Points in a box

Probability Level 3

$x$ , $y$ , and $z$ are each chosen uniformly and at random in the interval $[0,1]$

What is the probability that $x^2+y^2+z^2 < 1$ ?

Provide your answer to 3 decimal places.

The answer is 0.524.

1 solution

Geoff Pilling
Dec 9, 2016

Since, $x^2+y^2+z^2 < 1$ defines a unit sphere, and $x$ , $y$ , and $z$ are positive, the probability that $x^2+y^2+z^2 < 1$ is equivalent to the ratio of the volume of one eighth of a unit sphere to that of a unit cube.

So,

$P = \frac{\frac{1}{8} \cdot (\frac{4}{3} \pi \cdot 1^3) }{ 1^3} = \frac{\pi}{6} = \boxed{0.524}$

A nice problem. After a few moments thought it suddenly dawned on me to think geometrically and I solved it just as you did.

Paul Hindess - 4 years, 6 months ago

Points in a box

The answer is 0.524.

1 solution

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