Two points P and Q are taken at random on aline segment OA of length a. The probability that `PQgtb,` where `0ltblta`, is

Options

A. `(1-(c )/(a))^(2)`
B. `(1-(a)/(c ))^(2)`
C. `1-(c^(2))/(a^(2))`
D. `1-(a^(2))/(c^(2))`

Answer

A

Explanation

Let AB be a straight line segment of length a and let P and Q be two points on it such that AP=x and AQ=y. Then, we have to find the probability for `|PQ| gt c` i.e. `|x y|gt c`.
Clearly, `0 le x le a " and " 0 le y le a`.
We observe that there are two variables x and y satisfying `0 le x le a " and " 0 le y le a`. In the cartesian plane the inequalities `0 le x le a " and " 0 le y le a ` represent the region enclosed by the square having x=0, y=0, x=a and y=a as its sides. Clearly, area enclosed by this square is `a^(2)`.
Now, `| x y| gt c`
`implies x-y gt c " and " -(x-y) gt c`
`implies x-y gt " and " y-x gt c`
In order to find `P(|x-y| gt c)`, we require the area area of the region enclosed by `x-y gt c, y-x gt c, 0 le x le a " and " 0 le y le a`. Clearly, shaded region is Fig. 1 represents the area enclosed. We have,

We have,
Area of the shaded region = Area APQ + Area BRS `=(1)/(2)(a-c)^(2)+(1)/(2)(a-c)^(2)=(a-c)^(2)`

Hence, required probability
`=("Area enclosed by " |x-y| gt c, 0 le x le a " and " 0 le y le a)/(" Area enclosed by " 0 le x le a, 0 le y le a)`
`=("Area of the shaded region")/("Area of the square")=((a-c)^(2))/(a^(2))=(1-(c )/(a))^(2)`