The probability of winning(PROB) 2nd test case ,solving approach?

can some one check this,is this the right way of solving?

regarding the 2nd test case 2 3 4 1

since t4 is 1, the possible outcomes after chef discarding 1 ticket is
(1,3,4) with prob. (2/9)

prob. of winning is (1/8)+(4/8)(1/7)+(4/8)(3/7)(1/6)+(4/8)(3/7)(2/6)(1/5)+(4/8)(3/7)(2/6)(1/5)(1/4)

.(1/8) —>a winning ticket is picked up in the first turn

.(4/8)*(1/7)---->T-3 ticket in the first turn and a winning next

.(4/8)(3/7)(1/6)---->T-3 tickets in the first two turns followed by a winning ticket

.(4/8)(3/7)(2/6)*(1/5)----->T-3 tickets in the first three turns followed by a winning ticket

.(4/8)(3/7)(2/6)(1/5)(1/4)------>T-3 tickets in the first four turns followed by a winning ticket

the other two possible outcomes are (2,2,4) and (2,3,3)
when i do this type of solving, the total prob. i am getting is >1. how to approach this problem???

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