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## AP Board Class 10 Maths Chapter 13 Probability Optional Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 13 Probability Optional Exercise Book Answers

 AP Board Class 10 Maths Chapter 13 Probability Optional Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 13 Probability Optional Exercise Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 13 Probability Optional Exercise Books Solutions

 Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 10th Subject Maths Chapters Maths Chapter 13 Probability Optional Exercise Provider Hsslive

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## AP Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

### 10th Class Maths 13th Lesson Probability Optional Exercise Textbook Questions and Answers

Question 1.
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?
(ii) consecutive days?
(iii) different days?
Shyam and Ekta can visit the shop in the following combination:

(T Tu) (W, Tu) (Th, Tu) (F, Tu) (S, Tu) (Tu, W) (W, W) (Th, W) (F, W) (S, W) (Tu, Th) (W, Th) (Th, F) (F, Th) (S, Th) (Tu, F) (W, F) (Th, S) (F, F) (S, F) (Tu, S) (W, S) (Th, Th) (F, S) (S, S)
∴ Number of total outcomes = 5 × 5 = 52 = 25 [also from the above table]
i) Number of favourable outcomes to that of visiting on the same day
(Tu, Tu), (W, W), (Th, Th), (F, F), (S, S) = 5
∴ Probability of visiting the shop on the same day = No. of favourable outcomes No. of total outcomes = 525 = 15
ii) Number of outcomes favourable to consecutive days
(Tu, W), (W, Th), (Th, F), (F, S), (W, Tu), (Th, W), (F, Th), (S, F) = 8
∴ Probability of visiting the shop on consecutive days = 825
iii) If P(E) is the probability of visiting the shop on the same day,
then P(E⎯⎯⎯⎯) is the probability of visiting the shop not on the same day.
i.e., P(E⎯⎯⎯⎯) is the probability of visiting the shop on different days.
Such that P(E) + P(E⎯⎯⎯⎯) = 1
P(E⎯⎯⎯⎯) = 1 – P(E) = 1 – 15 = 45

Question 2.
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.
Number of red balls in the bag = 5 As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
∴ Blue balls = 5 × 2 = 10.
[!! Let the number of blue balls = x
Number of red balls = 5
Total no. of balls = x + 5
Total outcomes in drawing a ball at random = x + 5
Number of outcomes favourable to red ball = 5
∴ P(R) = 5𝑥+5
from the problem,
P(B) = 2 × 5𝑥+5 = 10𝑥+5
Also 5𝑥+5 + 10𝑥+5 = 1
[∵ P(E) + P(E⎯⎯⎯⎯) = 1]
⇒ 5+10𝑥+5 = 1
⇒ 15𝑥+5 = 1
⇒ x + 5 = 15]

Question 3.
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 of what it was before. Find x.
Number of black balls = x
Total number of balls in the box = 12
Probability of drawing a black ball = No. of favourable outcomes No. of total outcomes = 𝑥12 …….. (1)
When 6 more black balls are placed in the box, number of favourable outcomes to black ball becomes = x + 6.
Total number of balls in the box becomes = 12 + 6 = 18.
Now the probability of drawing a black ball become = 𝑥+618 …….. (1)
By Problem,
𝑥+618 = 2. 𝑥12
⇒ 𝑥+618 = 𝑥6
⇒ 6(x + 6) = 18(x)
⇒ 6x + 36 = 18x
⇒ 18x – 6x = 36
⇒ 12x = 36
⇒ x = 3612 = 3
Check:

and hence proved.

Question 4.
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 marbles in the jar.
Total number of marbles in the jar = 24.
Let the number of green marbles = x.
Then number of blue marbles = 24 – x.
Probability of drawing a green marbles = No. of favourable outcomes No. of total outcomes = 𝑥24
By problem,
𝑥24 = 23
⇒ 3x = 24 × 2
x = 24×23 = 16
∴ Number of green marbles = 16
Number of blue marbles = 24 – 1 = 8
∴ Probability of picking blue marble = 824 = 13
(OR)
P(B) = P(E) – P(G) = 1 – 23 = 13
[!! P(G) = 23
P(G) + P(B) = 1
∴ P(B) = 1 – P(G) = 1 – 23 = 13
Number of blue marbles in the jar = 13 × 24 = 8]

## Andhra Pradesh Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 13 Probability Optional Exercise exam preparation. The AP Board STD 10th Maths Chapter 13 Probability Optional Exercise Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 13 Probability Optional Exercise Books State Board syllabus with maximum efficiency.

## FAQs Regarding Andhra Pradesh Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbook Solutions

#### How to get AP Board Class 10th Maths Chapter 13 Probability Optional Exercise Textbook Answers??

Students can download the Andhra Pradesh Board Class 10 Maths Chapter 13 Probability Optional Exercise Answers PDF from the links provided above.

#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

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