SOLUTION: A sack contains 7 dimes, 5 nickels, 10 quarters. 8 coins are drawn at random, what is the probability of getting 4 dimes, 3 nickels, and 1 quarter?

Question 970783: A sack contains 7 dimes, 5 nickels, 10 quarters. 8 coins are drawn at random, what is the probability of getting 4 dimes, 3 nickels, and 1 quarter?

Found 2 solutions by jim_thompson5910, Boreal:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I am using the combination formula n C r = (n!)/(r!*(n-r)!)

So for example, 7 C 4 = (7!)/(4!*(7-4)!) = (7!)/(4!*3!) = 5040/(24*6) = 35

There are 7+5+10 = 22 coins total

Number of ways to draw 4 dimes = 7 C 4 = 35
Number of ways to draw 3 nickels = 5 C 3 = 10
Number of ways to draw 1 quarter = 10 C 1 = 10

Number of ways to draw 4 dimes, 3 nickels, and 1 quarter = 35*10*10 = 3,500

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Number of ways to draw 8 coins = 22 C 8 = 319,770

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Summary:

Number of ways to draw 4 dimes, 3 nickels, and 1 quarter = 3,500

Number of ways to draw 8 coins = 319,770

Divide the two values: (3,500)/(319,770) = 0.01094536698251 (this probability is approximate)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
30 coins
Look at 4 dimes, 3 nickels and 1 quarter as if they were done in that order.
(7/30)*(6/29*(5/28)*(4/27)*(5/26)*(4/25)*(3/24)*(10/23).
Now suppose they were done in opposite order, quarters first.
(10/30)*(5/29)*(4/28)*(3/27)*(7/26)*(6/25)*(5/24)*(4/23).
Note that the denominators are the same. So are the numerators. How many different ways are these formed?
Dimes 7C4=35 Nickels 5C3=10 Quarters=10 The product of these 3 is 3500.
The numerator product is 504000*3500
The denominator product is 2.36 * E11; divide top and bottom
Probability=0.0075
While it seems low, the probability of getting exactly 1 quarter is
(10/30)*(20/29)*(19/28)*(18/27)*(17/26)*(16/25)*(15/24)*(14/23) *10
only 0.16
For 3 nickels out of 30 coins when there are only 5, it is lower, and that has to be multiplied by 0.16.
Then we still have to deal with dimes.

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