Rd Sharma 2021 Solutions for Class 9 Maths Chapter 25 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 9 students for Maths Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 25 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 25.13:

#### Question 1:

A coin is tossed 1000 times with the following frequencies:

Head: 455, Tail: 545

Compute the probability for each event.

#### Answer:

The coin is tossed 1000 times. So, the total number of trials is 1000.

Let *A *be the event of getting a head and *B *be the event of getting a tail.

The number of times *A* happens is 455 and the number of times *B* happens is 545.

Remember the empirical or experimental or observed frequency approach to probability.

If *n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

#### Page No 25.13:

#### 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.

#### Answer:

The total number of trials is 500.

Let *A *be the event of getting two heads, *B* be the event of getting one tail and *C *be the event of getting no head.

The number of times *A* happens is 95, the number of times *B* happens is 290 and the number of times *C* happens is 115.

Remember the empirical or experimental or observed frequency approach to probability.

If *n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

#### Page No 25.13:

#### Question 3:

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

Outcome: | No head | One head | Two heads | Three heads |

Frequency: | 14 | 38 | 36 | 12 |

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.

#### Answer:

The total number of trials is 100.

Remember the empirical or experimental or observed frequency approach to probability.

If *n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A *be the event of getting two heads.

The number of times *A* happens is 36.

Therefore, we have

(ii) Let* B* be the event of getting three heads

The number of times *B* happens is 12.

Therefore, we have

(iii) Let *C* be the event of getting at least one head.

The number of times *C* happens is.

Therefore, we have

(iv) Let *D* be the event of getting more heads than tails.

The number of times *D* happens is.

Therefore, we have

(v) Let *E *be the event of getting more tails than heads.

The number of times *E* happens is.

Therefore, we have

#### Page No 25.13:

#### Question 4:

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

Number of girls in a family | 0 | 1 | 2 |

Number of families | 211 | 814 | 475 |

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

#### Answer:

The total number of trials is 1500.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A *be the event of having no girl.

The number of times *A* happens is 211.

Therefore, we have

(ii) Let* B* be the event of having one girl.

The number of times *B* happens is 814.

Therefore, we have

(iii) Let *C* be the event of having two girls.

The number of times *C* happens is 475.

Therefore, we have

(iv) Let *D* be the event of having at most one girl.

The number of times *D* happens is.

Therefore, we have

(v) Let *E *be the event of having more girls than boys.

The number of times *E* happens is 475.

Therefore, we have

#### Page No 25.13:

#### Question 5:

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays.

(i) he hits boundary

(ii) he does not hit a boundary.

#### Answer:

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A *be the event of hitting boundary.

The number of times *A* happens is 6.

Therefore, we have

(ii) Let* B* be the event of does not hitting boundary.

The number of times *B* happens is.

Therefore, we have

#### Page No 25.13:

#### Question 6:

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

Unit test: | I | II | III | IV | V |

Percentage of marks obtained: | 69 | 71 | 73 | 68 | 76 |

Find the probability that the student gets:

(i) more than 70% marks

(ii) less than 70% marks

(iii) a distinction

#### Answer:

The total number of trials is 5.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A *be the event of getting more than 70% marks.

The number of times *A* happens is 3.

Therefore, we have

(ii) Let* B* be the event of getting less than 70% marks.

The number of times *B* happens is 2.

Therefore, we have

(iii) Let *C* be the event of getting a distinction.

The number of times *C* happens is 1.

Therefore, we have

#### Page No 25.14:

#### 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:

Opinion: | Like | Dislike |

Number of students: | 135 | 65 |

Find the probability that a student chosen at random (i) likes Mathematics (ii) does not like it.

#### Answer:

The total number of trials is 200.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A *be the event of liking mathematics.

The number of times *A* happens is 135.

Therefore, we have

(ii) Let* B* be the event of disliking mathematics.

The number of times *B* happens is 65.

Therefore, we have

#### Page No 25.14:

#### 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, Fin the probability that the blood group of the student chosen is:

(i) A

(ii) B

(iii) AB

(iv) O

#### Answer:

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A*_{1}* *be the event that the blood group of a chosen student is A.

