Probability Revisited.
Gokul gave me a nice one some time back ...
If you take two REAL numbers between 0 to 10, what is the probability that the square of one number is greater than the other number ?
By the way...
If you take two REAL numbers between 0 to 10, what is the probability that the square of one number is greater than the other number ?
By the way...
Probability = No of Favourable Outcomes/ Total No of Outcomes.
Theoretically, both numberator and denominator are infinity. There's a better answer than infinity/infinity though :-)
posted by Praveer @ Thursday, August 26, 2004
14 Comments:
Is your question "square of one number greater than the other" or is it "square of one number greater than the square of the other"?
If it is the former then the answer is one. For any two real numbers, this holds (even when they are equal). If it's the latter it is not one because, it doesn't hold when a = b.
If a and b are two real numbers :
a^2 > b is the condition.
Now the answer is not 1. You are thinking on different lines probably.
For a=2.0 and b = 6.0 a^2 < b
For a=2.0 and b = 3.0 a^2 > b
Now, To rephrase the question, When a and b are two randomly choosen REAL numbers, between 0 and 10, what is the probability that a^2 > b.
Mind you I'm talking about REAL numbers not Natural Numbers !
This shud make it sufficiently clear !
Hey, then your question should have been probability a^2 > b. It simply says square of "one" number being greater than the other. I took P(a^2>b or b^2>a), which is trivially 1.
I agree I shud have used "first number" instead of "one number".
Hope its clear now.
Thanks for pointing it out Sundar.
On second thoughts Sundar ...
YOU ARE TOO MUCH OF AN OPEN MIND !!!
X(
Is the answer 1/3 ? i.e. integral x^2.dx between 0 - 10 divided by 1000. The numerator is the area under the parabola y = x^2 and the denominator is the area of the rectangle [(0,0),(10,0),(10,100),(100,0)].
Oops! I was wrong as usual. It should be atleast greater than 1/2 (the probability that a > b). I wrote a Perl script that does random selection of ordered pairs of (close to) real numbers a billion times and computes the probability. The probability I got was 0.7891 We need to slap some reason over this result, though!
Sundar
Sundar,
U are objective type questions(fill-in the blank type) are made for each other.
But thanks to the discussion with u I was too lazy to actually solve it. That indeed was an eye opener.
The probability expression is:
For 0<= a,b <=10:
P=Nr/Dr
Nr = Number of ordered pairs (a,b) where a^2 >b d
Dr = total number of ordered pairs (a,b).
The answer is 1 - sqrt(10)/15
0.78918148932210804453340709703782
Area1 = The coordinates of all the points on the right and/or under the curve x^2=y form an ordered pair (x,y) where x^2>y.
Area2 = The coordinates of all the points inside the square {(0,0)-(10,10)} form a pair(x,y) for which 0<=x,y<=10
So denominator is the Area1
And, Numerator is Area1 intersection Area2.
i.e. Nr = Integration(x^2 dx) from 0 to sqrt(10) + area of rectangle {(0,sqrt(10))-(10,10)}
Dr = 100
This gives Nr/Dr = .7891
As usual the comments outworded the post many times over ...
Great job, Praveer. Empirical study sometimes at least gives a direction to look into.
Sundar
An Academic speculated whether a bather is beautiful if there is no one in the forest to admire her. He hid in the bushes to find out, which vitiated his premise but made him happy.
Moral: Empiricism is more fun than speculation.
- Sam Weber
I think we verified each other :)
the answer i got is little different
no of possible cases = area of square(10,10)=100
no of favourable cases = area of rect(10,10-sqrt(10))
= area of rect(10,6.85)=68.5
plus
area of parabola y=x^2 with the limits (0,3.15)=6.30
so probability=(6.30 + 68.5)/100=.7485
Hey Anon,
The area of Parabola x^2 = y between 0 to sqrt(10) is ( (sqrt(10))^3 )/3 Thats between 9 & 10, NOT 6.3 as u have suggested.
my mistake
me dumbo...;)
Nothing to worry Anon, DUMB is your normal state :D
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