Question 950234
Let's call the number of blue balls {{{X}}} and the number of black balls {{{Y}}}.
The probability of choosing the first blue ball would then be,
{{{P(B1)=X/(X+Y)}}}
Similarly for the second through the fifth,
{{{P(B2)=(X-1)/(X+Y-1)}}}
{{{P(B3)=(X-2)/(X+Y-2)}}}
{{{P(B4)=(X-3)/(X+Y-3)}}}
{{{P(B5)=(X-4)/(X+Y-4)}}}
So then the probability of getting all 5 is,
{{{P=P(B1)*P(B2)*P(B3)*P(B4)*P(B5)}}}
{{{P=(X(X-1)(X-2)(X-3)(X-4))/((X+Y)(X+Y-1)(X+Y-2)(X+Y-3)(X+Y-4))}}}
That seems pretty daunting, let's take a guess and you'll see that it's not really.
Let's assume {{{Y=1}}}, then the equation becomes,
{{{P=(X(X-1)(X-2)(X-3)(X-4))/((X+1)X(X-1)(X-2)(X-3))}}}
{{{(X-4)/(X+1)=1/2}}}
{{{2(X-4)=X+1}}}
{{{2X-8=X+1}}}
{{{X=9}}}
Is that the only answer?
How about {{{Y=2}}}, then the equation becomes,
{{{P=(X(X-1)(X-2)(X-3)(X-4))/((X+2)(X+1)X(X-1)(X-2))}}}
Now it becomes more daunting.
{{{((X-3)(X-4))/((X+2)(X+1))=1/2}}}
{{{2(X^2-7X+12)=X^2+3x+2}}}
{{{X^2-17X-22=0}}}
This equation doesn't have an integer solution.
So there's one solution but I'm not sure it's unique.