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Question

In throwing a fair die, following are the probabilities of getting each face.
1 - k
2 - 2k
3 - 2k
4 - 3k
5 - 3k2
6 - 7k2+k
Expected value of the outcome = ___

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Solution

This is a continuation of the previous problem we solved. We want to find the expected value of outcome. It would be given by 1.p(1) + 2.p(2) + 3.p(3) + 4. p(4) + 5. p(5) + 6. p(6)
where p(i) is the probability of getting face ‘i’
expected value=(1×k)+(2×2k)+(3×2k)+(4×3k)+(5×3k2)+(6(7k2+k))=k+4k+6k+12k+15k2+42k2+6k=29k+57k2
If we add all the probabilities we should get 1, because the probability of sample space is 1.
k+2k+2k+ 3k+3k2+(7k^2 + k) = 1
10k2 + 9k - 1 = 0
k = -2 or k=110
Since probability can’t be negative k=110
Expectation or mean value =29×110+57×(110)2
=347100
= 3.47

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