GMAT probability question
July 30, 2006 2:49 PM   Subscribe

For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, what is the probability that the outcome will be heads *at least* 4 times?

Another fun GMAT question...
posted by VillageLion to Grab Bag (12 answers total)
Shouldn't this be == to the probability that the coin comes up heads 4 times (since you don't care what the 5th toss is and order is unimportant)?

So, .6 * .6 * .6 * .6 = .1295 ~= 13%
posted by beerbajay at 2:54 PM on July 30, 2006

Best answer: If you want to do it without using formulas, you could do the following:
p(all five are heads)=0.65
p(4 are heads)=0.64x0.4
because there are five different ways the second situation could happen, multiply that probability by five. Now add all probabilities together.
posted by needs more cowbell at 2:56 PM on July 30, 2006

P(4H and 1T, any combination)=0.6^4 * 0.4^1
There are 5 different ways to arrange 4 heads and 1 tail. (This is "5 choose 1".) So, the preceding number is multiplied by 5.
Total probability:
0.6^5 + 5 * 0.6^4 * 0.4^1
posted by CrunchyFrog at 2:57 PM on July 30, 2006

CrunchyFrog & nmc are correct. It totals up to 1053/3125 or about 33.7% chance.
posted by Wolfdog at 3:01 PM on July 30, 2006

For more information, look up "binomial distribution" in any introductory statistics textbook. In general, if an experiment is repeated n times, and we want the probability of r successes, when the probability of any individual experiment being a success is p, the probability is:
(n choose r) * p^r * (1-p)^(n-r)

where "n choose r" is:
r! (n-r)!
posted by CrunchyFrog at 3:01 PM on July 30, 2006

Why am I wrong?

The fifth toss is unimportant. Its probability is 1 that you will get some result.

Why does the order affect this probelm at all?
posted by beerbajay at 3:05 PM on July 30, 2006

/me goes to reread the WP article on BD...
posted by beerbajay at 3:07 PM on July 30, 2006

(note: IIRC it'll get harder to do it without using CrunchyFrog's method when the numbers are different--it's easy to figure out that there are 5 ways to have 1 tail and 4 heads, but it would be less obvious if you were dealing with 17 coin-flips and 4 instances of heads. I'm not sure what the GMAT expects you to know/be able to do. The easiness of this particular question may or may not be a fluke. )
posted by needs more cowbell at 3:08 PM on July 30, 2006

Er, using CrunchyFrog's formulas from his second response, that is.
posted by needs more cowbell at 3:10 PM on July 30, 2006

Beerbajay: The last toss is unimportant if the first four tosses are heads, yes. But you're ignoring the possibility that one of the first four tosses is tails, in which case the last toss *is* important.
posted by Johnny Assay at 3:12 PM on July 30, 2006

Ah yes of course. Forgive me father, for it has been several years since my last discrete math course.
posted by beerbajay at 3:15 PM on July 30, 2006

Response by poster: Thanks everyone!
posted by VillageLion at 3:53 PM on July 30, 2006

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