RD Sharma Class 9 solutions Chapter 25 Probability

RD Sharma Class 9 Solution Chapter 25 Probability Ex 25.1

Question 1.
A coin is tossed 1000 times with the following frequencies
Head : 455, Tail : 545.
Compute the probability for each event.
Solution:
Total number of events (m) 1000
(i) Possible events (m) 455
∴Probability P(A) = mn = 4551000
= 91200 = 0.455
(ii) Possible events (m) = 545
∴Probability P(A) = mn = 5451000 = 109200 = 0.545

Question 2.
Two coins are tossed simultaneously 500 times with the following frequencies of different
outcomes:
Two heads : 95 times
One tail : 290 times
No head: 115 times
Find the probability of occurrence of each of these events.
Solution:
Two coins are tossed together simultaneously 500 times
∴ Total outcomes (n) 500
(i) 2 heads coming (m) = 95 times
∴Probability P(A) = mn
= No.ofpossibleeventsTotalnumberofevents
= 95500 = 19100 = 0.19
(ii) One tail (m) = 290 times
∴Probability P(A) = mn = 290500 = 5801000 = 58100 = 0.58
(iii) No head (m) = 115 times
∴Probability P(A) = mn = 115500 = 23100 = 0.23

Question 3.
Three coins are tossed simultaneously 1oo times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.
Solution:
Three coins are tossed simultaneously 100 times
Total out comes (n) = 100

(i) Probability of 2 heads coming up (m) = 36
Probability P(A) = mn = 36100 = 0.36
(ii) Probability of 3 heads (m) = 12
ProbabilityP(A)= mn = 12100 = 0.12
(iii) Probability of at least one head coming up (m) = 38 + 36 + 12 = 86
Probability P(A) = mn = 86100 = 0.86
(iv) Probability of getting more heads than tails (m) = 36 + 12 = 48
Probability P(A) = mn = 48100 = 0.48
(v) Getting more tails than heads (m) = 14 + 38 = 52
Probability P(A) = mn = 52100 = 0.52

Question 4.
1500 families with 2 children were selected randomly and the following data were recorded:

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution:
Total number of families (n) = 1500

(i) Probability of a family having no girls (m) = 211
∴Probability P(A)= mn = 2111500 = 0.1406
(ii) Probability of a family having one girl (in) = 814
∴Probability P(A) = mn = 8141500 = 0.5426
(iii) Probability of a family having 2 girls (m) = 475
∴Probability P(A) = mn = 4751500 = 0.3166
(iv) Probability of a family having at the most one girls
∴m = 814 + 211 = 1025
∴Probability P(A) =mn = 10251500 = 0.6833
(v) Probability of a family having more girls than boys (m) = 475
∴Probability P(A) = mn = 4751500 = 0.3166

Question 5.
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that on a ball played:
(i) he hits boundary
(ii) he does not hit a boundary.
Solution:
Total balls played (n) 30
No. of boundaries = 6
(i) When the batsman hits the boundary = 6
∴m = 6
∴Probability P(A) = mn = 630 = 15 = 0.2
(ii) When the batsman does not hit the boundary (m) = 30 – 6 = 24
∴Probability P(A) = mn = 2430 = 45 = 0.8

Question 6.
The percentage of marks obtained by a student in monthly unit tests are given below:

Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Solution:
Percentage of marks obtain in

(i) Probability of getting more than 70% marks (m) = In unit test II, III, V = 3
Total unit test (n) = 5
∴Probability P(A) = mn = 35 = 0.6
(ii) Getting less then 70% marks = units test I and IV
∴m = 2
Total unit test (n) = 5
∴Probability P(A) = mn = 25 = 0.4
(iii) Getting a distinction = In test V (76 of marks)
∴m = 1
Total unit test (n) = 5
∴Probability P(A) = mn = 15 = 0.2

Question 7.
To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:

Find the probability that a student chosen at random
(i) likes Mathematics
(ii) does not like it.
Solution:
Total number of students (n) = 200

(i) Probability of students who like mathematics (m) = 135
∴Probability P(A) = mn = 135200 = 0.675
(ii) Probability of students who dislike mathematics (m) = 65
∴Probability P(A) = mn = 65200 = 0.325

Question 8.
The blood groups of 30 students of class IX are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O,
A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:
(i) A (ii) B (iii) AB (iv) O
Solution:
Total number of students of IX class = 30
No. of students of different blood groups
A AB B O
9 3 6 12

Question 9.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour
(in kg):
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Number of total bags (n) = 11
No. of bags having weight more than 5 kg (m) = 7
Probability P(A) = mn = 711

Question 10.
Following table shows the birth month of 40 students of class IX.

