Some balls are placed in a large box; the balls include one ball marke

Well ok Here goes nothing

the balls are marked in the following fashion: 1 - A ; 2 - B ; 3 - C ; ... so total number of balls that are there in the bag are 351 ( sum of AP from 1,2,3....,26 )

we need to exceed 1/30 as a probability ( say 0.033 for decimal crunchers )

ST 1 : l o u s y

here L is the least numbered one and Y is the most numbered one. If L lies in the constraints then all other letters lie as well. so lets se for L
there are 12 balls for L ( 12th letter )

so probability = 12/351 = 0.341. we can see that it exceeds 0.33, and all other letters will exceed as well, as the numerator increases.

sufficient

ST 2 : s k u n ( K comes twice but only one balled picked out, no need to count twice )

lowest value is K - 11th letter

prob = 11/351 = 0.31 which is lower than 0.33.

for S the 19/351 > 0.33 . since we cannot get a definite answer, insufficient.

Hence Answer = A

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