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Probability
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A box contains six $1 bills, three $5 bills, two $10 bills, and seven $20 bills. What is the expectation if one bill is selected?
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Well there are 18 bills in total... \[\large P(\text{1 dollar bill}) = \frac{6}{18}\] \[\large P(\text{5 dollar bill}) = \frac{3}{18}\] \[\large P(\text{10 dollar bill}) = \frac{2}{18}\] \[\large P(\text{20 dollar bill}) = \frac{7}{18}\] So it looks like the bigger probability of 7/18 would say that we have a better chance of pulling out a 20 dollar bill from this box
6*1 = 6 value 3*5 = 15 2*10= 20 7*20= 140 ---- total = 181 /18 10.0555555555556 I think 1 draw has a $10.05555 expectation
for a specific dollar there is a different outcome!!
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