Find the probability that a student was born in August.
Solution:
Total number of students (n) = 40

Number of students who born in Aug. (m) = 6
Probability P(A) = mn = 640 = 320

Question 11.
Given below is the frequency distribution table regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Find the probability of concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
Total number of days (n) = 30

Probability of cone, of S02 of the interval 0.12-0.16 (m) = 2
Probability P(A) = mn = 230 = 115

Question 12.
A company selected 2400 families at random and survey them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

If a family is chosen, find the probability that the family is:
(i)earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(ii)earning Rs 16000 or more per month and owning exactly I vehicle.
(iii)earning less than Rs 7000 per month and does not own any vehicle.
(iv)earning Rs 13000-16000 per month and owning more than 2 vehicle.
(v)owning not more than 1 vehicle.
(vi)owning at least one vehicle.
Solution:
Total number of families (n) = 2400

(i) Number of families earning income Rs 10000-13000 and owning exactly 2 vehicles (m) = 29
Probability P(A) = mn = 292400
(ii) Number of families earning income Rs 16000 or more having one vehicle (m) = 579
Probability P(A) = mn = 5792400
(iii) Number of families earning income less than Rs 7000 having no own vehicle (m) = 10
Probability P(A) = mn = 102400 = 1240
(iv) Number of families having X13000 to X16000 having more than two vehicles (m) = 25
Probability P(A) = mn = 252400 = 196
(v) Number of families owning not more than one vehicle (m)
= 10 + 1 + 2 + 1 + 160 + 305 + 533 + 469 + 579 = 2062
Probability P(A) = mn = 20622400 = 10311200
(vi) Number of families owning at least one vechile (m) = 2048 + 192 + 110 = 2356
Probability P(A) = mn = 23562400 = 589600

Question 13.
The following table gives the life time of 400 neon lamps:

A bulb is selected at random. Find the probability that the life time of the selected bulb is: (i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.
Solution:
Total number of neon lamps (n) = 400

A bulb is chosen:
(i)No. of bulbs having life time less than 400 hours (m) = 14
Probability P(A) = mn = 14400 = 7200
(ii)No. of bulbs having life time between 300 to 800 hours (m) = 14 + 56 + 60 + 86 + 74 = 290
Probability P(A) = mn = 290400 = 2940
(iii)No. of bulbs having life time at least 700 hours (m) = 74 + 62 + 48 = 184
Probability P(A) = mn = 184400 = 2350

Question 14.
Given below is the frequency distribution of wages (in Rs) of 30 workers in a certain factory:

A worker is selected at random. Find the probability that his wages are:
(i) less than Rs 150
(ii) at least Rs 210
(iii) more than or equal to 150 but less than Rs 210.
Solution:
A worker is selected.
(i)No. of workers having less than Rs 150 (m) = 3 + 4 = 7
Probability P(A) = mn = 730
(ii)No. of workers having at least Rs 210 (m) = 4 + 3 = 7
Probability P(A) = mn = 730
(iii)No. of workers having more than or equal to Rs 150 but less than Rs 210 = 5 + 6 + 5 = 16
Probability P(A) = mn = 1630 = 815

Probability Class 9 RD Sharma Solutions VSAQS

Question 1.
Define a trial.
Solution:
When we perform an experiment, it is called a trial of the experiment.

Question 2.
Define an elementary event.
Solution:
An outcome of a trial of an experiment is called an elementary event.

Question 3.
Define an event.
Solution:
An event association to a random experiment is said to occur in a trial.

