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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her [#permalink]
19 Dec 2004, 00:51
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Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?
Please post answer and step by step explanation on how to solve. Thanks in advance!
Suppose karen drew all dimes and nickels in 9 draws, then she is will left with 12 coins. the remaining coins are 6 quarters and 6 other two (could be any combinations of dimes and nickles). now if karen draws any coin, now she has a chance of 50% getting quarters.
I got the same answer as MA, but here is how I approached it. Since you want at least a 50% chance of getting a quarter, you want to be guaranteed at least a 6/12 chance. In order to be sure you would have to create the most difficult conditions to prevent you from getting a quarter, which would be if you drew non quarters as much as possible.
You start out with a 6/21 chance of drawing a quarter. You want a 6/12 chance, so you need to draw at least (21-12) or 9 coints to be sure. I hope this explanation helps.
I also agree with 9 as toddmartin explained. Saurabh, what you said here:
(15/21)^ n < .5
assumes that the coins are withdrawn with replacement and it would instead be the answer to the following question: Given the above coin breakdown/value and assuming that the coins are put back in her pocket after each draw, how many times will Karen have to draw a coin so that the % of not getting a quarter is less than 50%
Hi Paul the question dosen't asy anything about the with or without replacement. So i don't think any assumption is valid. May be we have to come up with the assumption after seeing the answer choices. Moreover if you re-rad your last statement:
"a coin so that the % of not getting a quarter is less than 50%"
It also tells you that" % of getting a quater is more than 50%"
both the statement are just NEGATION of each other.
well I migh be missing something so please do let me know about my apporach.
saurabh, I agree that statements are just negation of each other but what I mean is that the way you set up the formula is assuming that it is done with replacement.
When you are saying (15/21)^n, the mere fact of having "n" as exponent means that replacement is allowed.
Consider this simpler example. Let's say you have 3 coins with different values, after 3 draws, how many outcomes are possible:
with replacement: 3^3 --> 3^n
without replacement: 3!