SOLUTION: Nine marbles numbered from 1 to 9 are placed in a bag and three are drawn out at random without replacement. By first drawing a suitable tree diagram, find the probability that the

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Question 1121710: Nine marbles numbered from 1 to 9 are placed in a bag and three are drawn out at random without replacement. By first drawing a suitable tree diagram, find the probability that the sum of the numbers on the marbles is odd.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Let's see what a typed tree diagram looks like....

          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

The probability of an odd sum is (60+60+60+60)/504 = 240/504 = 10/21

The probability of an even sum is (80+80+24+80)/504 = 264/504 = 11/21

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.

A much faster path to the answer is using basic combinatorics.

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



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