A bowl contains 10 chips, of which 8 are marked $2 each and 2 are marked $5 each. Let a person choose, at random and with replacement, three chips from this bowl. If the person receive the sum of the resulting amounts, find his expectation and standard deviation of the amount he received. What is the probability the amount he will receive is within (+-) 1 standard deviation from the mean? E(X) = 2.6 Then I have no more idea.

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A bowl contains 10 chips, of which 8 are marked $2 each and 2 are marked $5 each. Let a person choose, at random and with replacement, three chips from this bowl. If the person receive the sum of the resulting amounts, find his expectation and standard deviation of the amount he received. What is the probability the amount he will receive is within (+-) 1 standard deviation from the mean? E(X) = 2.6 Then I have no more idea.

Mathematics
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Ah 20 more mins to submission due date.
Guess I will go up to my teacher office and submit now :(.
This can be solved using the binomial distribution. On each draw the probability of $5 is 8/10 = 0.8 and the probability of $2 is 2/10 = 0.2 The probability of getting 3 of $5 chips in three draws is given by: \[\large P(3\ $5)=3C3\times0.8^{3}\times0.2^{0}=0.512\] By similar calculation, we find that: P(2 $5) = 0.384 P(1 $5) = 0.096 However there are only two $2 chips, therefore P(0 $5) = 0 The expected value E is given by: \[\large E=$15\times0.512+$12\times0.384+$9\times0.096+$6\times0=$13.15\]

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O.!, not 2.6?
$2.60 is far too low for the expected amount.
That's my friend answer. hmm
Alright I go submit now. Thanks @kropot72 do continue the rest of the question so I could learn.
I have to log out now. Hopefully someone can continue with this question.
Thank you so much @kropot72
You're welcome :)

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