There are four balls of different colours and four boxes of the same colour . THe balls have to be put in the boxes of matching colours . Find the probability that all the four balls are not put in the right box .
Hello mathaddictImagine that the boxes are arranged in a line in front of you, and you've then got to arrange the balls, one in each box so that the colours match. There's only one possible way to do this.
Now figure out how many different ways there are altogether of putting the balls into the boxes, one in each. There are 4 ways of putting a ball into the first box, 3 ways for the second, and so on. So how many arrangements are possible altogether? Only one of these is correct. So what's the probability of that? So what's the probability that it's not correct?
Grandad
You need the concept of a derangement. Read these:
Derangement -- from Wolfram MathWorld
Derangements and Generalizations
Derangement - Wikipedia, the free encyclopedia
Robert M. Dickau - Derangements
etc.
If all four balls are not put in the right box, I understand that as meaning that all four balls are put in a wrong box. That's a derangement. Have you interpreted it as all four balls are not put in the right box therefore only some or none have been in the right box ....?Originally Posted by Grandad
(This is what I hate about these sorts of problems - for me I often find ambiguity in the wording .....)
Hello Mr FI agree that the wording of these problems can be a nightmare (I think we've had this sort of discussion recently in another context!), but I take it that the proposition "All the four balls are not put in the right box" is the negation of the proposition "All the four balls are put in the right box", which is logically equivalent to "There exists a ball that is not put in the right box". You may be right, but my interpretation has the advantage that it is much simpler to solve than yours - and I think mathaddict's solution of 23/24 is correct!
Grandad