SOLUTION: two digits are randomly selected without repition; compute the probability that their sum is is odd

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Question 475441: two digits are randomly selected without repition; compute the probability that their sum is is odd
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

EVEN + EVEN = EVEN
EVEN + ODD = ODD
ODD + EVEN = ODD
ODD + ODD = EVEN

So to have their sum odd, one has to be odd and the other even.

The digits are

0,1,2,3,4,5,6,7,8,9

5 are ODD and 5 are EVEN.

After we have chosen the first digit, there are only 9 digits 
left to choose from:

P[(odd 1st AND even 2nd) OR (even 1st and odd 2nd)] =

AND indicates multiplication and OR indicates ADDITION:

  P(odd 1st)×P(even 2nd) + P(even 1st)×P(odd 2nd) =

             (5/10)(5/9) + (5/10)(5/9) =

              (1/2)(5/9) + (1/2)(5/9) =

                   5/18  + 5/18 =

                       10/18 =

                        5/9

Edwin
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