You randomly pick two numbers from 1 to 9 (including 1 and 9). a number that is chosen once could be chosen again. if at least one of the two numbers is odd and less than 5, what is the probability that their sum will be less than 5?

Here's how I look at it:

-- If one number is odd and less than 5, then it must be either 1 or 3.
The probability that it was chosen out of 9 numbers is  (2/9) .

-- There is a 50/50 chance that the first number is 1 .
If it is, and their sum is less than 5, then the second number is 1, 2, or  3 .
The probability that it was chosen out of 9 numbers is  (3/9) .

-- There is also a 50/50 chance that the first number is 3 .
If it is, and their sum is less than 5, then the second number is  1 .
The probability that it was chosen out of 9 numbers is  (1/9) .

The probability of the whole thing is

         (2/9)  times  [ (0.5 x 3/9)  +  (0.5 x 1/9) ]  = 

         (2/9)  times  [ (1.5/9)  +  (0.5/9) ]  =

         (2/9)  times  [ 2/9 ]  =  4/81  =  about  4.94 percent (rounded)



I had no confidence in my answer, so I wrote down all  81  possible combinations
of  2  numbers between  1  and  9,  and I checked them out.

I found  4  pairs that meet all the conditions.
They are:

           1, 1
           1, 2
           1, 3
           3, 1 .

This confirms the long, agonizing solution.  It might have been easier to do it
this way the first time ... list all the possibilities and see how many are successful ...
but working it out the way I did sure helped me refresh my confidence in some
probability stuff, and I hope it did the same for you.