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# Class 10 RD Sharma Solutions – Chapter 13 Probability – Exercise 13.2

• Last Updated : 18 Mar, 2021

### 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 Ï€m2

Area of the rectangle = length Ã— breadth = 3 Ã— 2 = 6m2

Now,

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

= 0.25 Ï€ m2 / 6 m2

= Ï€ /24

Therefore, the probability that the tie will land inside the circle = Ï€/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 âˆ BOC = 45Â°. What is the probability that the spinner will land in the region X?

Solution:

Given,

âˆ BOC = 45Â°

Also, by the application of linear pair

âˆ AOC = 180 â€“ 45 = 135Â°

Area of circle of radius r = Ï€r2

Area of region x according to the figure= Î¸/360 Ã— Ï€r2

= 135/360 Ã— Ï€r2

= 3/8 Ã— Ï€r2

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 â€“ with radius 3

II circle â€“ with radius 7

III circle â€“ with radius 9

Their corresponding areas are :

Area of I circle = Ï€(3)2 = 9Ï€

Area of II circle = Ï€(7)2 = 49Ï€

Area of III circle = Ï€(9)2 = 81Ï€

Now, calculating,

Area of shaded region = Area of II circle â€“ Area of I circle

= 49Ï€ âˆ’ 9Ï€

= 40Ï€

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

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 oâ€™ 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 = side = a2 = 2 * 2 = 4 sq. units

Also,

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

= 4 * 1/4 Ï€r

Substituting the values of r , we get,

= Ï€ sq. unit

Therefore, area of shaded region = (4 – Ï€) sq. units

And, the probability of the point that is selected from the shaded region = (4 – Ï€)/4 = (1 – Ï€/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 âˆ†JAB?

Solution:

We know,

Length of sq side of JKLM = 6 units

Now, the area of the sq. JKLM = 6 = 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 =

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