Question 1152183
<pre>

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
=  {{{ highlight( 4/7 ) }}}

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


-----
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

-----
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.