Balls and Urns!

number of black balls : 2

number of white balls: 2

now, to have a black ball in the 'third' draw, we have only 4 possible ways:

1) WWB = $\frac { 2 }{ 4 }$ * $\frac { 1 }{ 3 }$ * $\frac { 2 }{ 2 }$ = $\frac{1}{6}$ .

2) WBB = $\frac { 2 }{ 4 } *\frac { 2 }{ 3 } *\frac { 3 }{ 4 }$ = $\frac { 1 }{ 4 }$

3)BWB = $\frac { 2 }{ 4 } *\frac { 2 }{ 5 } *\frac { 3 }{ 4 } = \frac { 3 }{ 20 }$

4)BBB = $\frac { 2 }{ 4 } *\frac { 3 }{ 5 } *\frac { 4 }{ 6 } =\frac { 1 }{ 5 } \\$

Adding these 4 fractions we get:

$\frac { 1 }{ 6 } +\frac { 1 }{ 4 } +\frac { 3 }{ 20 } +\frac { 1 }{ 5 } =\frac { 46 }{ 60 } =\frac { 23 }{ 30 }$

but $\frac { 23 }{ 30 } =\frac { \sqrt { a } }{ b } \Longrightarrow \frac { \sqrt { a } }{ b } =\frac { \sqrt { 529 } }{ 30 } \\ \therefore \quad a=529\\ \quad \quad b=\quad 30\\ thus,\quad a+\quad b\quad =\quad 529\quad +\quad 20\quad =\boxed{559}\quad$

simple use of tree diagram in probability

Write out the sample space: {W,W,W} {W,W,B} {W,B,W} {W,B,B} {B,W,W} {B,W,B} {B,B,W} {B,B,B} Since we want to know the Probability of the 3rd ball is drawn black, there are only 4 ways it can happen out of the 8 possible ways.

P{W,W,B}= 1/2 1/3 1 = 1/6 P{W,B,B}=1/2 2/3 3/4=1/4 P{B,W,B}=1/2 2/5 3/4=3/20 P{B,B,B}=1/2 3/5 4/6=1/5

Using the addition principle, sum all these up. You get 23/30 Since they want the answer in sqrt(a)/b, 23/30=sqrt(529)/30 Thus, a = 529 and b=30 and a+b=559