Question 165381: What is the probability that 15 rolls of a fair die will show three threes?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! since i wasn't sure what you wanted, i answered two ways.
first way is getting "at least" 3 threes.
second way is getting "exactly" 3 threes.
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probability of getting "at least" 3 threes.
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that should be 1 minus the sum of the probability of
a.))) getting exactly 0 threes.
b.))) getting exactly 1 three.
b.))) getting exactly 2 threes.
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the probability of getting exactly 0 threes in 15 rolls of the dice is:
(1) * (5/6)^15 = .064905472
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the probability of getting exactly 1 three in 15 rolls of the dice is:
(15) * (1/6)^1 * (5/6)^14 = .194716415
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the probability of getting exactly 2 threes in 15 rolls of the dice:
[(15*14)/(1*2)] * (1/6)^2 * (5/6)^13 = .27260298
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the sum of these probabilities is:
.532224866
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1 minus the sum of these probabilities is:
.467775134
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your answer for the probability of getting "at least" 3 threes is .467775134.
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probability of getting "exactly" 3 threes.
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[(15*14*13)/(1*2*3)] * (1/6)^3 * (5/6)^12 = .236255916
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probability of getting "exactly" 3 threes is .236255916.
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to help you to understand what is happening, i'll take a much simpler example that will allow me to show you the concepts in more detail.
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say you have 3 rolls of the dice.
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the probability of getting exactly 0 threes is: 1 * (5/6)^3 = .578703704
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the probability of getting exactly 1 three is: [(3)/(1)] * (1/6)^1 * (5/6)^2 = .347222222
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the probability of getting exactly 2 threes is: [(3*2)/(1*2)] * (1/6)^2 *
(5/6)^1 = .069444444
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the probability of getting exactly 3 threes is: [(3*2*1)/(1*2*3)] * (1/6)^3 = .00462963
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since this is all the possibilities, then all of the probabilities must add up to 1.
they do.
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looking at the probability of getting exactly 2 threes provides the following:
probability of getting 1 three on one roll of the dice is 1/6.
probability of getting 0 threes on one roll of the dice is 5/6.
if you have 3 rolls of the dice, you can only get a three 2 time out of the 3 rolls.
probability on the first roll is 1/6 (you get the three)
probability on the second roll is 1/6 (you get the three)
probability on the third roll is 5/6 (you don't get the three)
the probability is therefore (1/6) * (1/6) * (5/6) = (1/6)^2 * (5/6)^1.
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but that's not the total ways you can get eactly 2 threes.
if you let a = 1/6, and b = 5/6, you can easily see that there are 3 ways this can happen:
aab (this is the way we just looked at).
aba
baa
you get it on the first and the second roll, or you get it on the first and the third roll, or you get it on the second and the third roll.
since the probability of each of these ways of getting exactly 2 threes on 3 rolls of the dice is the same, then the probability of all of them occurring is 3 * (1/6)*(1/6)*(5/6) = 3 * (1/6)^2 * (5/6)^1
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another example:
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if you make the number of rolls = 5, and you want to get exactly 3 threes, then it gets more complicated but still no where near as complicated as 15 rolls.
probability of exactly 3 threes out of 5 rolls is [(5*4*3)/(1*2*3)] * (1/6)^3 * (5/6)^2.
this probability is: .032150206
this is how it happens:
the probability of it happening one way is (1/6)^3 * (5/6)^2
that's the probability of you getting a three on the first 3 rolls of the dice and not getting a three on the last 2 rolls.
we get (1/6) * (1/6) * (1/6) * (5/6) * (5/6) = (1/6)^3 * (5/6)^2 = .003215021
that's only one way it can happen though.
it can in 10 total ways.
that would be (5*4*3)/(1*2*3) = 10.
if you let a = 1/6, and b = 5/6, then here's how it can happen:
aaabb (the one we looked at)
aabab
aabba
abaab
ababa
abbaa
baaab
baaba
babaa
bbaaa
that's 10 different ways it can happen.
get it on the first 3 rolls
get it on the first 2 and the 4th
get it on the first 2 and the 5th
etc.....
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the formula for determining the number of possible ways it can happen is n factorial divided by x factorial divided by (n-x) factorial.
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for 3 out of 5,
n = 5
x = 3
n factorial = 5*4*3*2*1
x factorial = 3*2*1
(n-x) factorial = 2*1
n factorial / x factorial / (n-x) factorial = 120 / 6 / 2 = 20 / 2 = 10
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i use a short hand version.
instead of 5*4*3*2*1 / 3*2*1 / 2*1
i do 5*4*3 / 3*2*1
which becomes 60 / 6 = 10.
you get the same answer since the (2*1) cancels out.
i only mention it because i used it above and you might get confused if i didn't.
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i spent enough time analyzing this so i'm reasonably sure it's correct, but you should verify with others just to make sure.
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gonzo@gmx.us
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