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NCERT Solutions for Exercise 15.2 Class 10 Maths Chapter 15 - Probability

NCERT Solutions for Exercise 15.2 Class 10 Maths Chapter 15 - Probability

Edited By Sumit Saini | Updated on Jul 11, 2022 07:05 PM IST | #CBSE Class 10th
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CBSE Class 10th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT Solutions for Class 10 Maths exercise 15.2 – We will come across some terminologies like Experiment, Outcome, Event, etc. Any situation or phenomenon, such as tossing a coin, playing cards, rolling dice, and so on, can be used as an experiment.

Outcome: The outcome of an experiment, such as the number that appears on the dice after rolling, the side of the coin after flipping, and a card drawn from a pack of well-shuffled cards, and so on.

Event: The shuffling of all possible outcomes of an experiment, such as finding a king in an all-around rearranged deck of cards, getting heads or tails on a flipped coin, settling the score or odd numbers on dice, etc.

Sample space is the arrangement of every conceivable outcome or result, for example, getting heads or tails while flipping a coin.

NCERT solutions Class 10 Maths exercise 15.2- An elementary event is one that has only one outcome. The sum of the probabilities of all of an experiment's elementary events is one.

The probability (likelihood) of an occasion goes from 0 to 1, inclusive of 0 and 1. i.e., 1639051329430

Complementary events are the main two potential results of a solitary occasion (a single event). This is closely resembling flipping a coin and checking whether it lands on heads or tails.

1639051328987

where E is representing event and 1639051329658 is representing not E or complementary of the event E.

Along with Class 10 Maths chapter 15 exercise 15.2 the following exercises are also present.

Probability Class 10 Chapter 15 Exercise: 15.2

1 (i) Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day?

Answer:

Total possible ways Shyam and Ekta can visit the shop = 5\times5 = 25

(1) A case that both will visit the same day.

Shyam can go on any day between Tuesday to Saturday in 5 ways.

For any day that Shyam goes, Ekta will go on the same day in 1 way.

Total ways that they both go in the same day = 5\times1 = 5

\therefore P(both\ go\ on\ same\ day) = \frac{5}{25} = \frac{1}{5}

Q1 (ii) Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (ii) consecutive days?

Answer:

Total possible ways Shyam and Ekta can visit the shop = 5\times5 = 25

(2) The case that both will visit the shop on consecutive days.

Shyam can go on any day between Tuesday to Friday in 4 ways.

For any day that Shyam goes, Ekta will go on the next day in 1 way

Similarly, Ekta can go on any day between Tuesday to Friday in 4 ways.

And Shyam will go on the next day in 1 way.

(Note: None of the cases repeats since they are in a different order!)

Total ways that they both go in the same day = 4\times1+4\times1 =8

\therefore P(they\ go\ on\ consecutive\ days) = \frac{8}{25}

Q1 (iii) Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (iii) different days?

Answer:

Total possible ways Shyam and Ekta can visit the shop = 5\times5 = 25

(1) A case that both will visit the same day.

Shyam can go on any day between Tuesday to Saturday in 5 ways.

For any day that Shyam goes, Ekta will go on a different day in (5-1) = 4 ways.

Total ways that they both go in the same day = 5\times4 = 20

\therefore P(both\ go\ on\ different\ days) = \frac{20}{25} = \frac{4}{5}

2 (i) A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

Probability -A die is numbered in such a way that its faces show


What is the probability that the total score is (i) even?

Answer:

+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 3 4 4 5 5 8
3 4 5 5 6 6 9
3 4 5 5 6 6 9
6 7 8 8 9 9 12

Total possible outcomes when two dice are thrown = 6\times6=36

(1) Number of times when the sum is even = 18

\therefore P(sum\ is\ even) = \frac{18}{36} = \frac{1}{2}

Q2 (ii) A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

Probability -A die is numbered in such a way that its faces show


What is the probability that the total score is (ii) 6?

Answer:

+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 3 4 4 5 5 8
3 4 5 5 6 6 9
3 4 5 5 6 6 9
6 7 8 8 9 9 12

Total possible outcomes when two dice are thrown = 6\times6=36

Number of times when the sum is 6 = 4

\therefore P(sum\ is\ 6) = \frac{4}{36} = \frac{1}{9}

Q2 (iii) A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

Probability -A die is numbered in such a way that its faces show


What is the probability that the total score is (iii) at least 6?

