Respuesta :
If you know you have six books in total, then calculating the probability of the other two possible outcomes should be easy.
*P that 1 will have a red cover = 3/6 (which, if you simplify it, should be equal to 1/2)
*P that 1 will have a blue cover = 3/6 (which again, if you simplify it, should be equal to 1/2)
*P that 1 will have a red cover = 3/6 (which, if you simplify it, should be equal to 1/2)
*P that 1 will have a blue cover = 3/6 (which again, if you simplify it, should be equal to 1/2)
I think it has to be this, but please think about it. It's not so simple:
6 books, 3 red and 3 blue. The red books are indistinguishable from each other. Same for the blue ones.
Then, there are these many ways of putting them on the shelf (combinations with repetition):
6!/(3!3!)=6*5*4*3!/(3! 3*2*1) = 5*4 = 20 ways!
Now in order to get one red and one blue: there are only 2 ways:
RB or BR,
So the probability should be:
P = 2/20 = 1/10.
6 books, 3 red and 3 blue. The red books are indistinguishable from each other. Same for the blue ones.
Then, there are these many ways of putting them on the shelf (combinations with repetition):
6!/(3!3!)=6*5*4*3!/(3! 3*2*1) = 5*4 = 20 ways!
Now in order to get one red and one blue: there are only 2 ways:
RB or BR,
So the probability should be:
P = 2/20 = 1/10.