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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?

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  • $\begingroup$ Can you tell the answer? $\endgroup$ – Akash Roy Apr 19 '18 at 1:39
  • $\begingroup$ @AkashRoy Nevermind, the solution manual says it was a typo $\endgroup$ – Goldname Apr 19 '18 at 1:41
  • $\begingroup$ 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 $\endgroup$ – Akash Roy Apr 19 '18 at 1:44
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Well the question does seem to be rather poorly worded. In probability it is important to be precise on what is being conditioned on and the wording of the question was not. I took it to mean that

  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.)

  2. As you pulled each coin out of the vat you looked at it, and after 4 pulls you had 4 dimes.

If that is the case then yes your reasoning is correct.

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$\frac{2^5-1}{2^5}=\frac{31}{32}$ Is indeed the correct answer.

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