10 professional players have 10 different balls
probability of A being chosen among 4 balls drawn at random =
9C3/10C4 = 0.1
probability of A being first when chosen among 4 balls=
let us assume A,B,C,D are chosen
no of combinations of A being 1st = no of combinations of 3 other alphabets placed in 3 places
A _ _ _
=3 * 2* 1=6
no of combinations of 4 alphabets being placed in 4 boxes
So, probability of A being first when chosen among 4 balls=6/24 = 0.25
1.) probability of finally A being chosen among 1st place = 0.25*0.1 = 0.025 =1/40
2.)probability of A chosen among first 4 = 0.1
3.) probability of A not being chosen among first four = 1-0.1 = 0.9
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