Question 4.
Define probability of an event.
Solution:
In n trials of a random experiment if an event A happens m times, then probability of happening
of A is given by P(A) = mn

Question 5.
A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is 25, find the number of red balls in the bag
Solution:
No. of white balls = 4
Let number of red balls = x
Then total number of balls (n) = 4 white + x red = (4 + x) balls

Question 6.
A die is thrown 100 times. If the probability of getting an even number is 25. How many times an odd number is obtained?
Solution:
Total number of a die is thrown = 100
Let an even number comes x times, then probability of an even number = x100

Question 7.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Find the probability of getting at most two heads.
Solution:
Total number of three coins are tossed (n) = 200
Getting at the most 2 heads (m) = 72 + 77 + 28 = 177
Probability P(A) = mn = 177200

Question 8.
In the Q. No. 7, what is the probability of getting at least two heads?
Solution:
Total number of possible events = 200
No. of events getting at the least = 2 heads (m) = 23 + 72 = 95
Probability P(A) = mn = 95200 = 1940

RD Sharma Class 9 PDF Chapter 25 Probability MCQS

Mark the correct alternative in each of the following:

Question 1.
The probability of an impossible event is
(a) 1
(b) 0
(c) less than 0
(d) greater than 1
Solution:
The probability of an impossible event is 0 (b)

Question 2.
The probability on a certain event is
(a) 0
(b) 1
(c) greater than 1
(d) less than 1
Solution:
The probability of a certain event is 1 (b)

Question 3.
The probability of an event of a trial is
(a) 1
(b) 0
(c) less than 1
(d) more than 1
Solution:
The probability of an even of a trial is less than 1 (c)

Question 4.
Which of the following cannot be the probability of an event?
(a) 13
(b) 35
(c) 53
(d) 1
Solution:
The probability of an event is less than 1
53 i.e .123 is not the probability

Question 5.
Two coins are tossed simultaneously. The probability of getting atmost one head is
(a) 14
(b) 34
(c) 12
(d) 14
Solution:
Total number of possible events (n) = 2 + 2 = 4
Number of events coming at the most 1 head (m) 2 times + 1 times = 3
Probability P(A) = mn = 34 (b)

Question 6.
A coin is tossed 1000 times, if the probability of getting a tail is 3/8, how many times head is obtained?
(a) 525
(b) 375
(c) 625
(d) 725
Solution:
No. of times a coin is tossed (n) = 1000
Probability of getting a tail = 38
Let No. of tail come = x
Probability P(A) = mn=x1000
x1000 = 38
=> x1000=38 => 3X10008=3X125
=> x = 375
∴ No. of heads are obtained = 1000 – 375 = 625 (c)

Question 7.
A dice is rolled 600 times and the occurrence of the outcomes 1, 2, 3, 4, 5 and 6 are given below:

The probability of getting a prime number is
(a)13
(b)23
(c)4960
(d)39125
Solution:
Total number of times a dice is rolled (n) = 600

Now total number of times getting a prime number i.e. 2, 3, 5, (m) = 30 + 120 + 50 = 200
Probability P(A) = mn = 200600 = 13 (a)

Question 8.
The percentage of attendance of different classes in a year in a school is given below:

What is the probability that the class attendance is more than 75%?
(a) 16
(b) 13
(c) 56
(d) 12
Solution:
Percentage of attendance of different classes

Total attendance more than 75% (m) VIII,VII and VI = 3 classes
and total number of classes (n) = 6
Probability P(A) = 36 = 12.

Question 9.
A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at random.
The probability that the number on the coin is not a prime number, is
(a) 15
(b) 35
(c) 25
(d) 45
Solution:
Total number of coins (n) = 50
Prime numbers between 51 to 100 are 53, 59, 6, 67, 71, 73, 79, 83, 89, 97 = 10
Numbers which are not primes (m) = 50 – 10 = 40
Probability P(A) = mn = 4050 = 45(d)

Question 10.
In a football match, Ronaldo makes 4 pals from 10 penalty kids. The probability of converting a penalty kick into a goal by Ronaldo,is
(a) 14
(b) 16
(c) 13
(d) 25
Solution:
No. of penalty kicks (n) = 10
No. of goal scored (m) = 4
Probability of converting a penally Into goals P(A) = 410 = 25(d)
RD Sharma Class 9 solutions Chapter 25 Probability