The number of times *A*_{1} happens is 9.

Therefore, we have

(iii) Let* **A*_{2} be the event that the blood group of a chosen student is B.

The number of times *A*_{2} happens is 6.

Therefore, we have

(iii) Let *A*_{3}* *be the event that the blood group of a chosen student is AB.

The number of times *A*_{3} happens is 3.

Therefore, we have

(iv) Let* **A*_{4} be the event that the blood group of a chosen student is O.

The number of times *A*_{4} happens is 12.

Therefore, we have

#### Page No 25.14:

#### 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.

#### Answer:

The total number of trials is 11.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Let *A*_{1}* *be the event that the actual weight of a chosen bag contain more than 5 Kg of flour.

The number of times *A*_{1} happens is 7.

Therefore, we have

#### Page No 25.14:

#### Question 10:

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

Jan. | Feb | March | April | May | June | July | Aug. | Sept. | Oct. | Nov. | Dec. |

3 | 4 | 2 | 2 | 5 | 1 | 2 | 5 | 3 | 4 | 4 | 4 |

Find the probability that a student was born in August.

#### Answer:

The total number of trials is 40.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Let *A*_{1}* *be the event that the birth month of a chosen student is august.

The number of times *A*_{1} happens is 5.

Therefore, we have

#### Page No 25.14:

#### 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.

Conc. 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. days: | 4 | 8 | 9 | 2 | 4 | 3 |

Find the probability of concentration of sulphur dioxide in the interval

0.12-0.16 on any of these days.

#### Answer:

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Let *A*_{1}* *be the event that the concentration of sulphur dioxide in a day is 0.12-0.16 parts per million.

The number of times *A*_{1} happens is 2.

Therefore, we have

#### Page No 25.14:

#### 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:

Monthly income:(in Rs) |
Vehicles per family |
|||

0 | 1 | 2 | Above 2 | |

Less than 7000 7000-10000 10000-13000 13000-16000 16000 or more |
10 0 1 2 1 |
160 305 535 469 579 |
25 27 29 29 82 |
0 2 1 25 88 |

If a family is chosen, find the probability that family is:

(i) earning Rs10000-13000 per month and owning exactly 2 vehicles.

(ii) earning Rs 16000 or more per month and owning exactly 1 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.

#### Answer:

The total number of trials is 2400.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A*_{1}* *be the event that a chosen family earns Rs 10000-13000 per month and owns exactly 2 vehicles.

The number of times *A*_{1} happens is 29.

Therefore, we have

(ii) Let *A*_{2}* *be the event that a chosen family earns Rs 16000 or more per month and owns exactly 1 vehicle.

The number of times *A*_{2} happens is 579.

Therefore, we have

(iii) Let *A*_{3}* *be the event that a chosen family earns less than Rs 7000 per month and does not owns any vehicles.

The number of times *A*_{3} happens is 10.

Therefore, we have

(iv) Let *A*_{4}* *be the event that a chosen family earns Rs 13000-16000 per month and owns more than 2 vehicles.

The number of times *A*_{4} happens is 25.

Therefore, we have

(v) Let *A*_{5}* *be the event that a chosen family owns not more than 1 vehicle (may be 0 or 1). In this case the number of vehicles is independent of the income of the family.

The number of times *A*_{5} happens is

.

Therefore, we have

(vi) Let *A*_{6}* *be the event that a chosen family owns atleast 1 vehicle (may be 1 or 2 or above 2). In this case the number of vehicles is independent of the income of the family.

The number of times *A*_{6} happens is

.

Therefore, we have

#### Page No 25.15:

#### Question 13:

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

Life time (in hours) |
300-400 | 400-500 | 500-600 | 600-700 | 700-800 | 800-900 | 900-1000 |

Number of lamps: | 14 | 56 | 60 | 86 | 74 | 62 | 48 |

A bulb is selected of random, Find the probability that the the life time of the selected bulb is:

(i) less than 400

(ii) between 300 to 800 hours

(iii) at least 700 hours.

#### Answer:

The total number of trials is 400.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A*_{1}* *be the event that the lifetime of a chosen bulb is less than 400 hours.

