Question 1120141
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Let me reformulate the problem in more precise / (more correct) form:


<pre>
    Ten balls numbered 1 to 10 are in a jar. Jack reaches into the jar and randomly removes two of the balls. 
    What is the probability that the sum of the two numbers on the balls removed is even?
</pre> 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This phrase "Then Jill reaches into the jar and randomly removes a different ball." of the original condition is not necessary . . . 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;He FOR SURE will remove a different ball, since he does not replace the first one . . . 



<U>Solution</U>


<pre>
The number of all combinations of pairs of the balls is  {{{C[10]^2}}} = {{{(10*9)/2}}} = 45.


The favorable events happen when both the balls in a pair are even,  <U>OR</U> when both the balls in a pair are odd.



The even numbers from 1 to 10 are  2, 4, 6, 8  and  10.

The number of combinations where both balls are even is {{{C[5]^2}}} = {{{(5*4)/2}}} = 10.



The odd numbers from 1 to 9 are  1, 3, 5, 7, and 9.

The number of combinations where both balls are odd is {{{C[5]^2}}} = {{{(5*4)/2}}} = 10.


Thus of 45 possible outcomes, only 10 + 10 = 20 are favorable.


Hence, the probability under the question is  {{{20/45}}} = {{{4/9}}}.
</pre>

Solved.