SOLUTION: Consider a loaded die, in 10 independent throws the probability that an even number turns up 5 times is twice the probability that even number turns up 4 times. Prepare the probabi

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Question 1109861: Consider a loaded die, in 10 independent throws the probability that an even number turns up 5 times is twice the probability that even number turns up 4 times. Prepare the probability distribution of the even numbers and compute the mean and variance. Verify your results as.
1) μ = E(x) = np 2) Var(x) = σ^2 = npq
Where p = prob. of even number q = 1-p
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Consider a loaded die, in 10 independent thr
ows the probability that an even number turns up 5 times is twice the probability that even number turns up 4 times. Prepare the probability distribution of the even numbers and compute the mean and variance. Verify your results as.
1) mean = E(x) = np
2) Var(x) = sigma^2 = npq
Where p = prob. of even number q = 1-p
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P(x = 5) = 10C5p^5q^5
P(x = 4) = 10C4p^4q^6
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Equation:
10C5p^5q^5 = 2[10C4p^4q^6]
p/q = [2*10C4]/10C5 = 5/3
----
p = (5/3)q
p + q = 1
p + (3/5)p = 1
8/5 p = 1
p = 5/8
1) mean = n*p = 10(5/8) = 50/8 = 6.25
2) Var = npq = 6.25(3/8) = 2.344
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Cheers,
Stan H.
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Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
using the binomial probability formula, we have
:
Probability (P) ( 5 evens out of 10 throws) = 10C5 * p^5 * (1-p)^5
:
P (4 evens out of 10 throws) = 10C4 ^ p^4 * (1-p)^6
:
we are given that the loaded die P is
:
10C5 * p^5 * (1-p)^5 = 2 * 10C4 ^ p^4 * (1-p)^6
:
252 * p^5 * (1-p)^5 = 2 * 210 * p^4 * (1-p)^6
:
252p = 420 - 420p
:
p = 420/672 = 0.625
:
Note p is the probability of throwing an even number
:
q = 1 - 0.625 = 0.375
:
mean(u) = (summation of k=0 to 10 of (P(k out of 10) * k))
mean(u) = ((0 * 0) + (0.0009 * 1) + (0.0069 * 2) + (0.0306 * 3) + (0.0891 * 4) + (0.1782 * 5) + (0.2475 * 6) + (0.2357 * 7) + (0.1473 * 8) + (0.0546 * 9) + (0.0090 * 10)) = 6.2486
variance = E(x^2) - E(x)^2
E(x^2) = ((0 * 0^2) + (0.0009 * 1^2) + (0.0069 * 2^2) + (0.0306 * 3^2) + (0.0891 * 4^2) + (0.1782 * 5^2) + (0.2475 * 6^2) + (0.2357 * 7^2) + (0.1473 * 8^2) + (0.0546 * 9^2) + (0.0090 * 10^2)) = 41.3936
variance = 41.3936 - 39.0450 = 2.3486
:
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calculated mean = 6.2486 and var = 2.3486
:
check results
:
u = 10 * 0.625 = 6.25
var = 10 * 0.625 * 0.375 = 2.34375
:
answer checks
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