# 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...

Another fun GMAT question...

Best answer: If you want to do it without using formulas, you could do the following:

p(all five are heads)=0.6

p(4 are heads)=0.6

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(all five are heads)=0.6

^{5}p(4 are heads)=0.6

^{4}x0.4because 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(HHHHH)=0.6^5

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

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

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:

n!

----------

r! (n-r)!

posted by CrunchyFrog at 3:01 PM on July 30, 2006

(n choose r) * p^r * (1-p)^(n-r)

where "n choose r" is:

n!

----------

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

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

(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

posted by needs more cowbell at 3:08 PM on July 30, 2006

*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

posted by needs more cowbell at 3:10 PM on July 30, 2006

*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

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

posted by beerbajay at 3:15 PM on July 30, 2006

This thread is closed to new comments.

So, .6 * .6 * .6 * .6 = .1295 ~= 13%

posted by beerbajay at 2:54 PM on July 30, 2006