The number of times *A*_{1} happens is 14.

Therefore, we have

(ii) Let *A*_{2}* *be the event that the lifetime of a chosen bulb is in between 300 to 800 hours.

The number of times *A*_{2} happens is.

Therefore, we have

.

(iii) Let *A*_{3}* *be the event that the lifetime of a chosen bulb is atleast 700 hours.

The number of times *A*_{3} happens is.

Therefore, we have

#### Page No 25.15:

#### Question 14:

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

Wages (in Rs) | 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:

(i) less than Rs 150

(ii) at least Rs 210

(iii) more than or equal to 150 but less than Rs 210.

#### Answer:

The total number of trials is 30.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

(i) Let *A*_{1}* *be the event that the wages of a worker is less than Rs 150.

The number of times *A*_{1} happens is.

Therefore, we have.

(ii) Let *A*_{2}* *be the event that the wages of a worker is atleast Rs 210.

The number of times *A*_{2} happens is.

Therefore, we have.

(iii) Let *A*_{3}* *be the event that the wages of a worker is more than or equal to Rs 150 but less than Rs 210.

The number of times *A*_{3} happens is.

Therefore, we have

#### Page No 25.16:

#### Question 1:

*Mark the correct alternative in each of the following:*

The probability of an impossible event is

(a) 1

(b) 0

(c) less than 0

(d) greater than 1

#### Answer:

We have to find the probability of an impossible event.

Note that the number of occurrence of an impossible event is 0. This is the reason that’s why it is called impossible event.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Note that *n *is a positive integer, it can’t be zero. So, whatever may be the value of *n*, the probability of an impossible event is.

Hence the correct option is **(b)**.

#### Page No 25.16:

#### Question 2:

The probability of a certain event is

(a) 0

(b) 1

(c) greater than 1

(d) less than 0

#### Answer:

We have to find the probability of a certain event.

Note that the number of occurrence of an impossible event is same as the total number of trials. When we repeat the experiment, every times it occurs. This is the reason that’s why it is called certain event.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Note that *n *is a positive integer, it can’t be zero. So, the probability of an impossible event is.

Hence the correct option is **(b)**.

#### Page No 25.16:

#### Question 3:

The probability an event of a trial is

(a) 1

(b) 0

(c) less than 1

(d) more than 1

#### Answer:

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Note that *m *is always less than or equal to *n *and *n* is a positive integers, it can’t be zero. But, *m *is a non negative integer. So, the maximum value of probability of an event is, which is the probability of a certain event and the minimum value of it is 0, which is the probability of an impossible event. For any other events the value is in between 0 and 1.

Hence the correct option is **(c)**.

#### Page No 25.16:

#### Question 4:

Which of the following cannot be the probability of an event?

(a) $\frac{1}{3}$

(b) $\frac{3}{5}$

(c) $\frac{5}{3}$

(d) 1

#### Answer:

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Note that *m *is always less than or equal to *n *and *n* is a positive integers, it can’t be zero. But, *m *is a non negative integer. So, the maximum value of probability of an event is, which is the probability of a certain event and the minimum value of it is 0, which is the probability of an impossible event. For any other events the value is in between 0 and 1.

All the options except (c) satisfy the above criteria’s.

Hence the correct option is **(c)**.

#### Page No 25.16:

#### Question 5:

Two coins are tossed simultaneously. The probability of getting atmost one head is

(a) $\frac{1}{4}$

(b) $\frac{3}{4}$

(c) $\frac{1}{2}$

(d) $\frac{5}{4}$

#### Answer:

The random experiment is tossing two coins simultaneously.

All the possible outcomes are HH, HT, TH, and TT.

Let *A *be the event of getting at most one head.

The number of times *A *happens is 3.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

So, the correct choice is **(b)**.

#### Page No 25.16:

#### 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

#### Answer:

The total number of trials is 1000. Let *x *be the number of times a tail occurs.

Let *A *be the event of getting a tail.

The number of times *A* happens is *x*.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have.

But, it is given that. So, we have

Hence a tail is obtained 375 times.

Consequently, a head is obtainedtimes.

So, the correct choice is **(c)**.

