A box contains 5 white balls, 3 red balls and 2 green balls.
The balls are taken from box. With replacement and taking a total of 3 balls,
P(all balls have different colour)=(5/10)x(3/10)x(2/10) x 3! =0.18
This answer is the same as at the back of the book. My explanation is that a white, red or green ball must be taken and there are 3! ways of arranging the 3 scenarios.
Is this correct????
Without replacement and taking a total of 5 balls
My wrong method:
P(exactly 2 balls are white) = (5/10)x(4/9)x(5/8)x(4/7)x(3/6) x 5!
This probability cannot be correct since it is above 1. What is the loophole that I did not see???
I dont need the answer to this question (I can solve using counting total number of cases) but I want to know the loophole in my thinking.