Question 1186814
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There are A = 5 ways to get what we want, since there are 5 items in the set {1,2,3,4,5}. This set is repeated four times to represent the 4*5 = 20 balls.


For example, one possible desired outcome is selecting the four balls labeled "3". 
In other words, having this outcome: 3,3,3,3


There are n = 20 balls total and r = 4 of them are selected. Order doesn't matter, so we'll use the combination formula nCr


The notation *[Tex \Large _{n} C_{r}] is the same as C(n,r)


C(n, r) = (n!)/( r!*(n-r)! )
C(20, 4) = (20!)/( 4!*(20-4)! )
C(20, 4) = (20!)/( 4!*16! )
C(20, 4) = (20*19*18*17*16!)/( 4!*16! )
C(20, 4) = (20*19*18*17)/( 4! ) .....  the "16!" terms cancel
C(20, 4) = (20*19*18*17)/( 4*3*2*1 )
C(20, 4) = ( 116280 )/( 24 )
C(20, 4) = 4845
This represents the number of ways to select the four balls.
Let B = 4845



To summarize:
We have A = 5 ways to get what we want (having all four balls with the same number)
There are B = 4845 ways to select the four balls from a pool of 20


The probability of getting what we want is A/B = 5/4845 = <font color=red>1/969</font>. To reduce that fraction, I divided both parts by the GCF 5.


Use your calculator to find that 1/969 = 0.00103199174407 approximately. Rounding to the nearest millionth gets us <font color=red>0.001032</font>


Note: Rounding to the nearest millionths decimal place means the decimal answer will have 6 decimal places.  



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Answer as a fraction: <font color=red>1/969</font> (exact)
Answer as a decimal: <font color=red>0.001032</font> (approximate; rounded to 6 decimal places)
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