Class 10 RD Sharma Solutions - Chapter 13 Probability - Exercise 13.2

Question 1. Suppose you drop a tie at random on the rectangular region shown in fig. below. What is the probability that it will land inside the circle with diameter 1 m?

Solution:

Area of a circle with the specified radius 0.5 m = (0.5)² = 0.25 πm²

Area of the rectangle = length × breadth = 3 × 2 = 6m²

Probability = \frac{measure of specified region}{measure of whole region}

Now,

The probability that the tie will land inside the circle, = area of circle/area of rectangle 

= 0.25 π m² / 6 m²

= π /24 

Therefore, the probability that the tie will land inside the circle = π/24

Question 2. In the accompanying diagram, a fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z.? If ∠BOC = 45°. What is the probability that the spinner will land in the region X?

Solution:

Given,

∠BOC = 45°

Also, by the application of linear pair

∠AOC = 180 – 45 = 135° 

Area of circle of radius r = πr²

Area of region x according to the figure= θ/360 × πr²

= 135/360 × πr²

= 3/8 × πr²

x = \frac{area of x}{area of circle} \\ x = \frac{\frac{3}{8}\pi r^2}{\pi r^2} \\ x = \frac{3}{8}

Hence, The required probability that the spinner will land in the region X is 3/8.

Question 3. A target is shown in fig. below consists of three concentric circles of radii, 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region?

Solution:

Now, we have the following values

I circle – with radius 3

II circle – with radius 7

III circle – with radius 9

Their corresponding areas are : 

Area of I circle = π(3)² = 9π

Area of II circle = π(7)² = 49π

Area of III circle = π(9)² = 81π

Now, calculating, 

Area of shaded region = Area of II circle – Area of I circle

= 49π − 9π

= 40π

Now, the probability that it will land on the shaded region is given by,

\frac{area of shaded region}{area of third circle} \\ = \frac{40\pi}{81 \pi} \\ = \frac{40}{81}

Hence, the required probability that the dart will land on the shaded region is equivalent to 40/81.

Question 4. In the figure, points A, B, C and D arc the centres of four circles that each have a radius of length one unit. If a point is selected at random from the interior o’ square ABCD. What is the probability that the point will be chosen from the shaded region?

Solution:

Radius of each of the circles = 1 unit 

Therefore, 

side of the square ABCD = 2 units

Area of sq ABCD = side²  = a² = 2 * 2 = 4 sq. units

Also,

Area of four quadrants at A,B,C and D is given by

= 4 * 1/4 πr²  

Substituting the values of r , we get, 

= π sq. unit

Therefore, area of shaded region = (4 - π) sq. units

And, the probability of the point that is selected from the shaded region = (4 - π)/4 = (1 - π/4)

Question 5. In the figure, JKLM is a square with sides of length 6 units. Points A and B are the mid-points of sides KL and LM respectively. If a point is selected at random from the interior of the square. What is the probability that the point will be chosen from the interior of ∆JAB?

Solution:

We know,

Length of sq side of JKLM = 6 units

Now, the area of the sq. JKLM = 6²  = 36 sq. units

We have, A and B as the midpoints of sides KL and LM.

Now, 

AL = AK = BM = BL = 3 units

Therefore, 

Area of triangle AJK = (JK * AK) /2 = (6 * 3) / 2 = 9 sq. units

Area of triangle JMB = (JM * MB) /2 = (6 * 3) / 2 = 9 sq. units

Area of triangle LAB = (LA * LB) /2 = (3 * 3) / 2 = 9/2 sq. units

Sum of these areas = 9 + 9 + 9/2 = 45/2 sq units.

Area of triangle JAB = Area of sq JKLM - Area of all the three triangles 

= 36 - 45/2 = 72-45/2 sq. units 

= 27/2 sq. units  

Probability = Area of triangle JAB/ Area of sq JMLK 

= 27 /(2 * 36) = 3/8

Question 6. In the figure, a square dart board is shown. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square. What is the probability that it will land in the interior of the smaller square?

Solution:

Let us assume the side of the smaller sq to be a. 

Also, let the length of the side of sq ABCD be 3/2 * a

Area of the sq. ABCD = (3a/2)² = 9/4 a² sq. units

Therefore, 

Probability = \frac{a^2}{\frac{9}{4}a^2} \\ = a^2 * \frac{4}{9a^2} \\ = \frac{4}{9}