If a is to be chosen at random from the integers between 1

Using the old school method

SET A : 1 , 2 , 3 , 4, 5

SET B : 6 , 7, 8, 9 ,10

Now for the numbers for be even ( a + b ) we can have either both numbers are even or both numbers are odd

Keeping SET B number constant 6 and varying A ( 1 , 2 ,3 ,4 ,5 )

1 + 6 => 7 ( Odd ) , 2 + 6 => 8 (Even) , 3 + 6 => 9 ( Odd ) , 4 + 6 => 10 (Even) , 5 + 6 => 11 ( Odd )

So using above probability of 2/5 will be even .

Keeping SET B number constant 7 and varying A ( 1 , 2 ,3 ,4 ,5 )

1 + 7 => 8 ( Even) , 2 + 7 => 9 (Odd) , 3 + 7 => 10 (Even) , 4 + 7 => 11 (Odd) , 5 + 7 => 12 ( Even)

So using above probability of 3/5 will be even for set B .

Hence P(A even , B even ) + P ( A odd , B odd ) = 3/5 * 2 /5 *2 = 12 / 25

Reverse method:

1 - P (sum is odd)

3/5 * 3/5 + 2/5 * 2/5 = 13/25

1 - 13/25=12/25

C

In order for (a + b) to be even we must have either both a and b are odd or both a and b are even.

P(a and b are both odd) = 3/5 x 2/5 = 6/25

P(a and b are both even) = 2/5 x 3/5 = 6/25

So P(a + b = even) = 6/25 + 6/25 =12/25.

Answer: C

Total possibilities = 5*5 = 25

Favorable out come =
(1 & 7), (1&9), (2&6), (2&8), (2&10), (3&7), (3&9), (4&6), (4&8), (4&10), (5&7), (5&9) -> 12

12/25

Answer C

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