**NCERT Solutions Class 10 Maths Chapter 15 Probability** – Here are all the NCERT solutions for Class 10 Maths Chapter 15. This solution contains questions, answers, images, explanations of the complete Chapter 15 titled Probability of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 15 Probability. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 15 Probability in one place.

## NCERT Solutions Class 9 Maths Chapter 15 Probability

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 15 Probability , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 15 |

Chapter Name |
Probability |

### NCERT Solutions Class 9 Maths chapter 15 Probability

Class 9, Maths chapter 15, Probability solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.1:In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Ans :Number of times the batswoman hits a boundari = 6 Total number of balls played = 30 Number of times that the batwoman does not hit a boundary = 30 -6 = 24 p(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of times when she does not hit boundary }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{24}{30}=\frac{4}{5}\)

Q.2:1500 families with 2 children were selected randomly, and the following data were recorded:Compute the probability of a family,chosen at random,having (i) 2 girls (ii) 1 girl (iii) no girl Also check weather the sum of these probabilities is 1.

Ans :Total number of family=475+814+211=1500 (i) number of families of 2 girls =475 p1(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having 2 girls}} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{475}{1500}=\frac{19}{60}\) (ii) Number of families having 1 girl =814 p2(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having 1 girl }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{814}{1500}=\frac{407}{750}\) (iii) Number of families having no girl = 211 p3(she does not hit the boundary)=\( \begin{array}{c} \frac {{\text { Number of families having no girl }} }{ {\text { Total number of balls played }}}\end{array} \) \( =\frac{211}{1500}\) Sum of all these probabilities = \( =\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\) \( =\frac{475+814+211}{1500}\) \( =\frac{1500}{1500}=1\)

Q.3:Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.

Ans :Number of students born in the month of August = 6 Total number of students = 40 \( \mathrm{P} \text { (Students born in the month of August) }=\frac{\text { Number of students born in August }}{\text { Total number of students }}\) \( =\frac{6}{40}=\frac{3}{20}\)

Q.4:Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up,

Ans :Number Of times 2 heads come up = 72 Total number Of times the coins were tossed = 200 \( \mathrm{P}(2 \text { heads will come up })=\frac{\text { Number of times } 2 \text { heads come up }}{\text { Total number of times the coins were tossed }}\) \( =\frac{72}{200}=\frac{9}{25}\)

Q.5:An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:Suppose a family is chosen, find the probability that the family chosen 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 vehicles. (v) owning not more than 1 vehicle.

Ans :Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 +2+1+535 +29+1+2 + 469 +59+ 25 +1 + 579 + 82 +88 = 2400 (i) Number of families earning Rs 10000 - 13000 per month and owning exactly 2 vehicles = 29 Hence, required probability, p=\( \mathrm{P}=\frac{579}{2400}\) (ii) Number Of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579 Hence, required probability, p=\( \frac{10}{2400}=\frac{1}{240}\) (iii) Number of families earning less than Rs 7000 per month and dcmas not own any vehicle = 10 Hence, required probability, p= \(\frac{10}{2400}=\frac{1}{240}\) (iv) Number of families earning Rs 13000 - 16000 per month and owning more than 2 vehicles = 25 Hence, required probability, p= \( \frac{25}{2400}=\frac{1}{96}\) (v) Number of families owning not more than 1 vehicle = 10+ 160+0+ 305 + 535 +2+ 469 +1 + 579 =2062 Hence, required probability, p= \( \frac{2062}{2400}=\frac{1031}{1200}\)

## NCERT / CBSE Book for Class 9 Maths

You can download the NCERT Book for Class 9 Maths in PDF format for free. Otherwise you can also buy it easily online.

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### All NCERT Solutions Class 9

- NCERT Solutions for Class 9 English
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- NCERT Solutions for Class 9 Sanskrit

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