What is the probability that 15 tosses of a fair coin will show 6 tails?
2006-10-28 14:00:26 · 7 answers · asked by tifi50 1 in Science & Mathematics ➔ Mathematics
yes, fair coin is heads on one side; tails on the other.
2006-10-28 14:05:49 · update #1
okay, now i kind of get it - still trying to figure out the rationale behind it - i seem to kind of "get it" but not sure why on the formula. thank you. I've been working this all day and had to step away and come back to it.
2006-10-28 15:08:14 · update #2
This is a binomial probability poroblem. You'll use the formula
(nCr)(p^r)(q^(n-r)), where n is the number of trials (15), r is the number of "successes" (6), p is the probability of success (.5), and q is the probability of failure (.5). (nCr stands for combinations, which you can work out on any scientific calculator.) This works out to be appx. .1527404785
2006-10-28 14:31:09 · answer #1 · answered by dmb 5 · 0⤊ 0⤋
This must be a binomial distribution because you are replacing the coin (using the same one), there is the random variable (6 tails), they are independent trials, and you are dealing with a discrete value. We are making tail a success. Think about it this way: the probability is always a half, the probability of success is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 (or 0.5^6) x the probability of failure (because to only win 6 times you must have gotten a head 9 times) x the combinations (because you could get one wrong the first time or alternate or change the pattern...)
P(success) = p = 1/2
P(failure) = q = 1/2
Number of trials = n = 15
Number of successful trials (tail) = r = 6
P(X = r) = nCr(p^r)(q^n-r)
Therefore,
15C6 ((1/2)^6)((1/2)^9) = 0.1527 (4 s. f.)
I hope you have done this type of thing before...notify me if I was of help...or ask more
2006-10-28 14:55:15 · answer #2 · answered by Carrot, the Peanut 1 · 1⤊ 0⤋
See Martian answer. Its the right rationale.
The terms quoted are from the expansion of (p + q)^15 where p=0.5 and q=1 - p = 0.5
So the term that is being equated is 15C6p^6 q^9 where
15C6 = 15! x 6! / 9! = 15x14x13x12x11x10 / (6x5x4x3x2)
2006-10-28 18:20:02 · answer #3 · answered by Anonymous · 0⤊ 0⤋
9:15
2006-10-28 14:03:47 · answer #4 · answered by fuzzystuff511 2 · 0⤊ 0⤋
What is a FAIR coin - you mean one with a head or tails on either side ?
2006-10-28 14:02:31 · answer #5 · answered by casrcitizen 2 · 0⤊ 0⤋
3603600 (15x14x13x12x11x10). There are 15 ways to pick the first head, 14 ways to pick the second, etc.
2006-10-28 15:20:56 · answer #6 · answered by mjvande@pacbell.net 1 · 0⤊ 0⤋
{15!/[(15-6)!*6!]}*[(.5)^(6)]*[(0.5)^(15-6)]
=5005*(0.5)^15=0.152740447851
2006-10-28 14:55:15 · answer #7 · answered by AlexisEd 2 · 0⤊ 0⤋