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Three numbers are chosen at random from 1,2,3....10 without replacement. The probability that the minimum of the chosen numbers in 4 or the maximum is 8 is?


Options:
A .   1140
B .   410
C .   112
D .   35
E .   23
Answer: Option A
:
A

Probability of 4 being the minimum number of the three numbers selected = 6C210C3 
Probability of 8 being the maximum number of the three numbers selected = 7C210C3 
The probability of 4 being the minimum number and 8 being the maximum number = 310C3 
Therefore, required probability is 6C210C3  + 7C210C3310C31140 



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