Problem Solving

DavoodBeater wrote:The question is from https://www.urch.com/forums/gre-math/108 ... -help.html
However there is no answer for this question and I am eager to know the answer.
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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
(C) 0.61
(D) 0.67
(E) 0.74

IMO: E No OA

The person who posted this question to a GRE forum (claiming they saw it on a real GRE test) was clearly making a joke. The problem isn't from the GRE at all; it was posted five years ago to an advanced math forum, as google reveals:

mathforum.org/kb/message.jspa?messageID=3227348&tstart=0

There's a solution there using Markov chains; the solver wrote a computer program to arrive at an answer. I arrived at the same answer quite quickly, using basic stats, however. Still, I'd stress there is absolutely *no way* you could ever see a question even remotely like this on the GMAT; I wrote up my solution (based on estimation) below because I find the problem interesting, and because you said you were curious about the answer, but GMAT test takers can stop reading now!

-We can approximate this as a 'binomial experiment' - that is, one with two possible outcomes each time - by combining two of the probabilities together. Say we have a 0.6 chance of gaining 1 person each week, and a 0.4 chance of losing half a person each week - this is an approximation for a finite experiment, but all should roughly even out;

-With a large sample, a binomial distribution will be approximately a normal distribution, so we can use the normal distribution to estimate our probabilities here;

-The 'expected value' of the number of people after 104 weeks is 41.6; that is, over the long run, we expect, on average, to gain 0.6*1 - 0.4*0.5 = 0.4 people each week (we have a 60% chance of gaining one person, and a 40% chance of losing half a person). So, if the experiment were to run for n weeks, where n is some large number, we'd expect the number of people to be very close to 0.4n;

-This value, 41.6, is the median (and mean) of our distribution, and about half the time we should have more people than that, and about half the time we should have fewer people. The question asks about 40 people, however. The standard deviation of our binomial distribution is 5 (using the formula for the standard deviation of a binomial distribution = root(n*p*(1-p))), so we want to know the probability that the number of people is 1.6/5 = 0.32 standard deviations below the mean or higher. Consulting a standard normal table, we find this probability to be about 0.62.

-Since we're estimating, we wouldn't expect to get an exact answer using the above, but only to be very close to the answer. So 0.61 looks like our best bet, and it's also the approximate answer arrived at by the guy who enumerated all the possibilities on his computer.

Even to apply the above, you need to refer to a stats table, and you need to use math that never appears on the GMAT - properties of binomial distributions, expected value, etc. Your best bet is to move on to simpler questions!