Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 23 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 7 students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 23 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 281:

#### Question 1:

A coin is tossed 300 times and we get head : 136 times and tail : 164 times.

When a coin is tossed at random, What is the probability of getting (i) a head, (ii) a tail?

#### Answer:

Total number of trials = 300

Number of times a head is obtained = 136

Number of times a tail is obtained= 164

(i) Probability of getting head =$\frac{\mathrm{Numbe}r\mathrm{of}\mathrm{times}\mathrm{heads}\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}$=$\frac{136}{300}=\frac{34}{75}$

(ii) Probability of getting a tail =$\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{tails}\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}$=$\frac{164}{300}=\frac{41}{75}$

#### Page No 281:

#### Question 2:

Two coins are tossed simultaneously 200 times and we get two heads: 58 times, one head: 83 times: 0 head: 59 times.

When two coins are tossed at random, what is the probability of getting (i) 2 heads, (ii) 1 head, (iii) 0 head?

#### Answer:

Total number of trials = 200

Number of times 2 heads are obtained = 58

Number of times one head is obtained = 83

Number of times no head is obtained = 59

(i) Probability of getting 2 heads = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{have}\mathrm{been}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}$ = $\frac{58}{200}=\frac{29}{100}$

(ii) Probability of getting 1 head = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{has}\mathrm{been}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}$ = $\frac{83}{200}$

(iii) Probability of getting 0 head = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{head}\mathrm{has}\mathrm{not}\mathrm{been}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}$ = $\frac{59}{200}$

#### Page No 281:

#### Question 3:

A dice is thrown 100 times and the outcomes are noted as given below:

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

Frequency | 21 | 14 | 18 | 15 | 23 | 9 |

When a dice is thrown at random, what is the probability of getting a (i) 3, (ii) 6, (iii) 6, (iv) 1?

#### Answer:

Total number of trials = 100

Number of times 3 is obtained = 18

Number of times 6 is obtained = 9

Number of times 4 is obtained = 15

Number of times 1 is obtained = 21

$\left(\mathrm{i}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}3=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}3\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{18}{100}=\frac{9}{50}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}6=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}6\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}4=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}4\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{15}{100}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}1=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{21}{100}$

#### Page No 281:

#### Question 4:

In a survey of 100 ladies it was found that 36 like coffee while 64 dislike it.

Out of these ladies, one is chosen at random. What is the probability that the chosen lady (i) likes coffee, (ii) dislikes coffee?

#### Answer:

Total number of ladies surveyed = 100

Ladies who like coffee = 36

Ladies who do not like coffee = 64

$\left(\mathrm{i}\right)\mathrm{Probability}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{lady}\mathrm{who}\mathrm{likes}\mathrm{coffee}=\frac{\mathrm{Number}\mathrm{of}\mathrm{ladies}\mathrm{who}\mathrm{like}\mathrm{coffee}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ladies}}=\frac{36}{100}=\frac{9}{25}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{Probability}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{lady}\mathrm{who}\mathrm{dislikes}\mathrm{coffee}=\frac{\mathrm{Number}\mathrm{of}\mathrm{ladies}\mathrm{who}\mathrm{dislike}\mathrm{coffee}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ladies}}=\frac{64}{100}=\frac{16}{25}$

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