# RD Sharma Solutions Class 10 Probability Exercise 13.2

## RD Sharma Solutions Class 10 Chapter 13 Exercise 13.2

### RD Sharma Class 10 Solutions Chapter 13 Ex 13.2 PDF Free Download

Q.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 radius 0.5 m

A circle = $(0.5)^{2}$ = $0.25\pi m^{2}$

Area of rectangle = 3 x 2 = $6m^{2}$

Probability (geometric) = $\frac{measure of specified region part}{measure of whole region}$

The probability that tie will land inside the circle with diameter 1m

= $\frac{area of circle}{area of rectangle}$

= $\frac{0.25\pi m^{2}}{6m^{2}}$

= $\frac{\pi }{24}$

Q.2: In the accompanying diagram, a fair spinner is placed at the center O of the circle. Diameter AOB and radius OC divide the circle into three regions labeled X, Y and Z.��If $\angle BOC$ = 45�. What is the probability that the spinner will land in the region X?

Solution:

Given,

$\angle BOC$ = 45�

$\angle AOC$ = 180 – 45 = 135�

Area of circle = $\pi r^{2}$

Area of region x = $\frac{\Theta }{360}\times \pi r^{2}$

= $\frac{135}{360}\times \pi r^{2}$

= $\frac{3}{8}\times \pi r^{2}$

The probability that the spinner will land in the region

X = $\frac{Area of region x}{Total area of circle}$

X= $\frac{\frac{3}{8}\pi r^{2}}{\pi r^{2}}$

X= $\frac{3}{8}$

Q.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:

1st circle – with radius 3

2nd circle – with radius 7

3rd circle – with radius 9

Area of 1st circle = $\pi(3)^{2}$ = $9\pi$

Area of 2nd circle= $\pi (7)^{2}$� = $49\pi$

Area of 3rd circle = $\pi (9)^{2}$ = $81\pi$

Area of shaded region = Area of 2nd circle – area of 1st circle

= $49\pi – 9\pi$

= $40\pi$

Probability that it will land on the shaded region =� $\frac{area of shaded region}{area of third circle}$

= $\frac{40\pi }{81\pi }$ = $\frac{40}{81}$

Q.4: In below fig. points A, B, C and D are the centers of four circles that each has a radius of length one unit. If a point is selected at random from the interior of square ABCD. What is the probability that the point will be chosen from the shaded region?

Solution:

Radius of circle = 1 cm

Length of side of square =1+ 1= 2 cm

Area of square = 2 x 2= $4cm^{2}$

Area of shaded region = area of a square – 4 x area of the quadrant

= $4-4\times (\frac{1}{4})\times \pi (1)^{2}$

= $(4-\pi ) cm^{2}$

Probability that the point will be chosen from the shaded region = $\frac{Area of shaded region}{Area of square ABCD}$

= $\frac{4-\pi }{4}$

= $1 – \frac{\pi }{4}$

Since geometrical probability,

P(E) = $\frac{Measure of specified part of region}{Measure of the whole region}$

Q.5: In the fig. below, JKLM is a square with sides of length 6 units. Points A and B are the midpoints 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 triangle JAB?

Solution:

JKLM is a square with sides of length 6 units. Points A and B are the midpoints of sides KL and ML, respectively. If a point is selected at random from the interior of the square.

We have to find the probability that the point will be chosen from the interior of $\Delta JAB$.

Now,

Area of square JKLM is equal to $6^{2}$ = 36 sq.units

Now, we have

$ar\left ( \Delta KAJ \right ) = \frac{1}{2} \times AK \times KJ$

= $\frac{1}{2} \times 3 \times 6$

= 9 $unit^{2}$

$ar\left ( \Delta JMB \right ) = \frac{1}{2} \times JM \times BM$

= $\frac{1}{2} \times 6 \times 3$

= 9 $unit^{2}$

$ar\left ( \Delta AJB \right ) = \frac{1}{2} \times AL \times BL$

= $\frac{1}{2} \times 3 \times 3$

=� $\frac{9}{2}� \;\; unit^{2}$

Now, an area of the triangle AJB

$ar\left ( \Delta AJB \right ) = 36 � 9 � 9 – = \frac{9}{2}$

= $\frac{27}{2} \;\; unit^{2}$

We know that:

Probability = $\frac{Number \; of \; favourable \; events}{Total number\; of \;events}$

= $\frac{\frac{27}{2}}{36}$

= $\frac{27}{2 \times 36}$

= $\frac{3}{8}$

Hence, the probability that the point will be chosen from the interior of $\Delta AJB = \frac{3}{8}$.

Q.6: In the fig. below, a square dartboard 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, the length of the side of smaller square = a

The a length of a side of bigger square = 1.5a

Area of smaller square = $a^{2}$

Area of bigger square = $(1.5)^{2}a^{2}$ = $2.25a^{2}$

Probability that dart will land in the interior of the smaller square = $\frac{Area of smaller square}{Area of bigger square}$

= $\frac{a^{2}}{2.25a^{2}}$

= $\frac{1}{2.25}$

= $\frac{4}{9}$

Geometrical probability,

P(E) = $\frac{Measure of specified region part}{Measure of whole region}$

#### Practise This Question

Sarosh comes home from the office at 8 o' clock. The angle between the hour hand and minute hand when he reaches home is: