Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x+y) is divisible by 4?

Explanation

Group the first 10 numbers by their remainder modulo 4: R_0 = \{4,8\}, R_1 = \{1,5,9\}, R_2 = \{2,6,10\}, R_3 = \{3,7\}. The sum is divisible by 4 if we choose two from R_0 (1 way), two from R_2 (3 ways), or one each from R_1 and R_3 (3 \times 2 = 6 ways). Total favorable ways = 1+3+6=10. Total ways to choose 2 numbers is C(10,2) = 45. Probability = \frac{10}{45} = \frac{2}{9}.

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