RD Sharma Solutions Class 9 Probability Exercise 25.1

RD Sharma Solutions Class 9 Chapter 25 Exercise 25.1

RD Sharma Class 9 Solutions Chapter 25 Ex 25.1 Download

Q1.  A coin is tossed 1000 times with the following sequence: Head: 455, Tail=:545. Compute the probability of each event

 

Answer:

It is given that the coin is tossed 1000 times. The number of trials is 1000

Let us denote the event of getting head and of getting tails be E and F respectively. Then

Number of trials in which the E happens= 455

So, Probability of E=   \(\frac{Number of event heads}{Total no of trials}\)

i.e.  P(E) \(=\frac{455}{1000}\) =0.455

Similarity, the probability of the event getting a tail = \(\frac{Number of tails}{Total no of trials}\)

i.e.  P(F) \(=\frac{545}{1000}\) =0.545

 

Q2. Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:

TWO HEADS: 95 times

ONE HEADS: 290 times

NO HEADS: 115 times

Find the probability of occurrence of each of these events

 Answer:

Probability (E) = \(\frac{number of trials in which events happen}{Total no of trials}\)

P(Getting two heads) \(= \frac{95}{500}\)=0.19

P(Getting one tail) \(= \frac{290}{500}\)=0.58

P(Getting no head) \(= \frac{115}{500}\)=0.23

 

Q3.  Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

OUTCOME NO HEAD ONE HEAD TWO HEAD THREE HEAD
FREQUENCY 14 38 36 12

 If the three coins are tossed simultaneously again, compute the probability of:

1.  heads coming up

2. heads coming up

3. At least one Head coming up

4. Getting more Tails than Heads 

5. Getting more heads than tails

 

ANS:

1: Probability of 2 Heads coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{36}{100}\) =0.36

2. Probability of 3 Heads coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{12}{100}\) =0.12

3. Probability of at least one head coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{38+36+12}{100}\)

\(=\frac{86}{100}\) =0.86

4. Probability of getting more Heads than Tails=   \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{36+12}{100}\)

\(=\frac{48}{100}\)

5. Probability of getting more tails than heads=   \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{14+38}{100}\)

\(=\frac{52}{100}\) =0.52

Q4. 1500 families with 2 children were selected randomly, and the following data were recorded:

 

No of girls in a family 0 1 2
No of girls 211 814 475

 

If a family is chosen at random, compute the probability that it has:

1. No girl

2. 1 girl

3. 2 girls

4. At most one girl

5. More girls than boys

  

Answer

1. Probability of having no girl in a family= \(\frac{no of families having no girl}{Total no of families}\)

\(=\frac{211}{1500}\) =0.1406

2. Probability of having 1 girl in a family= \(\frac{no of families having  1 girl}{Total no of families}\)

\(=\frac{814}{1500}\)

\(=\frac{407}{750}\) =0.5426

3. Probability of having 2 girls in a family= \(\frac{no of families having 2 girls}{Total no of families}\)

\(=\frac{475}{1500}\) =0.3166

4. Probability of having at the most one girl= \(\frac{no of families having at the most one girl}{Total no of families}\)

\(=\frac{211+814}{1500}\)

\(=\frac{1025}{1500}\) =0.6833

5. Probability of having more girls than boys= \(\frac{no of families having more girls than boys}{Total no of families}\)

\(=\frac{475}{1500}\) =0.31

Q5. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that:

1. He hit a boundary

2. He did not hit a boundary.

 Answer

Number of times the batsman hits a boundary= 6

Total number of balls played= 30

Number of times the batsman did not hit a boundary= 30-6= 24

1. Probability that the batsman hits a boundary=  \(\frac{Number of times he hit a boundary}{Total no of balls}\)

\(=\frac{6}{30}\)

\(=\frac{1}{5}\)

2. Probability that the batsman does not hit a boundary

\(\frac{Number of time he did not hit a boundary}{Total no of balls}\)

\(=\frac{24}{30}\)

\(=\frac{4}{5}\)

