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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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