SOLUTION: 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. Random

Algebra ->  Probability-and-statistics -> SOLUTION: 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. Random      Log On


   

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.
Randomly select 5 coins from the bag. What is the probability that your selection is worth at least $1.00? (Note that the only way to make $1 with 5 coins is with four (4) quarters and one (1) other coin.)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
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.

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)

%28++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2825N-4%29%29%2F5%21 + expr%28%28%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%29%2F4%21%29%2A75N  

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

%28++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2825N-4%29%29%2F5%21 + expr%28%28%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2875N%29%29%2F4%21%29 

We get a least common denominator of 5! by multiplying the second fraction
by 5%2F5

%28++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2825N-4%29%29%2F5%21 + expr%28%28%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2875N%295%29%2F%284%215%29%29 

%28++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2825N-4%29%29%2F5%21 + expr%28%28%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%28375N%29%29%2F5%21%29



Factor the numerator:

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%28%2825N-4%29+%2B+%28375N%29%29%2F5%21

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%2825N-4+%2B+375N%29%2F5%21

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%28400N-4%29%2F5%21


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

C(100N,5) = %28%28100N%29%28100N-1%29%28100N-2%29%28100N-3%29%28100N-4%29%29%2F5%21

So the probability is

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%28400N-4%29%2F5%21÷%28%28100N%29%28100N-1%29%28100N-2%29%28100N-3%29%28100N-4%29%29%2F5%21

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%28400N-4%29%2F5%21×5%21%2F%28%28100N%29%28100N-1%29%28100N-2%29%28100N-3%29%28100N-4%29%29

++%2825N%29%2825N-1%29%2825N-2%29%2825N-3%29%2A%28400N-4%29%2Fcross%285%21%29×cross%285%21%29%2F%28%28100N%29%28100N-1%29%28100N-2%29%28100N-3%29%28100N-4%29%29








The first fraction is %2825N%29%2F%28100N%29, exactly 1%2F4.  
If N is large the next three are also very nearly %2825N%29%2F%28100N%29
or 1%2F4, and the last fraction is very nearly %28400N%29%2F%28100N%29, or 4.
So that is very nearly

%281%2F4%29%281%2F4%29%281%2F4%29%281%2F4%29%284%29

%281%2F4%29%281%2F4%29%281%2F4%29%281%2Fcross%284%29%29%28cross%284%29%29

%281%2F4%29%5E3

1%2F64

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

Edwin