Question 1121710
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Let's see what a typed tree diagram looks like....<br><pre>

          O   3 odds makes an ODD sum; probability (5/9)(4/8)(3/7) = 60/504
        /
      O - E   2 odds 1 even makes an EVEN sum; probability (5/9)(4/8)(4/7) = 80/504
    /
  O - E - O   2 odds 1 even makes an EVEN sum; probability (5/9)(4/8)(4/7) = 80/504
        \
          E   1 odd 2 evens makes an ODD sum; probability (5/9)(4/8)(3/7) = 60/504

          E   3 evens makes an EVEN sum; probability (4/9)(3/8)(2/7) = 24/504
        /
      E - O   2 evens 1 odd makes an ODD sum; probability (4/9)(3/8)(5/7) = 60/504
    /
  E - O - E   2 evens 1 odd makes an ODD sum; probability (4/9)(5/8)(3/7) = 60/504
        \
          O   1 even 2 odds makes an EVEN sum; probability (4/9)(5/8)(4/7) = 80/504</pre>
The probability of an odd sum is (60+60+60+60)/504 = 240/504 = 10/21<br>
The probability of an even sum is (80+80+24+80)/504 = 264/504 = 11/21<br>
Tree diagrams are useful in solving relatively simple probability problems; and they are a good aid in visualizing how the answer is obtained.  But they become extremely awkward very quickly as the problem get more complicated.<br>
A much faster path to the answer is using basic combinatorics.<br>
To get an odd sum when drawing 3 of the 9 marbles, you have to choose either 3 of the 5 odd and 0 of the 4 even, or 1 of the 5 odd and 2 of the 4 even.  The probability of doing that is<br>
{{{(C(5,3)*C(4,0)+C(5,1)*C(4,2))/C(9,3)
 = (10*1+5*6)/84 = 40/84 = 10/21}}}<br>
To get an even sum when drawing 3 of the 9 marbles, you have to choose either 0 of the 5 odd and 3 of the 4 even, or 2 of the 5 odd and 1 of the 4 even.  The probability of doing that is<br>
{{{(C(5,0)*C(4,3)+C(5,2)*C(4,1))/C(9,3)
 = (1*4+10*4)/84 = 44/84 = 11/21}}}<br>