Two random variables

Calculus Level 4

Two numbers are chosen uniformly at random on the interval $[0,2$ ].

What is the probability that their product is less than one, to 3 decimal places?

Hint: Try to first find the closed form of this probability.

The answer is 0.597.

1 solution

Geoff Pilling
Apr 3, 2017

The probability is given by the area under the curve $y = \dfrac{1}{x}$ which is inside the box defined by $0 < x < 2$ and $0 < y < 2$ divided by the area of the box itself. Or,

$P = \dfrac{1}{4}(1+ \int_{0.5}^{2}\dfrac{1}{x}dx)$
$P = \dfrac{1}{4}(1 + ln2 - ln(.5))$
$P = \dfrac{1 + 0.693 + 0.693}{4}$
$P = \boxed{0.597}$

Potential follow-up: If $x,y,z$ are all chosen uniformly at random over $[0,2]$ then what is the probability that $xyz \lt 1$ ?

Brian Charlesworth - 4 years, 2 months ago

Then we can imagine 3 axes X,Y,Z.Then we have to find the volume enclosed by the hyperbolic plane.

Am I right Sir ?

Kushal Bose - 4 years, 2 months ago

Yes, that's exactly right. I haven't done the calculation yet; could be a bit tricky.

Brian Charlesworth - 4 years, 2 months ago

I'm getting a probability of

$\dfrac{1 + 3\ln(2) + \frac{9}{2}(\ln(2))^{2}}{8} \approx 0.655$ ,

in case you want a value to compare to.

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth – I'm curious... Did you need to integrate over $\ln (x)$ to arrive at that result?

Geoff Pilling - 4 years, 2 months ago

@Geoff Pilling – My approach involved integrating over $\dfrac{\ln(z)}{z}$ , which was convenient. I'm not 100% sure of my answer, though. I took slices of $xy = \dfrac{1}{z}$ and then added them up.

Brian Charlesworth - 4 years, 2 months ago

@Geoff Pilling – If we, (Kushal included), can agree on a value then this would make for a great level 5 problem. I'll be interested to see what value you come up with.

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth – In the above problem there are two areas (1) A rectangle and (2) Area under hyperbola.

For three variables imagine the rectangle as parallelepiped and area under hyperbola becomes a hyperbolic volume.

So, total area $A_1=1/2 \times 2 \times 2=2$ and $A_2=2 \ln(3) \times 2$ i.e. base-area into height.

And I am getting the same probability.

Kushal Bose - 4 years, 2 months ago

@Kushal Bose – The logic looks good... When I get a chance I'll sit down and see if I come up with the same with a "slightly" different approach...

Geoff Pilling - 4 years, 2 months ago

Two random variables

The answer is 0.597.

1 solution

0 pending reports