Question

A coin of diameter $\dfrac{1}{2}$ units is tossed randomly onto the rectangular cartesian plane, the probability that the coin does not intersect any line whose equation is of the form \[x = k\] , is
A.$\dfrac{1}{{\sqrt 2 }}$
B.$1 - \dfrac{1}{{\sqrt 5 }}$
C.$\dfrac{1}{4}$
D.$\dfrac{1}{2}$

Answer 1

Hint: Here we use the concept of probability and x-y plane.


Required Formula: ${\text{Probability of an event = }}\dfrac{{{\text{favourable cases}}}}{{{\text{Total cases}}}}$


Complete step by Solution:
Given: Diameter of a coin tossed is $\dfrac{1}{2}$ $ \Rightarrow radius = \dfrac{{diameter}}{2} = \dfrac{1}{4}$
Equation of the line is \[x = k\].
 Consider two adjacent lines \[x = k{\text{ }}and\;x = k + 1\]
We need line equations of the form \[x = k{\text{ }}and\;x = k + 1\].
 According to the question, For the favorable event, the centre of the coin should fall into a shaded region.

Distance between two adjacent lines \[x = k{\text{ }}and\;x = k + 1\] is $1$.
 As radius is $\dfrac{1}{4}$ from both adjacent lines, If the coin falls exactly in the shaded region it will not intersect the two lines.
\[ \Rightarrow {\text{Required probability = }}\dfrac{{{\text{Area of shaded portion}}}}{{{\text{Distance between two adjacent lines}}}}{\text{ = }}\dfrac{{\dfrac{{\text{1}}}{2}}}{1}{\text{ = }}\dfrac{{\text{1}}}{2}\]
Hence, Option choice D is the correct answer.

Note: In such types of questions which involves concept probability having an idea about the formula is needed. Sometimes probability questions may be accompanied with x-y planes. Graphing helps to solve the question. Follow the conditions in the question. Frame the equations accordingly to get the required value.