Rd Sharma 2019 Solutions for Class 7 Math Chapter 25 Data Handling Iv Probability are provided here with simple step-by-step explanations. These solutions for Data Handling Iv Probability are extremely popular among Class 7 students for Math Data Handling Iv Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 7 Math Chapter 25 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 25.6:

#### Question 1:

A coin is tossed 1000 times with the following frequencies:

Head: 445, | Tail: 555 |

#### Answer:

Total number of times a coin is tossed = 1000

Number of times a head comes up = 445

Number of times a tail comes up = 555

(i) Probability of getting a head = $\frac{\mathrm{Number}\mathrm{of}\mathrm{heads}}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{445}{1000}=0.445$

(ii) Probability of getting a tail = $\frac{\mathrm{Number}\mathrm{of}\mathrm{tails}}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{555}{1000}=0.555$

#### Page No 25.6:

#### Question 2:

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

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

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

*a*/

*an*.

(i) 3

(ii) 5

(iii) 4

(iv) Even number

(v) Odd number

(vi) Number less than 3.

#### Answer:

Total number of trials = 100

Number of times "1" comes up = 21

Number of times "2" comes up = 9

Number of times "3" comes up = 14

Number of times "4" comes up = 23

Number of times "5" comes up = 18

Number of times "6" comes up = 15

(i) Probability of getting 3

=$\frac{\mathrm{frequency}\mathrm{of}3}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{14}{100}=0.14$

(ii) Probability of getting 5

= $\frac{\mathrm{frequency}\mathrm{of}5}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{18}{100}=0.18$

(iii) Probability of getting 4

=$\frac{\mathrm{frequency}\mathrm{of}4}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{23}{100}=0.23$

(iv) Frequency of getting an even no. = Frequency of 2 + Frequency of 4 + Frequency of 6 = 9+ 23 + 15 = 47

Probability of getting an even no. =$\frac{\mathrm{frequency}\mathrm{of}\mathrm{even}\mathrm{no}.}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{47}{100}=0.47$

(v) Frequency of getting an odd no. = Frequency of 1 + Frequency of 3 + Frequency of 5 = 21+ 14 + 18 = 53

Probability of getting an odd no. =$\frac{\mathrm{frequency}\mathrm{of}\mathrm{odd}\mathrm{no}.}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{53}{100}=0.53$

(vi) Frequency of getting a no. less than 3 = Frequency of 1 + Frequency of 2= 21 + 9 = 30

Probability of getting a no. less than 3

=$\frac{\mathrm{frequency}\mathrm{of}\mathrm{no}.\mathrm{less}\mathrm{than}3}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{trails}}=\frac{30}{100}=0.30$

#### Page No 25.6:

#### Question 3:

A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?

#### Answer:

No. of socks in the box = 4

Let B and W denote black and white socks respectively.

Then we have:

S = {B,B,W,W}

If a white sock is picked out, then the total no. of socks left in the box = 3

No. of white socks left = 2-1 =1

Probability of getting a white sock

=$\frac{\mathrm{Number}\mathrm{of}\mathrm{white}\mathrm{socks}\mathrm{left}\mathrm{in}\mathrm{the}\mathrm{box}}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{socks}\mathrm{left}\mathrm{in}\mathrm{the}\mathrm{box}}=\frac{1}{3}$

#### Page No 25.6:

#### Question 4:

Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:

Outcome: | Two heads (HH) |
One head (HT or TH) |
No head (TT) |

Frequency: | 105 | 275 | 120 |

#### Answer:

Number of trials = 500

Number of outcomes of two heads (HH) = 105

Number of outcomes of one head (HT or TH) = 275

Number of outcomes of no head (TT) = 120

(i) Probability of getting two heads =$\frac{\mathrm{frequency}\mathrm{of}\mathrm{getting}2\mathrm{heads}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trails}}=\frac{105}{500}=\frac{21}{100}$

(ii) Probability of getting one head =$\frac{\mathrm{frequency}\mathrm{of}\mathrm{getting}1\mathrm{head}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trails}}=\frac{255}{500}=\frac{11}{20}$

(iii) Probability of getting no head =$\frac{\mathrm{frequency}\mathrm{of}\mathrm{getting}\mathrm{no}\mathrm{head}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trails}}=\frac{120}{500}=\frac{6}{25}$

#### Page No 25.7:

#### Question 1:

An unbiased coin is tossed once, the probability of getting head is

(a) $\frac{1}{2}$ (b) 1 (c) $\frac{1}{3}$ (d) $\frac{1}{4}$

#### Answer:

Tossing a coin, either we get a head (H) or a tail (T). So, the probability of getting a head is $\frac{1}{2}$.

Hence, the correct option is (a).

