You have six books on yourself. you notice that three of them have red covers, and three of them have blue covers, if you choose 2 books at random what is the probability that one will have a red cover and one will have a blue cover?

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)

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.

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.