Question 1122439


a)  P(blue and blue) = (5/8)*(4/7) = 20/56 = {{{ highlight(5/14) }}}, or about 0.3571
b)  P(blue | blue) = P(blue and blue) / P( at least one blue) = (20/56) / (50/56)   = 20/50 = {{{ highlight(2/5) }}} or 0.40.

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Here is the probability distribution for the four raw outcomes:

1.  P(B and B) =  20/56
2.  P(R and B) =  15/56
3.  P(B and R) =  15/56
4.  P(R and R) =   6/56

Part(b) is equivalent to asking "once you know you are in one of the cases 1...3, what is the probability you have case 1?"    There are  20+15+15 = 50 possible outcomes in cases 1...3, and 20 of them are case 1.