#### Page No 25.7:

#### Question 2:

There are 10 cards numbered from 1 to 10. A card is drawn randomly. The probability of getting

an even numbered card is

(a) $\frac{1}{10}$ (b) $\frac{1}{5}$ (c) $\frac{1}{2}$ (d) $\frac{2}{5}$

#### Answer:

The number on the cards are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

The even numbers on the cards are 2, 4, 6, 8, 10.

∴ Probability of getting an even numbered card = $\frac{\mathrm{Number}\mathrm{of}\mathrm{even}\mathrm{numbered}\mathrm{card}}{\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{with}\mathrm{numbers}\mathrm{from}1\mathrm{to}10}$

$=\frac{5}{10}=\frac{1}{2}$

Hence, the correct option is (c).

#### Page No 25.7:

#### Question 3:

A dice is rolled. The probability of getting an even prime is

(a) $\frac{1}{6}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{5}{6}$

#### Answer:

The possible numbers on a dice are 1, 2, 3, 4, 5, 6.

There is only one even prime number which is 2.

∴ Probability of getting an even prime = $\frac{\mathrm{Number}\mathrm{of}\mathrm{even}\mathrm{prime}\mathrm{numbers}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}\mathrm{on}\mathrm{the}\mathrm{dice}}$$=\frac{1}{6}$

Hence, the correct option is (a).

#### Page No 25.7:

#### Question 4:

There are 100 cards numbered from 1 to 100 in a box. If a card is drawn from the box and

the probability of an event is $\frac{1}{2}$, then the number of favourable cases to the event is

(a) 20 (b) 25 (c) 40 (d) 50

#### Answer:

Here, $\frac{50}{100}=\frac{1}{2}$.

So, if the the probability of an event is $\frac{1}{2}$, then the number of favourable cases has to be 50.

Hence, the correct option is (d).

#### Page No 25.7:

#### Question 5:

When a dice is thrown, the total number of possible outcomes is

(a) 6 (b) 1 (c) 3 (d) 4

#### Answer:

The number on the faces of a dice are 1, 2, 3, 4, 5, and 6.

∴ Number of possible outcomes = 6

Hence, the correct option is (a).

#### Page No 25.7:

#### Question 6:

There are 10 marbles in a box which are marked with the distinct numbers from 1 to 10.

A marble is drawn randomly. The probability of getting prime numbered marble is

(a) $\frac{1}{2}$ (b) $\frac{2}{5}$ (c) $\frac{9}{3}$ (d) $\frac{3}{10}$

#### Answer:

The numbers marked on the marbles are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

Here, the prime numbers (favourable outcomes) are 2, 3, 5, and 7.

∴ Number of favourable outcomes = 4

Therefore

Probability of getting prime numbered marble = $\frac{4}{10}=\frac{2}{5}$

Hence, the correct option is (b).

#### Page No 25.7:

#### Question 7:

The probability of getting a red card from a well shuffled pack of cards is

(a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (c) $\frac{3}{4}$ (d) $\frac{1}{3}$

#### Answer:

There are 52 cards in a standard deck. There are four different suits Diamonds (red), Clubs (black)

, Hearts (red), and Spades (black) each containing 13 cards.

∴ Number of red cards (favourable outcomes) = 13 + 13 = 26

Therefore

Probability of getting a red card = $\frac{26}{52}=\frac{1}{2}$

Hence, the correct option is (b).

#### Page No 25.7:

#### Question 8:

A coin is tossed 100 times and head is obtained 59 times. The probability of getting a tail is

(a) $\frac{59}{100}$ (b) $\frac{41}{100}$ (c) $\frac{29}{100}$ (d) $\frac{43}{100}$

#### Answer:

Number of all possible outcomes = 100

Number of head obtained = 59

Number of tail obtained (favourable outcomes) = 100 − 59 = 41

Therefore

Probability of getting a tail = $\frac{41}{100}$

Hence, the correct option is (b).

#### Page No 25.7:

#### Question 9:

A dice is tossed 80 times and number 5 is obtained 14 times. The probability of not getting the number 5 is

(a) $\frac{7}{40}$ (b) $\frac{7}{80}$ (c) $\frac{33}{40}$ (d) None of these

#### Answer:

Probability of getting 5 = $\frac{14}{80}=\frac{7}{40}$

Therefore

Probability of not getting 5 = $1-\frac{7}{40}=\frac{33}{40}$

Hence, the correct option is (c).

#### Page No 25.7:

#### Question 10:

A bag contains 4 green balls, 4 red balls and 2 blue balls. If a ball is drawn from the bag, the

probability of getting neither green nor red ball is

(a) $\frac{2}{5}$ (b) $\frac{1}{2}$ (c) $\frac{4}{5}$ (d) $\frac{1}{5}$

#### Answer:

The probability of getting neither green nor red ball is equal to the probability of getting blue balls.

Number of blue balls = 2

Total number of balls = 4 + 4 + 2 = 10

Therefore

Probability of getting neither green nor red ball = $\frac{2}{10}=\frac{1}{5}$

Hence, the correct option is (d).

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