# Expected Value and Standard Error question?

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A letter is drawn 1,000 times, at random, from the word A R A B I A. There are two offers:

(A) You win \$1 if the number of A’s among the draws is 10 or more above the expected number.

(B) You win \$1 if the number of B’s among the draws is 10 or more above the expected number.

I know I should make box models for each, but how would they look like? Do each letter count for different numbers? So in this case, I wouldn’t be able to use the “short-cut” method for finding the S.D.? For the first one I’m thinking three boxes with “1” which counts for the letter “A” (since there’s 3 of 6 letters), so then “0” would count for the other numbers? Is this right? Thanks!

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• cinéphile

A letter is drawn 1,000 times, at random, from the word A R A B I A. There are two offers:

These are BOTH Binomial distributions and can be solved using a Binomial calculator …

http://stattrek.com/online-calculator/binomial.asp…

(A) You win \$1 if the number of A’s among the draws is 10 or more above the expected number.

expected = np = 1000(3/6) = 500

Now, using the Binomial calculator …

P(X ≥ (500+10) = 0.274

(B) You win \$1 if the number of B’s among the draws is 10 or more above the expected number.

expected = np = 1000(1/6) = 167

Now, using the Binomial calculator …

P(X ≥ (167+10) = 0.201

Your decision should be to choose game A since the probability of 10 or more above expected is 0.274

hope that helped