There are 24 blue marbles and 1 red marbles in a bag and a marble is picked at random. Within the bag, all the blue marbles are picked out until a single marble is left. What are the chances that the marble left in the bag is the red one?
+128483
Look at this problem in a more simple manner....suppose that there are 4 books on a shelf, but we want one of the books to always be the last one.
There are 4! arrangements possible = 4! = 24........ but we're only interested in the possible arangements of three of them, since the 4th book is always "given" in the order.
And the possible arrangements of the other three books is 3! = 6
So, the arrangements we're interested in / the total arrangemets = 6/ 24 = 25% chance that the book we want to come last, in fact, will come last.
So having the possible total arrangements of 25 marbles is 25!.....but since the red one always comes last.....we can arrange the other 24 in 24! ways.
So the probability that the red one comes last is just 24! / 25! = .04= 4%
+128483
Look at this problem in a more simple manner....suppose that there are 4 books on a shelf, but we want one of the books to always be the last one.
There are 4! arrangements possible = 4! = 24........ but we're only interested in the possible arangements of three of them, since the 4th book is always "given" in the order.
And the possible arrangements of the other three books is 3! = 6
So, the arrangements we're interested in / the total arrangemets = 6/ 24 = 25% chance that the book we want to come last, in fact, will come last.
So having the possible total arrangements of 25 marbles is 25!.....but since the red one always comes last.....we can arrange the other 24 in 24! ways.
So the probability that the red one comes last is just 24! / 25! = .04= 4%
