1. 2 fruits can be selected out of 6 fruits in ⁶C₂ ways. Also 1 apple and 1 mango can be selected in 3 X 3 ways. So, the required probability = 9 / ⁶C₂ = 3/5.

2. Let A = Event that a sum of 5 occurs, B = event that a sum of 7 occurs and C = event that neither a sum of 5 nor a sum of 7 occurs.

So, P(A) = 4/36 = 1/9, P(B) = 6/36 = 1/6 and P(C) = 26/36 = 13/18

Thus P(A occurs before B) = P [ A or (C∩A) or (C∩C∩A) or ......] = P(A) + P(C) P(A) + P(C)² P(A) + .......

 = (1/9) + (13/18)(1/9) + (13/18)² (1/9) + ...... = (1/9) [ 1 + (13/18) + (13/18)² + ......]

= (1/9) / 1 - (13/18) = 2/5