Start with 5 white and 3 black

1) P(BB) \(= \frac{3}{8}*\frac{2}{7}=\frac{6}{56}\)

P(BW) = \(\frac{3}{8}*\frac{5}{7}=\frac{15}{56}\)

throw back the white ball then P(drawing black) \(=\frac{2}{7}\)

2) So Prob of getting BWB \(\frac{15}{56}*\frac{2}{7}= \frac{15}{196}\)

3) The prob of getting WBB is the same as BWB

P(WW) \(\frac{5}{8}*\frac{4}{7}=\frac{20}{56}\)

Throw both balls back and the prob of then getting BB is 6/56

4) P( WWBB) = (20/56)*(6/56) = \frac{15}{392}

So the prob of ending up with 2 black balls is \(\frac{15}{56}+\frac{15}{196}+\frac{15}{196}+\frac{15}{392}= \frac{45}{98}\)

.