Q6. The percentage of marks obtained by a student in the monthly unit tests are given below:

UNIT TEST I II III IV V
PERCENTAGE OF MARK OBTAINED 69 71 73 68 76

 

Find the probability that the student gets

1. More than 70% marks

2. Less than 70% marks

3. A distinction

 

Answer:

1: Let E be the event of getting more than 70% marks

No of times E happens=3

Probability(Getting more than 70%)=   \(\frac{Number of times student got more than 70%}{Total no of exams taken}\)

\(=\frac{3}{5}\) =0.6

2. Let F be the event of getting less than 70% marks

No of times F happen= 2

Probability(Getting less than 70%)=    \(\frac{Number of times student got less than 70%}{Total no of exams taken}\)

\(=\frac{2}{5}\) =0.4

3. Let G be the event of getting distinction

No of times G happen= 1

Probability(Getting distinction)=    \(\frac{Number of times student got distinction}{Total no of exams taken}\)

\(=\frac{1}{5}\) =0.2

Q7. To know the opinion of the students about Mathematics, a survey of 200 students were conducted. The data was recorded in the following table

Opinion                                                       Like                                                        Dislike

Number of students                               135                                                         65

Find the probability that student chosen at random:

1. Likes Mathematics

2. Does not like it.

Answer

1. Probability that a student likes mathematics= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{135}{200}\) = 0.675

2. Probability that a student does not like mathematics= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{65}{200}\) = 0.325

Q8. The Blood group table of 30 students of class IX is 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:

1: A

2: B

3: AB

4: O

 

ANSWER

 

BLOOD GROUP A B O AB TOTAL
NUMBER OF STUDENTS 9 6 12 3 30

1. Probability of a student having blood group A= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{9}{30}\) = 0.3

2. Probability of a student having blood group B= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{6}{30}\) = 0.2

3. Probability of a student having blood group AB= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{3}{30}\)

= \(\frac{1}{10}\) = 0.10

4. Probability of a student having blood group O= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{12}{30}\) = 0.4

Q9. Eleven bags of wheat flour, each marked 5kg, 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.

 

ANSWER:

 

Number of bags weighing more than 5kgs= 7

Total no of bags= 11

Probability of having more than 10kgs of rice= \(\frac{no of bags weighing more than 5kg}{Total no of bags}\)

= \(\frac{7}{11}\) =0.63

Q10. The following table show the birth month of 40 students in class IX:

JAN FEB MAR APR MAY JUNE JULY AUG SEP OCT NOV DEC
3 4 2 2 5 1 2 5 3 4 4 4

 

Find the probability that a student is born in October

 

ANSWER:

1. Probability that a student is born in the month of October= \(\frac{no of students born in October}{Total no of students}\)

= \(\frac{6}{40}\)

= \(\frac{3}{20}\) = 0.15

Q11. Given below is the frequency distribution table regarding the concentration of \(SO_{2}\)in the air in parts per million of a certain city for 30 days.

 

Concentration of \(SO_{2}\) 0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24
No of days 4 8 9 2 4 3

 

Find the probability of the concentration of \(SO_{2}\) in the interval 0.12-0.16 on any of these days.

Answer:

Total no of days: 30

Probability of concentration of \(SO_{2}\) in interval 0.12-0.16= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{2}{30}\)

= \(\frac{1}{15}\) = 0.06

 

Q12. A company selected 2400 families at random and surveys them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed below

  

VEHICLES PER FAMILY:

Monthly Income 0 1 2 Above 2
Less than 7000 10 160 25 0
7000-10000 0 305 27 2
10000-13000 1 535 29 1
13000-16000 2 469 29 25
16000 above 1 579 82 88

 

If a family is chosen at random find the probability that the family is:

1. Earning Rs 10000 – 13000 per month and owning exactly 2 vehicles.

2. Earning Rs 16000 or more per month and owning exactly 1 vehicle.

3. Earning less than Rs 7000 per month and does not own any vehicle.

4. Earning Rs 13000 – 16000 per month and owning more than 2 vehicles.

5. Owning not more than 1 vehicle.

6.Owning at least one vehicle

 

