Given ,

10 professional players have 10 different balls

A,B,C,D,E,F,G,H,I

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

=4*3*2*1=24

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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