Students can Download Class 10 Maths Chapter 14 Probability Ex 14.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 14 Probability Ex 14.2

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?

Answer:

Number of all possible out comes

= 5 × 5 = 25

n(S) = 25

(i) Let E_{1} be a event favourable to visit same day for shop [Tue, Tue; Wed, Wed; Th, Thu; Fri, Fri; Sat, Sat]

n(E_{1}) = 5

Probability of event both visit shop same day

(ii) Let E_{2} be a event favourable to visit shop both on consecutive days

E_{2} = {T, W; W, T; W, Th; Th, W; Th, F; F, Th; F, S; S, F} n(E_{2}) = 8

Probability of a event favourable to visit both consecutive days

(iii) Probability of both will visit shop different days = 1 – Probability both visit shop same day

Question 2.

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 in the two throws:

What ist the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

Solution:

The complete table as follows

∴ Number of all possible outcomes = 36

(i) For total score being even:

Favourable outcomes = 18

[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8,12]

∴ The required probability = \(\frac{18}{36}=\frac{1}{2}\)

(ii) For total score being 6 :

In list of scores, we have four 6’s.

∴ Favourable outcomes = 4

∴ Required probability = \(\frac{4}{36}=\frac{1}{9}\)

(iii) For toal score being at least 6:

The favourable scores are : 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12

∴ Number of favourable outcomes = 15

∴ Required probability = \(\frac{15}{36}=\frac{5}{12}\)

Question 3.

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 the number of blue balls = x.

∴ total number of balls = 5 + x.

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

P (drawing blue ball) = \(\frac{x}{x+5}\)

x = 2 × 5

x = 10

Hence the number of blue balls in a bag is 10.

Question 4.

A box contains 12 balls out of whichxare 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.

Solution:

∵ The total number of balls in the box = 12

∴ Number of possible outcomes = 12

Case – I: For drawing a black ball

Number of favourable outcomes = x

∴ Probability of getting a black ball = \(\frac{x}{12}\)

Case – II: When 6 more black balls are added

Then, the total number of balls = 12 + 6 = 18

⇒ Number of possible outcomes = 18

Now, the number of black balls = (x + 6)

∴ Number of favourable outcomes = (x + 6)

∴ Required probability = \(\frac{x+6}{18}\)

According to the given condition,

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

⇒ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x

⇒ 36x – 12x = 72 ⇒ 24x = 72

⇒ x = \(\frac{72}{24}\) = 3

Thus, the required value of x is 3.

Question 5.

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 \(\frac{2}{3}\) Find the number of blue balls in the jar.

Answer:

Total number of marbles in the jar = 24

Let the number of blue marbles is x and the number of green marbles is 24 – x

Probability of marble drawn is green

24 – x = 16

x = 24 – 16

x = 8

∴ Number of blue marble is 8.

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