Sum of Two points

Two points are selected at random in the interval 0 x 1 0\le x\le1 . Determine the probability that the sum of their squares is less than 1.

1 2 \frac{1}{2} π 3 \pi - 3 π 6 \frac{\pi}{6} 1 3 \frac{1}{3} π 4 \frac{\pi}{4} 2 3 \frac{2}{3}

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2 solutions

Let the two numbers be x , y x , y on two co-ordinate axes.
0 x , y 1 0 \le x,y \le 1
The area of the entire region A = 1 1 = 1 A = 1 \cdot 1 = 1 .
x 2 + y 2 1 , 0 x , y x^{2} + y^{2} \le 1 , 0 \le x,y denotes the region inside a quarter circle of radius 1.
Area of this region A 1 = π 4 A_{1} = \dfrac{\pi}{4}
Probability = A 1 A = π 4 = \dfrac{A_{1}}{A} = \dfrac{\pi}{4}

Nice work.

Hana Wehbi - 5 years ago
Hana Wehbi
May 26, 2016

Let x x and y y denote the random variables associated with the given points. Since equal intervals are assumed to have equal probabilities, the density functions of x x and y y are given respectively by:

f 1 f_{1} ( x ) (x) = 1 w h e n 0 x 1 1\ when\ 0\le x\le 1 ; otherwise, it is 0 0

f 2 f_{2} ( y ) (y) = 1 w h e n 0 y 1 1\ when\ 0\le y\le 1 ; otherwise, it is 0 0

Since x x and y y are independent, the joint density function is given by:

f ( x , y ) = f 1 ( x ) f 2 ( y ) f(x,y)= f_{1}(x)f_{2}(y) = 1 w h e n 0 x 1 1\ when\ 0\le x\le 1 and 0 y 1 \ 0\le y\le 1 ; otherwise, it is 0 0

It follows that the required probability is given by: P ( x 2 + y 2 1 ) = P(x^{2}+y^{2}\le 1)= R d x d y \large\iint_Rdxdy

where R \large R is th region defined by x 2 + y 2 1 x^{2}+y^{2}\le 1 for non-negative x x and y y , which is a quarter of a circle of radius 1.

We see now the area of R \large R is π 4 \frac{\pi}{4} , where the r = r a d i u s = 1 r= radius=1 and the area of the circle is π r 2 \pi*r^{2}

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