Probability of choosing two real numbers $a$ and $b$ from $[1,4]$ such that $ ab>4$.

Question

What is the probability that when you pick two real numbers from the closed interval $[1,4]$, their product is greater than 4?

I tried to solve it with integration but I couldn't get the right answer. And I think that this problem can be solved without integration.

Are we choosing them from $[1,4]$ according to a uniform distribution? — πr8, Jan 12 '16 at 19:25

score 4 · Accepted Answer · answered Jan 12 '16 at 19:33

4

On the figure below, I have plotted the curve $x * y = 4$, as well as the square $[1, 4] \times [1, 4]$.

plot http://puu.sh/mtgbV/cce4527223.png

Any point above the red curve and within the square has coordinates with product $> 4$. Any point on or below the red curve and within the square has coordinates with product $<= 4$. Thus the probability you are looking for is the partition of the square above the red curve, over the total area of the square.

answered Jan 12 '16 at 19:33

Aegis

353

I tried to calculate the area by integrating y=4/x from 1 to 4 and then subtracting that area from the whole square's area and then to calculate the partition of the square above the curve over the whole square, but I didn't get the correct answer. – user265554 Jan 12 '16 at 19:35
I got it now. I was dumb and forgot that the bottom of the square wasn't on the x-axis :D – user265554 Jan 12 '16 at 19:51
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@user265554 So you knew how to do the problem. I tell my students when they make this kind of mistake that they got the mathematics right but the arithmetic wrong. – Ethan Bolker Jan 12 '16 at 19:54

sinbadh · Answer 2 · 2016-01-12T19:33:25.740

2

Supposing independence, this yields:

$P(XY\le4)=\int_1^4\int_1^{4/x}f_{X,Y}(x,y)dydx=\int_1^4\int_1^{4/x}\frac{1}{3}\frac{1}{3}dydx=\frac{8\log 2}{9}-\frac{1}{3}$.

Then, $P(XY>4)=1-P(XY\le 4)=\frac{4}{3}-\frac{8\log 2}{9}\sim 0.717$

edited Jan 12 '16 at 19:33

answered Jan 12 '16 at 19:29

sinbadh

7,521

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$f_{X,Y}$ should be $\frac{1}{9}$, no? – πr8 Jan 12 '16 at 19:30
Yes. I confused with $F_{X,Y}$ – sinbadh Jan 12 '16 at 19:33

score 2 · Answer 3 · answered Jan 12 '16 at 19:29

2

Hint. Draw the picture of $xy=4$ in the square with corners $(1,1)$ and $(4,4)$.

answered Jan 12 '16 at 19:29

Ethan Bolker

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Is there a differencw between xy=4 and y=4/x? I tried to calculate the area under the curve y=4/x from 1 to 4, and then to subtract it from the square with area 16 to get the area of the points which are above the curve y=4/x, and then to calculate the probability but I couldn't get the right answer. – user265554 Jan 12 '16 at 19:33

Probability of choosing two real numbers $a$ and $b$ from $[1,4]$ such that $ ab>4$.

3 Answers3