SOLUTION: If you throw a pair of dice eight times, what is the probability that you will get "doubles" more then once?

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Question 601511: If you throw a pair of dice eight times, what is the probability that you will get "doubles" more then once?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The probability of getting doubles is p = 6/36 = 1/6 = 0.1667

Let's compute the probability of getting doubles x = 0 and x = 1 time. To find this, you would use the binomial probability distribution formula.


P(X = x) = (n ncr x)*(p)^(x)*(1-p)^(n-x)
P(X = 0) = (8 ncr 0)*(0.1667)^(0)*(1-0.1667)^(8-0)
P(X = 0) = (8 ncr 0)*(0.1667)^(0)*(0.8333)^(8-0)
P(X = 0) = (1)*(0.1667)^(0)*(0.8333)^8
P(X = 0) = (1)*(1)*(0.232494)
P(X = 0) = 0.232494

P(X = x) = (n ncr x)*(p)^(x)*(1-p)^(n-x)
P(X = 1) = (8 ncr 1)*(0.1667)^(1)*(1-0.1667)^(8-1)
P(X = 1) = (8 ncr 1)*(0.1667)^(1)*(0.8333)^(8-1)
P(X = 1) = (8)*(0.1667)^(1)*(0.8333)^7
P(X = 1) = (8)*(0.1667)*(0.2790035137489)
P(X = 1) = 0.372079


Now add these final results:

0.232494 + 0.372079

0.604573


So P(X <= 1) = 0.604573



Now subtract this from 1 to find P(X > 1)

P(X > 1) = 1 - P(X <= 1)

P(X > 1) = 1 - 0.604573

P(X > 1) = 0.395427


So the probability that you will get "doubles" more than once is roughly 0.395427

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If you throw a pair of dice eight times, what is the probability that you will get "doubles" more then once?
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# of possible doubles: 6
Probability of a double: 6/36 = 1/6
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Binomial Problem with n = 8 and p(double) = 1/6
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P(2<= x <=8) = 1 - P(0<= x <=1) = 1 - binomcdf(8,1/6,1) = 0.3953
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Cheers,
Stan H.