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?
2 answers:
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.
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 )
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