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10056 - What is the Probability ?

Posted: Fri Sep 13, 2002 8:57 pm
by nejhum
hi all,

i have solved the prolem of probabiliy using the formula

ans = (1-p)^i-1 * p / ( 1 - (1-p)^n )

but in case of big value of n and i it gives me 0.0000 result

is this o.k.

why this is WA???

Posted: Fri Sep 13, 2002 10:40 pm
by prom
Test for you
1
100 0 4

Posted: Sat Sep 14, 2002 4:17 am
by nejhum
thank you

i am now solved the problem

Posted: Sat Sep 14, 2002 4:17 am
by nejhum
thank you

i have now solved the problem

why WA??

Posted: Wed Oct 23, 2002 3:54 am
by Pedrinho UFPE
Hi people!

I'm tired having WA in this question!!
I real dont know why!
The case of p = 0.000000 is ok

My code is below:

[cpp]

#include <stdio.h>
#include <math.h>

double N, n, p;

void readdata() {
scanf("%lf%lf%lf",&N,&p,&n);
}

void process() {
if(p == 0) {
printf("%.4lf\n",0);
return;
} printf("%.4lf\n",pow(1.0-p,n-1) * p / (1.0 - pow(1-p,N)));
}

void main() {
#ifndef ONLINE_JUDGE
freopen("10056.in","r",stdin);
freopen("10056.out","w",stdout);
#endif
int ncasos;
scanf("%d",&ncasos);
while(ncasos--) {
readdata();
process();
}
}

[/cpp]

Please anyone know what is the problem?!!
Thanks Very much!!

solved

Posted: Wed Oct 23, 2002 4:28 am
by Pedrinho UFPE
I got AC.
The problem was precision error.
I have to print p + EPSILON...

Thanks all :-)

Posted: Mon Jun 30, 2003 2:33 pm
by szymcio2001
Your mistake is here:

[cpp]
printf("%.4lf\n",pow(1.0-P,I-1.0)/(6.0*(1.0-pow(1.0-P,N))));
[/cpp]

It should be as follow:

[cpp]
printf("%.4lf\n",pow(1.0-P,I-1.0)/((1/P)*(1.0-pow(1.0-P,N))));
[/cpp]

Because we throw "a dice like thing" not a dice, so it can have more or less than 6 sides.

And a little hint: it's prettier to write while(S--) than for(;S;S--) :)

10056

Posted: Fri Apr 23, 2004 6:14 pm
by dark man
can any one help me with some good inputs and outputs of 10056. T dont know what may be the problem with this simple problem.
Thanks in advance.

Check this

Posted: Sat Jun 19, 2004 9:05 am
by Rajib
I think the problem is easy. But what may be the problem in your code is, there may have some kind of precisional error :-? . Well, check this if it can solve your problem.

Input:
6
100 0.000001 1
100 0.000001 99
999 0.00023 591
2 0.0 1
915 0.166666 1
23 0.1111 13
Output:
0.0001
0.0001
0.0010
0.0000
0.1667
0.0290
Wish you to get AC :D

somebody save me

Posted: Tue Jun 21, 2005 8:05 pm
by gvs
I'm getting the following output for the following input:

[Input]:
6
100 0.000001 1
100 0.000001 99
999 0.00023 591
2 0.0 1
915 0.166666 1
23 0.1111 13

[Output]:
0.0099
0.0099
0.0010
0.0000
0.1667
0.0290

can anybody please help me.
I am using float variables, and using the pow function in math.h.
Also, i am using the formula probability=p*((1-p)^(i-1))/(1-(1-p)^n)

thanks i got AC

Posted: Tue Jun 21, 2005 8:07 pm
by gvs
well, it seems, the following output was good enuf... it was an error in my mail client...
I finally got AC.
:D

I don't now where a mistake.

Posted: Sun Oct 16, 2005 11:59 am
by StatujaLeha
Hi!

I use the formula in first post for solving. This code print the result. Where is a mistake?

Code: Select all

                if((p != 0)&&(p != 1))
		{
			q = 1.0 - p;
			printf("%.4lf\n",(pow(q,I - 1)*p)/(1.0 - pow(q,N)));
		}
		else
		{
			if(p == 0)
				printf("%.4lf\n",0.0000);
			else//p == 1
			{
				if(N == 1)
					printf("%.4lf\n",1.0000);
				else
					printf("%.4lf\n",0.0000);
			}
		}

Posted: Sun Oct 16, 2005 12:34 pm
by ReiVaX18
Just got AC, but my program gives slightly different output:
0.0100
0.0100
0.0010
0.0000
0.1667
0.0290

My program gives the same answer

Posted: Sun Oct 16, 2005 3:36 pm
by StatujaLeha
My program gives the same answer, but I get WA. Can you explain your algorithm?

Posted: Sun Oct 16, 2005 6:32 pm
by ReiVaX18
First, if the probability is zero, then the answer is zero also.

If not, for the player k to get success, the first k-1 should fail ( probability (1-p)^(k-1) ) and k should get success ( probability p ) or the n players have to fail ( probability (1-p)^n ) and then another time k-1 fails and a success or two times n fails and k-1 fails and a success or etc...

So if you calc the probabilities and sum them:

p*(1-p)^(k-1) + p*(1-p)^(k-1) * (1-p)^(n) + p*(1-p)^(k-1) * (1-p)^(2n) + ...

this is equal to

p*(1-p)^(k-1) * ( sum( i=0 to infinity, (1-p)^(i*n) ) )

the last sum is a geometric sum, so you aply the formula and get

p*(1-p)^(k-1) / ( 1 - (1-p)^n )

Note that if p = 0 the denominator is also 0, so this is why the case p = 0.0 is excluded.