Answer: 

1. The probability that the family is earning 10000-13000 and is having exactly 2 vehicles

=\(\frac{ No of families having 10000-13000 income and 2 vehicles }{ Total no of families }\)

= \(\frac{29}{2400}\)

 

2. The probability that the family is earning 16000 or more and is having exactly 1 vehicle

=\(\frac{ No of families having 16000 or more income and 1 vehicle }{ Total no of families }\)

= \(\frac{579}{2400}\)

3. The probability that the family is earning less than 7000 and is having no vehicle

=\(\frac{ No of families having less than 7000 income and no vehicle }{ Total no of families }\)

= \(\frac{10}{2400}\) = \(\frac{1}{240}\)

4. The probability that the family is earning 13000-16000 and is having more than 2 vehicles

=\(\frac{ No of families having 13000-16000 income and more than 2 vehicles }{ Total no of families }\)

= \(\frac{25}{2400}\) = \(\frac{1}{96}\)

5. The probability that the family is having not more than one vehicle

=\(\frac{ No of families having not more than 1 vehicle}{ Total no of families }\)

= \(\frac{10+0+1+2+1+160+305+535+469+579}{1200}\)

= \(\frac{2062}{2400}\) = \(\frac{1031}{1200}\)

6. The probability that the family is having atleast one vehicle

=\(\frac{ No of families having atleast 1 vehicle}{ Total no of families }\)

= \(\frac{160+305+535+469+579+25+27+29+29+82+0+2+1+25+88}{1200}\)

= \(\frac{2356}{2400}\) = \(\frac{589}{600}\)

Q13. The following table gives the life time of 400 neon lamps:

Life time 300-400 400-500 500-600 600-700 700-800 800-900 900-1000
bulbs 14 56 60 86 74 62 48

A bulb is selected at random. Find the probability that the lifetime of a selected bulb:

1. Less than 400 hrs

2. between 300-800 hours

3. Atleast 700 hours

 

Answers:

Total number of bulbs= 400

1. Probability that the life of the selected bulb is less than 400hrs

=\(\frac{ No of bulbs having life less than 400hrs }{ Total no of bulbs }\)

= \(\frac{14}{400}\) = \(\frac{7}{200}\)

2. Probability that the life of the selected bulb is between 300-800hrs

=\(\frac{ No of bulbs having life less than 400hrs }{ Total no of bulbs }\)

= \(\frac{14+56+60+86+74}{200}\)

= \(\frac{290}{400}\) = \(\frac{29}{40}\)

3. Probability that the life of the selected bulb is atleast 700hrs

=\(\frac{ No of bulbs having life atleast 700 hrs }{ Total no of bulbs }\)

= \(\frac{74+62+48}{400}\)

= \(\frac{184}{400}\) = \(\frac{23}{50}\)

Q14. Given below is the frequency distribution of wages (in Rs) of 30 workers in certain factory:

Wages 110-130 130-150 150-170 170-190 190-210 210-230 230-250
No of workers 3 4 5 6 5 4 3

 

A worker is selected at random. Find the probability that his wages are:

1. Less than Rs.150

2. Atleast Rs.210

3. More than or equal to 150 but less than 210

  

Answer

Total number of workers=30

1. Probability that the worker wages are less than Rs.150=

=\(\frac{ No of workers having wages below Rs.150}{ Total no of workers }\)

= \(\frac{3+4}{30}\) = \(\frac{7}{30}\)

2. Probability that the worker wages are atleast Rs.210=

=\(\frac{ No of workers having wages below Rs.210}{ Total no of workers }\)

= \(\frac{4+3}{30}\) = \(\frac{7}{30}\)

3. Probability that the worker wages are more than or equal to 150 but less than 210

=\(\frac{ No of workers having wages more than Rs.150 but less than Rs.210 }{ Total no of workers }\)

= \(\frac{5+6+5}{30}\) = \(\frac{16}{30}\) = \(\frac{8}{15}\)