You have some coins in your pocket: 3 quarters, 2 dimes, and 4 pennies. What’s the probability that 3 coins taken randomly will be worth more than 50 cents?

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- Jul 9th 2010, 12:22 AMlifeunderwaterCoins Probability
You have some coins in your pocket: 3 quarters, 2 dimes, and 4 pennies. What’s the probability that 3 coins taken randomly will be worth more than 50 cents?

- Jul 9th 2010, 12:29 AMundefined
Note that if you have less than two quarters, it's not possible to get more than 50 cents; and if you have two or more quarters, then you're guaranteed to have more than 50 cents. And you can use $\displaystyle \displaystyle \binom{9}{3}$ as the denominator, just be careful to consider the individual coins as distinct; you could also use $\displaystyle \displaystyle \binom{9}{3,2,4}$ as the denominator if you count differently.

Edit: After more reflection, using $\displaystyle \displaystyle \binom{9}{3,2,4}$ is overly complicated, because not all combinations are equally likely, hence you would need to use weighted terms somehow.. - Jul 9th 2010, 12:35 AMlifeunderwater
Thank you for the quick response, but until now I don't how to calculate the probability for this question!!

- Jul 9th 2010, 12:42 AMundefined
Okay, so say you label the quarters A,B,C. Say you have exactly two quarters. Then the third coin can be any of 6 other coins. There are $\displaystyle \displaystyle \binom{3}{2}=3$ ways to have exactly two quarters. The other option is having exactly three quarters; there is only one way for that.

So $\displaystyle \displaystyle P=\frac{3\cdot6+1}{\displaystyle \binom{9}{3}}$ - Jul 9th 2010, 12:54 AMlifeunderwater
Sorry to disturb you again!

But I have another question that I asked not understandable replies.

The question is>>>

If you randomly take a positive integer and divide it by 23, what’s the probability that the remainder is 13?

Thank You very much in advance!!! - Jul 9th 2010, 01:05 AMundefined
My answer might be a little difficult to understand. :(

I believe the question is not well-formed, because there is no uniform probability distribution over the positive integers. What this means is, there is no sensible way to define "randomly take a positive integer" that I know of, if we require each positive integer to be equally likely to be chosen. It's because we are selecting an element from an infinite set (a countably infinite set).

What we can do though is think in terms of limits. So say we choose randomly an integer between 1 and n inclusive. Then as n goes to infinity, the probability that our number is disible by 13 approaches 1/23. This is because there are 23 remainders possible, in {0, 1, ... , 22}, and for large n they are all approximately equally likely. (And if n is a multiple of 23, then the probability is exactly 1/23.)