Really hard probability question - please help!

[cambridgedove]

If we have an empty room, and every week there is a 0.6 probability for a person to join, 0.2 probability for a person to leave, and 0.2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks?

 

(A) 0.42

(B) 0.55

© 0.61

(D) 0.67

(E) 0.74

 

Please explain how you do it, thanks!

[cambridgedove]

Nobody can help me?

[vgp]

One of the hardest probability qestions I\ve ever seen. It is almost impossible to solve it in 2 or even 10 minutes if one is not a genious. Before trying to solve it, is there any known number of people who can join the room?

[cambridgedove]

I don't think the question states "if any known number of people can join the room," but this is a real question I encountered last month when I took my GRE math. Nobody in the GRE math forum can solve this so far; that's why I posted in the GMAT math section.

[vgp]

You are still missing something: please confirm that there should be “only one person can join the room each week”

[selytch]

it seems like a perpetual gmat/gre joke.

identical post @ mathforum from 9/04

 

 

Math Forum Discussions

 

it's either approximation to normal distribution, programming or markov's chain.

none of which are withing gmat scope.

 

from wikipedia

----

The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed. Therefore the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.

 

----

which mean as the other guy on the forums says

after approximation of 0.4 chance 1/2 person leaving the room (0.2 for staying, 0.2 for leaving)

0.6*1-0.4*1/2=0.4 (net change)

0.4*104=41.6 (mean after 104 weeks)

stddev=5

(σ = (p(1 − p)n)^1/2)

 

1.6/5=0.32 (0.32 std dev above mean)

which corresponds to probability 0.5+0.1255=~0.63 on normal standard distribution graph

(http://sweb.cz/business.statistics/normal.jpg)

 

which would be closest to C. (again, this is not my solution).

seeing such problem on GMAT I would pick C anyway without wasting my time on it. just because I like letter C.

[iamnotnader]

Hello! This actually looks pretty simple. Here is a solution the way I see it. Correct me if I'm wrong!

 

So let's consider the possibilities for wanting exactly x people in the room after 104 weeks. This can happen in the following ways:

 

1) x joins, 0 leaves, 104-x nc

2) x+1 joins, 1 leaves, 104-x-2 nc

3) x+2 j, 2 l, 104-x-4 nc

...

104) 104joins, 104-x leaves, 0 nc

 

So if we want the probability of exactly k people in the room after 104 weeks, we need to sum the following:

Sum[104!/((k+i)!*(i)!*(104-k-2*i)!) * .6^(k+i) * .2^(i) * .2^(104-k-2*i), i,0,Floor[(104-k)/2]]

To convince yourself of this, look up multinomial distribution:

Multinomial distribution - Wikipedia, the free encyclopediaand try to see it like this. We must also observe that we have a trinomial distribution which means we must multiply by 104!/(joins! leaves! nc!).

 

So if we want the probability for all k > 40, we simply sum over k from k = 40 to k = 104 since the number of weeks is finite. The solution is the following:

 

Sum[sum[104!/((k+i)!*(i)!*(104-k-2*i)!) * .6^(k+i) * .2^(i) * .2^(104-k-2*i), i, 0, Floor[(104-k)/2]], k, 40, 104]

 

which I compute with a little java program:

publicclass Problem

{

publicstaticvoid main(String[] args)

{

double sum = 0;

for (int k = 40; k

for (int i = 0; i

sum += fac(104)/(fac(k+i)*fac(i)*fac(104-k-2*i))*Math.pow(.6,k+i)*Math.pow(.2,i+104-k-2*i);

System.out.println(k+i + " " + i + " " + (104-k-2*i));

}

}

System.out.println(sum);

}

publicstaticdouble fac(int n)

{

if (n == 0 || n == 1)

return 1;

return n * fac(n-1);

}

}

 

and get about: .6063669447425515 which is about .61 or "C". Sorry it's not the easiest problem out there but hopefully your heart can rest easy knowing the exact answer and generally how to do it ;) This may be the kind of problem that wants to see what you do when you can't possibly know the answer.

[Brent Hanneson]

This question is WAYYYYYYYYYYYYYY out of scope for the GMAT.

 

Cheers,

Brent

[Nar]

iamnotnader you missed one point: when the room is empty you can't decrease the number of people to negative values.

 

the answer must be bigger than 0.61

 

[Nar]

the good answer is E = 0.749401

 

notice that you can't make a person leave when the room is empty

 

p(n,m) is the probability to find exactly n persons after m weeks

 

p(n,m) = 0.6*p(n-1,m-1)+0.2*p(n,m-1)+0.2*p(n+1,m-1) with m>n>0

 

p(0,m) = 0.2^m

p(m,m)= 0.6^m

Edited December 2, 2012 by Nar

[Tristar]

Here's what I think on this problem:

 

Since the net change out of 100 actions (weeks) is 40 people in the room, the given probabilities need to be adjusted by a factor of (100/104) = 0.961538 (approx.) So, we have: 0.6 * 0.961538 = 0.5769 (approx.). This is closer to answer B, which is 0.55.

[Nar]

0.61 is the answer Edited June 4, 2015 by Nar

[Nar]

.....