#### Page No 25.16:

#### Question 7:

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

Outcome |
1 | 2 | 3 | 4 | 5 | 6 |

Frequency |
200 | 30 | 120 | 100 | 50 | 100 |

The probability of getting a prime number is

(a) $\frac{1}{3}$

(b) $\frac{2}{3}$

(c) $\frac{49}{60}$

(d) $\frac{39}{125}$

#### Answer:

The total number of trials is 600.

Let *A *be the event of getting a prime number (2, 3 and5).

The number of times *A* happens is.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

So, the correct choice is **(a)**.

#### Page No 25.16:

#### Question 8:

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

Class: |
X | IX | VIII | VII | VI | V |

Attendance: |
30 | 62 | 85 | 92 | 76 | 55 |

What is the probability that the class attendance is more than 75%?

(a) $\frac{1}{6}$

(b) $\frac{1}{3}$

(c) $\frac{5}{6}$

(d) $\frac{1}{2}$

#### Answer:

The total number of trials is 6.

Let *A *be the event that the attendance of a class is more than 75%.

The number of times *A* happens is 3 (for classes’ VIII, VII and VI).

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

So, the correct choice is **(d)**.

#### Page No 25.16:

#### 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) $\frac{1}{5}$

(b) $\frac{3}{5}$

(c) $\frac{2}{5}$

(d) $\frac{4}{5}$

#### Answer:

The total number of trials is 50.

Let *A *be the event that the number on the picked coin is not a prime.

The prime’s lies in between 51 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. They are 10 in numbers. Therefore the numbers lies between 51 and 100 and which are not primes are in numbers.

So, the number of times *A* happens is 40.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

So, the correct choice is **(d)**.

#### Page No 25.16:

#### Question 10:

In a football match, Ronaldo makes 4 goals from 10 penalty kicks. The probability of converting a penalty kick into a goal by Ronaldo, is

(a) $\frac{1}{4}$

(b) $\frac{1}{6}$

(c) $\frac{1}{3}$

(d) $\frac{2}{5}$

#### Answer:

The total number of trials is 10.

Let *A *be the event that Ronaldo makes a goal in a penalty kick.

The number of times *A* happens is 4.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

So, the correct choice is **(d)**.

#### Page No 25.17:

#### Question 11:

*Mark the correct alternative in each of the following:*

Three biased coins were tossed 800 times simultaneously. The outcomes are given in the following table :

Outcome : No head One Head Two Head

Frequency : 120 280 * x*

If the probability of occurrence of two heads is thrice that of all heads, then the value of *x.*

(a) 150

(b) 240

(c) 300

(d) 340

#### Answer:

Three biased coins are tossed 800 times.

∴ Total number of trials = 800

Let *E* be the event of occurrence of two heads and *F* be the event of occurrence of all heads.

∴ Probability of occurrence of two heads = P(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{trials}\mathrm{of}\mathrm{occurrence}\mathrm{of}\mathrm{two}\mathrm{heads}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{x}{800}$ .....(1)

Now,

Number of trials of occurrence of all heads (or three heads)

= Total number of trials − Number of trials of occurrence of no head − Number of trials of occurrence of one head − Number of trials of occurrence of two heads

= 800 − 120 − 280 − *x*

= 400 − *x*

∴ Probability of occurrence of all heads = P(*F*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{trials}\mathrm{of}\mathrm{occurrence}\mathrm{of}\mathrm{all}\mathrm{heads}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{400-x}{800}$ .....(2)

It is given that,

Probability of occurrence of two heads = 3 × Probability of occurrence of all heads

$\therefore \frac{x}{800}=3\times \left(\frac{400-x}{800}\right)$ [From (1) and (2)]

$\Rightarrow x=1200-3x$

$\Rightarrow 4x=1200$

$\Rightarrow x=\frac{1200}{4}=300$

Thus, the value of *x* is 300.

Hence, the correct answer is option (c).

