Uniformly at Random

Probability Level pending

Two variables, $a$ and $b$ , are chosen uniformly at random.

$a$ is chosen on the interval $[0,5]$

$b$ is chosen on the interval $[3,13]$

For example, $a$ could be $1.342$ and $b$ could be $8.03328$

What is the probability that $a>b$ ?

The answer is 0.04.

1 solution

Geoff Pilling
Dec 9, 2016

The only way $a$ could be greater than $b$ is if both variables are between $3$ and $5$ .

$P(3<a<5) = 0.4$

$P(3<b<5) = 0.2$

And, if they are both between $3$ and $5$ there is a $50/50$ chance that $a>b$ .

So,

$P = (0.4) \cdot (0.2) \cdot \frac{1}{2} = \boxed{0.04}$

I did it the same way but had a niggling doubt so I tried the integral approach as well:

$\dfrac{1}{5} \times \dfrac{1}{10} \displaystyle \int_{3}^{5} \int_{3}^{x} dy dx = \dfrac{1}{50} \int_{3}^{5} (x - 3) dx =$

$\dfrac{1}{50} \left(\dfrac{x^{2}}{2} - 3x\right)_{3}^{5} = \dfrac{1}{50}\left(\left(\dfrac{25}{2} - 15\right) - \left(\dfrac{9}{2} - 9\right)\right) = \dfrac{1}{50}(8 - 6) = \dfrac{1}{25}$ ,

confirming your more intuitive approach.

Brian Charlesworth - 4 years, 6 months ago

Ah, nice confirmation! :-)

Geoff Pilling - 4 years, 6 months ago

Exactly my approach!

Paul Hindess - 4 years, 6 months ago

I wanted to assign 50-50 chance too but same as Brian I thought that that would be 'too' convenient for one of Geoff's question. So I made a number line 0/---/3/---x---/---y---/5/---/13 and my answer would be xy/(5-0)(13-3) = xy/50. xy would be the area of half a square with side 2, so xy/50 = (2^2/2)/50 = 0.04. Still bad at integration after all this while.

Saya Suka - 4 years, 6 months ago

Uniformly at Random

The answer is 0.04.

1 solution

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