Question 534682
Consider a large bag of coins, consisting of only quarters ($0.25), dimes ($0.10), and nickels ($0.05). 40% of the coins are nickels, 35% are dimes, and 25% are quarters.
<pre>

You must draw either 5 quarters of 4 quarters and a nickel or dime.

Suppose there are 100N coins consisting of 40N nickels, 35N dimes, and 25N
quarters.  Note: there are 75N non-quarters

The number of successes equals

the number of ways to draw 5 quarters is C(25N,5)

plus

the number of ways to draw 4 quarters and 1 non-quarter is C(25N,4)·(75N)

So the number of successes is

C(25N,5) + C(25N,4)·(75N)

{{{(  (25N)(25N-1)(25N-2)(25N-3)(25N-4))/5!}}} + {{{expr(((25N)(25N-1)(25N-2)(25N-3))/4!)*75N}}}  

Put the (75N) in the numerator of the second fraction

{{{(  (25N)(25N-1)(25N-2)(25N-3)(25N-4))/5!}}} + {{{expr(((25N)(25N-1)(25N-2)(25N-3)(75N))/4!)}}} 

We get a least common denominator of 5! by multiplying the second fraction
by {{{5/5}}}

{{{(  (25N)(25N-1)(25N-2)(25N-3)(25N-4))/5!}}} + {{{expr(((25N)(25N-1)(25N-2)(25N-3)(75N)5)/(4!5))}}} 

{{{(  (25N)(25N-1)(25N-2)(25N-3)(25N-4))/5!}}} + {{{expr(((25N)(25N-1)(25N-2)(25N-3)(375N))/5!)}}}

{{{(  (25N)(25N-1)(25N-2)(25N-3)(25N-4) + (25N)(25N-1)(25N-2)(25N-3)(375N))/5!)}}}

Factor the numerator:

{{{  (25N)(25N-1)(25N-2)(25N-3)*((25N-4) + (375N))/5!}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*(25N-4 + 375N)/5!}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*(400N-4)/5!}}}


The number of possible ways is the number of ways to select any 
5 coins from the 100N coins:

C(100N,5) = {{{((100N)(100N-1)(100N-2)(100N-3)(100N-4))/5!}}}

So the probability is

{{{  (25N)(25N-1)(25N-2)(25N-3)*(400N-4)/5!}}}÷{{{((100N)(100N-1)(100N-2)(100N-3)(100N-4))/5!}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*(400N-4)/5!}}}×{{{5!/((100N)(100N-1)(100N-2)(100N-3)(100N-4))}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*(400N-4)/cross(5!)}}}×{{{cross(5!)/((100N)(100N-1)(100N-2)(100N-3)(100N-4))}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*(400N-4)/((100N)(100N-1)(100N-2)(100N-3)(100N-4))}}}

{{{  (25N)(25N-1)(25N-2)(25N-3)*4(100N-1)/((100N)(100N-1)(100N-2)(100N-3)(100N-4))}}}

{{{
(expr(25N/100N))
(expr((25N-1)/(100N-1)))
(expr((25N-2)/(100N-2)))
(expr((25N-3)/(100N-3)))
(expr((400N-4)/(100N-4))) }}}


The first fraction is {{{(25N)/(100N)}}}, exactly {{{1/4}}}.  
If N is large the next three are also very nearly {{{(25N)/(100N)}}}
or {{{1/4}}}, and the last fraction is very nearly {{{(400N)/(100N)}}}, or 4.
So that is very nearly

{{{(1/4)(1/4)(1/4)(1/4)(4)}}}

{{{(1/4)(1/4)(1/4)(1/cross(4))(cross(4))}}}

{{{(1/4)^3}}}

{{{1/64}}}

That's not exact, but it is very close if N is large.

Edwin</pre>