Probability of a coin stack being greater than a value? What's wrong with my reasoning?

Basic probability question.

Consider a pile of 9 coins where each could either be 1 cent or 10 cents and the distribution of the coin combinations is uniform. Knowing that the upper 4 coins are all 10 cents, what is the probability that the total value is greater than 50 cents?

My reasoning was simply that we have 5 coins leftover and we needs at least 10 more cents to get to 50 cents. We have a total of $2^5$ combinations for the remaining 5 coins. Our sample space size is $2^5-1$ because the only way which wouldn't work out is if we get all pennies. So the probability should be $\frac{2^5-1}{2^5}$

What's wrong here?

• Can you tell the answer? – Akash Roy Apr 19 '18 at 1:39
• @AkashRoy Nevermind, the solution manual says it was a typo – Goldname Apr 19 '18 at 1:41
• Actually I thought your reasoning was correct, that is why I was asking for the answer lol. It is one of the basic probability questions – Akash Roy Apr 19 '18 at 1:44

1. Each of the 9 coins was pulled out of a large vat w an equal number of pennies and dimes, so that with each pull, the probability of a coin being a dime is 50% and that the values of the coins pulled out are mutually independent of one another. (That is quite different from getting a pile of $n$ coins and where half are pennies and half are dimes.)
$\frac{2^5-1}{2^5}=\frac{31}{32}$ Is indeed the correct answer.