Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3The probability that x1 + x2 + x3 is odd, isThe probability that x1, x2, x3 are in an arithmetic progression, is

For, x1 + x2 + x3 = oddCase 1 : One odd, two even(OEE) or (EOE) or (EEO)Total number of ways = 2 x 2 x 3 + 1 x 3 x 3 + 1 x 2 x 4 = 29.Case II: All three oddNumber of ways = 2 x 3 x 4 = 24∴ Favorable ways = 53∴ Required probability=533×5×7=53105Here, 2x2=x1+x3⇒ x1+x3= even So, either x1 and x3 are both odd or both even.Hence, number of favorable ways=2C1⋅4C1+1C1⋅3C1=11.Therefore, required probability is 11105.