1. ## Probability help

Suppose that a point $\displaystyle (X_1,X_2,X_3)$ is chosen at random, that is, in accordance with the uniform probability density function over the following set S:
$\displaystyle S = {(x_1,x_2,x_3) : 0 \leq x_1 \leq 1, for i = 1,2,3}$
$\displaystyle Determine P[(X_1-\frac{1}{2})^2+(X_2-\frac{1}{2})^2+(X_3-\frac{1}{2})^2 \leq \frac{1}{4}$

Not sure how to do this and I have many more similar problems so if someone could give me some help here I may be able to do them all.

Thanks.

2. Originally Posted by Len
Suppose that a point $\displaystyle (X_1,X_2,X_3)$ is chosen at random, that is, in accordance with the uniform probability density function over the following set S:
$\displaystyle S = {(x_1,x_2,x_3) : 0 \leq x_1 \leq 1, for i = 1,2,3}$
$\displaystyle Determine P[(X_1-\frac{1}{2})^2+(X_2-\frac{1}{2})^2+(X_3-\frac{1}{2})^2 \leq \frac{1}{4}$

Not sure how to do this and I have many more similar problems so if someone could give me some help here I may be able to do them all.

Thanks.
$\displaystyle 0 \leq x_1$, $\displaystyle 0 \leq x_2$, $\displaystyle 0 \leq x_3$ defines a unit cube.

$\displaystyle \left(X_1-\frac{1}{2}\right)^2 + \left(X_2-\frac{1}{2}\right)^2 + \left(X_3-\frac{1}{2}\right)^2 \leq \frac{1}{4}$ defines a sphere with centre at $\displaystyle \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right)$ and radius $\displaystyle r = \frac{1}{2}$. This sphere is inside the cube.

The probability you want is equal to the proportion of volume of the cube occupied by the sphere, that is, $\displaystyle \frac{V_{\text{sphere}}}{V_{\text{cube}}}$.