Answer:

+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 3 4 4 5 5 8
3 4 5 5 6 6 9
3 4 5 5 6 6 9
6 7 8 8 9 9 12

Total possible outcomes when two dice are thrown = 6\times6=36

Number of times when the sum is at least 6, which means sum is greater than 5 = 15

\therefore P(sum\ is\ atleast\ 6) = \frac{15}{36} = \frac{5}{12}

Q3 A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Answer:

Let there be the number of blue balls in the bag.

Number of red balls = 5

Thus, the total number of balls = total possible outcomes = 5+x

P(getting\ a\ red\ ball) = \frac{5}{5+x}

And, P(getting\ a\ blue\ ball) = \frac{x}{5+x}

According to question,

P(getting\ a\ blue\ ball) = P(getting\ a\ red\ ball)

\\ \frac{x}{5+x} = 2.\left (\frac{5}{5+x} \right )

\implies x = 2.5 = 10

Therefore, there are 10 blue balls in the bag.

Q4 A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double what it was before. Find x.

Answer:

Total number of balls in the bag = 12

Number of black balls in the bag = x

\therefore P(getting\ a\ black\ ball) = \frac{x}{12}

According to the question,

6 more black balls are added to the bag.

\therefore Total number of balls = 12 + 6 = 18

And, the new number of black balls = x+ 6

\therefore P'(getting\ a\ black\ ball) = \frac{x+6}{18}

Also, P' = 2\times P

\implies \frac{x+6}{18} = 2\left (\frac{x}{12} \right )

\\ \implies \frac{x+6}{18} = \frac{x}{6} \\ \implies x+6 = 3x \\ \implies 2x = 6

\implies x =3

The required value of x is 3

Q5 A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 Find the number of blue balls in the jar.

Answer:

Let x be the number of blue marbles in the jar.

\therefore Number of green marbles in the jar = 24-x

According to question,

P(getting\ a\ green\ marble) = \frac{24-x}{24} = \frac{2}{3}

\\ \implies 24-x = 2\times8 \\ \implies x = 24-16 = 8

\therefore Number of blue marbles in the jar is 8

Also Read| Probability Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 15.2

  • Exercise 15.2 Class 10 Maths, depends on PROBABILITY and higher ramifications of the likelihood of an event.

  • NCERT book Class 10 Maths chapter 15 exercise 15.2 assists us with testing our essential idea of probability by addressing a portion of the hard and extensive inquiries identified with it.

  • NCERT syllabus Class 10 Maths chapter 15 exercise 15.2 sets us up for the new sorts of problems that are to come in our higher classes related to probability.

This Story also Contains
  1. Probability Class 10 Chapter 15 Exercise: 15.2
  2. Benefits of NCERT Solutions for Class 10 Maths Exercise 15.2
  3. NCERT Solutions for Class 10 Subject Wise

Also, See:

NCERT Solutions for Class 10 Subject Wise

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Frequently Asked Questions (FAQs)

1. What are the benefits of NCERT solutions for Class 10 Maths exercise 15.2 ?

This exercise has tough question which tests the capability of our understanding and problem solving.

2. The total number of outcomes for two dice to be rolled at once is?

Since each dice has 6 different outcomes, and we are rolling two dice at once.

So, the total number of outcomes = 6×6 = 36

3. Two coins are flipped in air. The Probability of one head and one tail is?

Since each dice has 6 different outcomes, and we are rolling two dice at once.

So, the total number of outcomes = 6×6 = 36

4. A sample space is run and we end up with only three events. The first and the second event has a probability of 0.45 and 0.23 respectively. find the probability of the third event?

total probability of all events is = 1

probability of first and the second event is 0.45 and 0.23

probability of the third event is = 1-(0.45+0.23)=0.32

5. A bag contains 3 types of balls. 4 red balls, 5 green balls and 6 blue balls. Find the Probability of getting a green ball in a draw of balls?

total probability of all events is = 1

probability of first and the second event is 0.45 and 0.23

probability of the third event is = 1-(0.45+0.23)=0.32

6. Write the sample space for tossing three coins at once?

The sample space of tossing three coins at once is:

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}  

7. The Probability of a red ball to be drawn is 0.45. The bag contains 20 balls, find the total number of balls in the bag?

Probability of a red ball to be drawn = 0.45 

Total no of balls = 20

Total no of red balls = 20 × 0.45 = 9

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sai raja 01 September,2024
show the previous year exams ?

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hello Zaid,

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best of luck!

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i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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