#### Page No 25.17:

#### Question 12:

*Mark the correct alternative in each of the following:*

An unbiased dice was rolled 800 times simultaneously. The frequencies of the various outcomes are given in the table below :

Outcome : 1 2 3 4 5 6

Frequency : 150 200 100 75 125 150

When the dice is rolled, the probability of getting a number which is a perfect square is

$\left(\mathrm{a}\right)\frac{9}{32}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\frac{11}{32}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\frac{13}{32}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\frac{15}{32}$

#### Answer:

It is given that an unbiased dice was rolled 800 times.

∴ Total number of trials = 800

Let *E* be the event of getting a number on the dice which is a perfect square.

Now, 1 and 4 are perfect squares among the outcomes 1, 2, 3, 4, 5 and 6.

∴ Number of trials of getting a number which is perfect square on the dice

= Frequency of getting 1 or 4 on the dice

= 150 + 75

= 225

So, P(Getting a number which is perfect square) = P(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{trials}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{number}\mathrm{which}\mathrm{is}\mathrm{perfect}\mathrm{square}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{225}{800}=\frac{9}{32}$

Thus, the probability of getting a number which is a perfect square is $\frac{9}{32}$.

Hence, the correct answer is option (a).

#### Page No 25.17:

#### Question 1:

Fill In The Blanks

If an experiment does not produce the same outcomes every time but the outcomes in a trial is one of the several possible outcomes, then it is called an ________ experiment.

#### Answer:

If an experiment does not produce the same outcomes every time but the outcomes in a trial is one of the several possible outcomes, then it is called an ____random____ experiment.

#### Page No 25.17:

#### Question 2:

Fill In The Blanks

An outcome of a trial of a random experiment is called an _________ event.

#### Answer:

An outcome of a trial of a random experiment is called an ____elementary____ event.

#### Page No 25.17:

#### Question 3:

Fill In The Blanks

A collection of two or more possible outcomes (elementary events) of a trial is called a ________ event.

#### Answer:

A collection of two or more possible outcomes (elementary events) of a trial is called a ____compound____ event.

#### Page No 25.17:

#### Question 4:

Fill In The Blanks

In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability, that the person has a high school certificate is _______.

#### Answer:

Number of people in the sample study = 642

Number of people having high school certificate = 514

∴ P(Person selected at random has a high school certificate)

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{having}\mathrm{high}\mathrm{school}\mathrm{certificate}}{\mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{in}\mathrm{the}\mathrm{sample}\mathrm{study}}$

= $\frac{514}{642}$

= $\frac{257}{321}$

Thus, the probability that the person selected at random has a high school certificate is $\frac{257}{321}$.

In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability, that the person has a high school certificate is $\overline{)\frac{257}{321}}$.

#### Page No 25.17:

#### Question 5:

Fill In The Blanks

80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of following frequency table:

Life time (in hrs): 300 500 700 900 1000

Frequency : 10 12 23 25 10

One bulb is selected at random from the lot. The probability that its life is 1150 hours is __________.

#### Answer:

Number of bulbs in the lot = 80

Number of bulbs having life time 1150 hours = 0

∴ P(Bulb selected at random has life time 1150 hours)

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{bulbs}\mathrm{having}\mathrm{life}\mathrm{time}1150\mathrm{hours}}{\mathrm{Number}\mathrm{of}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{lot}}$

= $\frac{0}{80}$

= 0

Thus, the probability that a bulb selected at random from the lot has life time 1150 hours is 0.

One bulb is selected at random from the lot. The probability that its life is 1150 hours is ______0______.

#### Page No 25.17:

#### Question 6:

Fill In The Blanks

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat patato chips. If a child is selected at random, the probability that he/she does not like eat patato chips, is _________.

#### Answer:

Number of children in the survey = 364

Number of children who liked to eat potato chips = 91

∴ Number of children who did not liked to eat potato chips

= Number of children in the survey − Number of children who liked to eat potato chips

= 364 − 91

= 273

Now,

P(A child selected at random does not like to eat potato chips)

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{children}\mathrm{who}\mathrm{did}\mathrm{not}\mathrm{liked}\mathrm{to}\mathrm{eat}\mathrm{potato}\mathrm{chips}}{\mathrm{Number}\mathrm{of}\mathrm{children}\mathrm{in}\mathrm{the}\mathrm{survey}}$

= $\frac{273}{364}$

= $\frac{3}{4}$

Thus, the probability that a child selected at random does not like to eat potato chips is $\frac{3}{4}$.

