SOLUTION: a biased coin has 1 in 10 chance of landing heads. If tossed 400 times, what is the estimated chance of getting exactly 40 heads?

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Question 207919: a biased coin has 1 in 10 chance of landing heads. If tossed 400 times, what is the estimated chance of getting exactly 40 heads?
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
a biased coin has 1 in 10 chance of landing heads. If tossed 400 times, what is the estimated chance of getting exactly 40 heads?
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It's binomial with n=400, p = 0.1, x = 40.
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P(x+40) = 400C40(0.1)^40*(0.9)^360 = 0.0664
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Cheers,
stan H.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
probability of getting a head is .1
if tossed 400 times the probability of getting exactly 40 heads would be ?????
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to make this simpler, we need to make the numbers a lot smaller.
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let the number of tosses = 3
what is the probability of getting exactly 1 head?
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probability of getting a head on the first toss would be:
.1 * .9 * .9 = .081
probability of getting a head on the second toss would be:
.9 * .1 * .9 = .081
probability of getting a head on the third toss would be:
.9 * .9 * .1 = .081
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the number of possible occurrences of getting x heads out of n tosses is given by the formula:
n! / (x!*(n-x)!
in this particular case:
n = 3
x = 1
the formula becomes 3!/(1!*2!)
this becomes 3*2*1/1*2*1 which results in 3.
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this can be seen because the number are small:
the possible number of occurrences are:
htt
tht
tth
there are no other possible number of occurrences
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so for this very simple case, the probability of getting exactly 1 head out of 3 tosses is the probability of getting exactly 1 head in one occurrence times the number of possible occurrences.
this equals:
.1 * .9 * .9 = the probability of getting exactly one head on one occurrence.
* 3 = the number of possible occurrences
(.1*.9*.9) * (.3) equals to .243
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extending this simple example, let's do getting exactly 2 heads out of 4 tosses.
n = 4
x = 2
the answer is the probability of getting exactly one of the possible occurrences times the number of possible occurrences. This always assumes that the probability of each occurrence is exactly the same.
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the probability of getting exactly one of the occurrences is .1 * .1 * .9 * .9
the possible number of occurrences are 4! / (2!*(4-2)!) = 4! / (2!*2!)
this equals 4*3*2*1/1*2*1*2 which equals 6
the total probability is (.1 * .1 * .9 * .9) * (6) which equals .0486
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let's see how this works.
the total possible number of occurrences are:
hhtt
htht
htth
thht
thth
tthh
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each one of these has a probability of .1 * .1 * .9 * .9 = .0081
6 * .0081 = .0486
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we can now make a generalization.
the probability of x occurrences out of n tosses is the probability of one of those occurrences times the number of possible occurrences.
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the probability of exactly one of those occurrence = p%5Ex+%2A+q%5E%28n-x%29
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let's see how this fits the two examples we just did.
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first was the probability of getting exactly 1 head out of 3 tosses when the probability of getting one head is .1.
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n = 3
x = 1
probability of getting exactly one head on one occurrence is (.1)^1 * (.9)^2 = .081
number of possible occurrences is n!/(x!*(n-x)!) = 3!/(1!*2!) = 3*2*1/1*2*1 = 3
probability is .081 * 3 = .243 which is what we calculated earlier.
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second was the probability of getting exactly 2 heads out of 4 tosses when the probability of getting one head is .1.
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n = 4
x = 2
probability of getting exactly one head on one occurrence is (.1)^2 * (.9)^2 = .0081
number of possible occurrences is n!/(x!*(n-x)!) = 4!/(2!2!) = 4*3*2*1/1*2*1*2 = 6
probability is .0081 * 6 = .0486 which is what we calculated earlier.
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we'll do one more just to see if this formula works ok in all cases.
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let's assume 5 tosses and we want to get exactly 3 heads with the probability of getting 1 head on 1 toss = .1
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probability of getting exactly 3 heads in one occurrence is:
(.1)^3 * (.9)^2 = .00081
number of possible occurrences is:
5! / (3!*(5-3)!) = 5! / (3!*2!) = 5*4*3*2*1/3*2*1*2*1 = 10
probability is 10 * .00081 = .0081
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the possible number of occurrences are:
hhhtt
hhtht
hhtth
hthht
hthth
htthh
thhht
thhth
ththh
tthhh
each one of these has the same probability of occurrence therefore the total probability of occurrence is 10 * .00081 = .0081.
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we can extrapolate from this to solve your problem.
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your problem is:
400 tosses.
what is the probability of getting exactly 40 heads if the probability of getting one head is .1
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n = 400
x = 40
probability of 1 occurrence is %28.1%29%5E40%2A%28.9%29%5E%28400-40%29 = %28.1%29%5E40%2A%28.9%29%5E360
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the total possible number of occurrences is 400%21%2F%2840%21%2A%28400-40%29%21%29 = 4001%2F%2840%21%2A360%21%29
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this is beyond the capability of my calculator so we have to improvise.
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i'll cheat and use excel.
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that didn't work directly either, so i had to cheat even more by breaking the problem into sections that resulted in numbers that were within the capabilities of excel.
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i succeeded as far as i know which resulted in numbers that excel could handle.
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the answer is:
probability of getting exactly 40 heads out of 400 tosses when the probability of getting one head is .1 is .06635047
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the possible number of occurrences was:
1.97034E+55
the probability of getting exactly one occurrence of exactly 40 heads out of 400 tosses was:
3.36747E-57
the total probability was the first number times the second number which was:
0.06635047
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the method used was as follows in a simple example.
you want to get 15! / (8!)*(7!)
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without reducing your calculations would involve:
15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 / (8*7*6*5*4*3*2*1)*(7*6*5*4*3*2*1)
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since the 8*7*6*5*4*3*2*1 in the numerator and the denominator would cancel each other out anyway, you restructure the problem as:
15*14*13*12*11*10*9 / 7*6*5*4*3*2*1
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if these number are still too big for your calculator, you restructure the problem again as follows:
((15*14*13*12)/(7*6*5*4)) * ((11*10*9)/(3*2*1))
this results in two numbers that are more manageable and can be multiplied together later on to get the result you want.
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a very simple example that we can calculate is:
10*9*8*7/4*3*2*1 = 5040/24 = 210
breaking this up into 2 parts, we get:
10*9/4*3 = 90/12 = 7.5
8*7/2*1 = 56/2 = 28
multiplying the piece parts out, we get:
7.5 * 28 = 210
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that's the technique i used and it worked.
at least i think it did.
there are too many numbers in there for me to totally verify.
i'll assume they worked on faith that excel did what it was supposed to do and handled the numbers ok if it didn't complain that it couldn't.
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i did verify with smaller numbers that the technique worked so i'm reasonably confident.
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