Question 333206: Consider a die, which has the face of the 1-spot and the 5-spot, painted red. The remaining faces of the die are white. You roll this die 12 times. It either lands with a white face up, or a red face up.
a. What is the probability of getting exactly 7 red faces in these 12 rolls?
b. What is the probability of getting at least 4 white faces in 12 rolls?
Statics homework is kicking my butt, help would be greatly appreciated. Thanks again.
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and is calculated by !})
For part a, your number of trials, is 12, your desired number of successes, is 7, and your probability of success on an individual trial, is 
So
\ =\ \left(12\cr\ 7\right\)\left(\frac{1}{3}\right)^7\left(\frac{2}{3}\right)^5)
The other problem is worked the same way, except your number of trials, is 12, your desired number of successes, is 4, and your probability of success on an individual trial, is  , and you want to compute:
)
I'll leave you alone to spend some quality time with your calculator
John
My calculator said it, I believe it, that settles it
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Consider a die, which has the face of the 1-spot and the 5-spot, painted red. The remaining faces of the die are white. You roll this die 12 times. It either lands with a white face up, or a red face up.
P(red on each throw) = 2/6 = 1/3
P(white on each throw) = 4/6 = 2/3
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The following are binomial problems with n = 12:
a. What is the probability of getting exactly 7 red faces in these 12 rolls?
P(x = 7) = 12C7*(1/3)^7*(2/3)^5 = binompdf(12,1/3,7) = 0.0477
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b. What is the probability of getting at least 4 white faces in 12 rolls?
P(4<= x <=12) = 1 - binomcdf(12,1/3,3) = 0.6069
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Cheers,
Stan H.
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