7 coins - probability over a dollar

by bloynoys
Tags: coins, dollar, probability
 P: 827 In this type of questions priors must always be stated otherwise you produce something like the "two envelopes problem" as an extreme case.
 Sci Advisor P: 1,167 Maybe you can simplify your calculations a bit by using the fact that you need at least two quarters to have a total to add up to $1.00 . That leaves one option : 5 dimes. Then you can have 3 quarters , and 3^4=81 cases, or 4 quarters , and certainty. So your winning cases of total larger than$1.00 are (using _exact_ number of quarters): 0 quarters: 0 ways , 3^7 non-ways 1 quarter : 0 ways 3^6 non-ways. 2 quarters: 1 way ( 5 10's) 3 quarters: only complicated cases 4 quarters: 3^3 ways 5 quarters: 3^2 ways, etc. So it comes down to having four non-quarter coins adding up to $0.25-or-more. P: 506  Quote by Bacle2 ... So it comes down to having four non-quarter coins adding up to$0.25-or-more.
This step could be done using with generating function, ie expand $(x+x^5+x^{10})^4$
Mentor
P: 13,609
 Quote by Bacle2 2 quarters: 1 way ( 5 10's)
That's not a solution because the total is exactly \$1.00. The OP asks for "probability that a person has over a dollar in coins."

 Quote by bpet This step could be done using with generating function, ie expand $(x+x^5+x^{10})^4$
Even easier, don't break the problem down at all. Just use the generating function to form $(x+x^5+x^{10}+x^{25})^7$ (expanded) and be done with it. Generating functions are magical.

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