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 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?


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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?

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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?

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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?

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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?

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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?

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