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like eat potato chips, is $\overline{)\frac{3}{4}}$.

#### Page No 25.17:

#### Question 7:

Fill In The Blanks

In Q.No. 5, the probability that a bulb selected at random from the the lot has life less than 900 hours is _________.

#### Answer:

Number of bulbs in the lot = 80

Number of bulbs having life time less than 900 hours

= Number of bulbs having life time 300 hours + Number of bulbs having life time 500 hours + Number of bulbs having life time 700 hours

= 10 + 12 + 23

= 45

∴ P(Bulb selected at random has life time less than 900 hours)

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{bulbs}\mathrm{having}\mathrm{life}\mathrm{time}\mathrm{less}\mathrm{than}900\mathrm{hours}}{\mathrm{Number}\mathrm{of}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{lot}}$

= $\frac{45}{80}$

= $\frac{9}{16}$

Thus, the probability that a bulb selected at random from the lot has life time less than 900 hours is $\frac{9}{16}$.

The probability that a bulb selected at random from the the lot has life less than 900 hours is $\overline{)\frac{9}{16}}$.

#### Page No 25.17:

#### Question 8:

Fill In The Blanks

Two coins are tossed 1000 times and outcomes are recorded as below:

Numbers of heads: 2 1 0

Frequency: 200 550 250

The probability of getting at most one head is _______.

#### Answer:

Two coins are tossed 1000 times.

∴ Total number of trials = 1000

Let *E* be the event of getting at most one head on the two coins.

Number of trials for getting at most one head

= Frequency of getting 0 heads + Frequency of getting 1 head

= 250 + 550

= 800

∴ Probability of getting at most one head = P(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{trials}\mathrm{for}\mathrm{getting}\mathrm{at}\mathrm{most}\mathrm{one}\mathrm{head}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{800}{1000}=\frac{4}{5}$

Thus, the probability of getting at most one head is $\frac{4}{5}$.

The probability of getting at most one head is $\overline{)\frac{4}{5}}$.

#### Page No 25.18:

#### Question 9:

Fill In The Blanks

In a medical examination of students of a class, the following blood groups are recorded:

Blood group: A AB B O

Number of students: 10 13 12 5

A student is selected at random from the class. The probability that he/she has blood group B, is __________.

#### Answer:

Total number of students in the class = 10 + 13 + 12 + 5 = 40

Number of students having blood group B = 12

∴ P(A student selected at random has blood group B)

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{having}\mathrm{blood}\mathrm{group}\mathrm{B}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}\mathrm{in}\mathrm{the}\mathrm{class}}$

= $\frac{12}{40}$

= $\frac{3}{10}$

Thus, the probability that a student selected at random from the class has blood group B is $\frac{3}{10}$.

A student is selected at random from the class. The probability that he/she has blood group B, is $\overline{)\frac{3}{10}}$.

#### Page No 25.18:

#### Question 10:

Fill In The Blanks

A die is thrown 1000 times and the outcomes were recorded as follows:

Outcomes: 1 2 3 4 5 6

Frequency: 180 150 160 170 150 190

If the die is thown once more, then the probability that its show 5 is _________.

#### Answer:

It is given that a die is thrown 1000 times.

∴ Total number of trials = 1000

Let *E* be the event that the die shows 5.

Number of trials of getting 5 on the die = Frequency of getting 5 on the die = 150

∴ P(Die shows the number 5) = P(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{trials}\mathrm{of}\mathrm{getting}5\mathrm{on}\mathrm{the}\mathrm{die}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{150}{1000}=\frac{3}{20}$

Thus, the probability that die shows 5 is $\frac{3}{20}$.

If the die is thrown once more, then the probability that its show 5 is $\overline{)\frac{3}{20}}$.

#### Page No 25.18:

#### Question 1:

Define a trial.

#### Answer:

What is the meaning of trial?

The word trial means a test of performance, qualities, or suitability.

