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?
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 as the denominator, just be careful to consider the individual coins as distinct; you could also use as the denominator if you count differently.
Edit: After more reflection, using is overly complicated, because not all combinations are equally likely, hence you would need to use weighted terms somehow..
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!!!
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.)