Assuming 01, 02, etc. are considered two digit numbers:



P(even number formed from random selection of two digits, without replacement) = P(even number chosen for ten's digit AND even number chosen for one's digit) +

P(odd number chosend for ten's digit AND even number chosen for one's digit)



= (4/7)(3/6) + (3/7)(4/6)

=  12/42 + 12/42

=  24/42

=  



This matches the proportion of even digits in the set to the total number of digits in the set.  Coincidence?





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Expanding the problem to 3 digit numbers you will find P(even) = 4/7 

Expanding to 4 digit numbers also gives P(even) = 4/7



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The above cases assume 0 as a leading digit counts, if 0 is disallowed as a leading digit then for two-digit cases P(even) = (3/7)(3/6)+(3/7)(4/6) = 1/2.

  

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