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#### Page No 283:

#### 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 283:

#### 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 283:

#### 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 283:

#### 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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