HELP!

a)

There are 8 ways to choose 2 balls out of 8 total.

There are 3 ways to choose 2 black balls out of 3 black.

The probability is 3/8.

 

b)

The probability of drawing two black balls on the first draw is 3/8​, as you already calculated.

If you don't draw two black balls on the first draw, then you must draw at least one white ball. When you put the white ball back and draw again, you have a 82​ chance of drawing another white ball, and a 83​ chance of drawing a black ball.

Therefore, the probability of drawing two black balls, including the possibility of drawing two black balls on the first draw, is:

(3/8) + (5/8) * (3/8) = 9/16

 

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     \(\frac{15}{56}+\frac{15}{196}+\frac{15}{196}+\frac{15}{392}= \frac{45}{98}\)