**Definition:**

Any particular performance of a random experiment is called a trial. That is, when we perform an experiment it is called a trial of the experiment.

By experiment or trial, we mean a random experiment unless otherwise specified. Where you are required to differentiate between a trial and an experiment, consider the experiment to be a larger entity formed by the combination of a number of trials.

To illustrate the definition, let us take examples:

1. In the experiment of tossing 4 coins, we may consider tossing each coin as a trial and therefore say that there are 4 trials in the experiment.

2. In the experiment of rolling a dice 5 times, we may consider each rolls as a trial and therefore say that there are 5 trials in the experiment.

Note that rolling a dice 5 times is same as rolling 5 dices each one time. Similarly, tossing 4 coins is same as tossing one coin 4 times.

#### Page No 25.18:

#### Question 2:

Define an elementary event.

#### Answer:

What are the meanings of elementary event?

The word elementary means simple, non decomposable into elements or other primary constituents and the word event means something that result.

**Definition:**

An elementary event is any single outcome of a trial. Elementary events are also called simple events.

To illustrate the definition, let us take examples:

1. In the experiment of tossing a coin, the possible outcomes H and T. Any one outcome like H is called an elementary event.

2. In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Any one outcome like 4 is called an elementary event.

Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.

#### Page No 25.18:

#### Question 3:

Define an event.

#### Answer:

What are the meanings of event?

The word event means something that result.

**Definition:**

An event is a collection of outcomes of a trial of a random experiment.

To illustrate the definition, let us take examples:

1. When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT. Any one outcome like HH is called an event (elementary event). The collections like {HH, HT}, {HH, HT, TT} etc are all events (compound event).

2. In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Any one outcome like 4 is called an event (elementary event). The collections like {1, 2}, {1, 2, 3}, {2, 5, 6}, {2, 3, 4, 5} etc are all events (compound events).

Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.

#### Page No 25.18:

#### Question 4:

Define probability of an event.

#### Answer:

The probability of an event denotes the relative frequency of occurrence of an experiment’s outcome, when repeating the experiment.

**Definition:**

The empirical or experimental definition of probability is that if *n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials, then the probability of happening of event *A *is denoted byand is given by

To illustrate the definition, let us take examples:

1. When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT. The total number of trials is 4. Let *A *be the event of occurring exactly two heads. The number of times *A *happens is 1. So, the probability of the event *A *is

2. In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Let *A *be the event of occurring a number greater than 3. The total number of trials is 6. The number of times *A *happens is 3. So, the probability of the event *A *is

Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.

#### Page No 25.18:

#### Question 5:

A big contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is $\frac{2}{5}$, find the number of red balls in the bag.

#### Answer:

The number of white balls is 4. Let the number of red balls is *x. *Then the total number of trials is.

Let *A *be the event of drawing a white ball.

The number of times *A* happens is 4.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have.

But, it is given that. So, we have

Hence the number of red balls is.

#### Page No 25.18:

#### Question 6:

A die is thrown 100 times. If the probability of getting an even number is $\frac{2}{5}$. How many times an odd number is obtained?

#### Answer:

The total number of trials is 100. Let the number of times an even number is obtained is *x*.

Let *A *be the event of getting an even number.

The number of times *A* happens is *x*.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have.

But, it is given that. So, we have

Hence an even number is obtained 40 times. Consequently, an odd number is obtainedtimes.

#### Page No 25.18:

#### Question 7:

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome |
3 heads | 2 heads | 1 head | No head |

Frequency |
23 | 72 | 77 | 28 |

Find the probability of getting at most two heads.

#### Answer:

The total number of trials is 200.

Let *A *be the event of getting at most two heads.

The number of times *A* happens is.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have.

#### Page No 25.18:

#### Question 8:

In Q.No. 7, what is the probability of getting at least two heads?

#### Answer:

The total number of trials is 200.

Let *A *be the event of getting atleast two heads.

The number of times *A* happens is.

Remember the empirical or experimental or observed frequency approach to probability.

*n* be the total number of trials of an experiment and *A* is an event associated to it such that *A* happens in *m*-trials. Then the empirical probability of happening of event *A *is denoted byand is given by

Therefore, we have

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