major defect. One phone is selected at random from the lot. What is the probability that it is

(i) acceptable to Varnika?

(ii) acceptable to the trader?

Solution

Given, total number of mobile phones 

n(S) = 48 
(i) Let E₁ = Event that Varnika will buy a mobile phone 

= Varnika buy only, if it is good mobile 

∴ n(E₁)= 42 
∴ P(E₁) = n(E₁ )/n(S) = 42/48 = 7/8 

(ii) Let E₂ = Event that trader will buy only when it has no major defects 

= Trader will buy only 45 mobiles 

∴ n(E₂) = 45 
∴ P(E₂) = n(E₂)/n(S) = 45/48 = 15/16 

41. A bag contains 24 balls of which x are red, 2x are white and 3x am are. A ball is selected at random. What is the probability that it

(i) not red?
(ii) white

Solution

Given that, A bag contains total number of balls = 24 A bag contains number of red bails = 24

A bag contains number of white balls = 2x and a bag contains number of blue balls = x

By condition, x + 2x + 3x = 24

⇒ 6x = 24

∴ x = 4
∴ Number of red balls = x = 4 
Number of white balls = 2x = 2 ×4 = 8 

and number of blue balls = 3x = 3 × 4 = 12

So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.

⇒ n(S) = 24

(i) Let E₁ = Event of selecting a ball which is not red i.e., can be white or blue, 
∴ n(E₁) = Number of white balls + Number of blue balls 
⇒ n(E₁) = 8 + 12 = 20 
∴ Required probability = n(E₁)/n(S) = 20/24 = 5/6 

(ii) Let E₂ = Event of selecting a ball which is white 

∴ n(E₂) = Number of white balls = 8 
So, required probability = n(E₂)/n(S) = 8/24 = 1/3 

42. At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each

player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize?

(ii) the second player wins a prize, if the first has won?

Solution

Given that,, at a fete, cards bearing numbers 1 to 1000 one number on one card, are put in a

box. Each player selects one card at random and that card is not replaced so, the total number of outcomes are n(S) = 1000
If the selected card has a perfect square greater than 500, then player wins a prize. 

(i) Let E₁ = Event first player wins a prize = Player select a card  which is a perfect square greater than 500

= { 529, 576, 625, 676, 729, 784, 841, 900, 961}

(ii) First, has won i.e., one card is already selected, greater than 500, has a perfect square. Since, repeatition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining cards is  999. 

∴ Total number of remaining outcomes, n(S') = 999 

Let E₂ = Event the second player wins a prize, if the first has won. 

= Remaining cards has a perfect square greater than 500 are 8. 
∴ n(E₂) = 9 - 1 = 8 
So, required probability  = n(E₂)/n(S') = 8/999