• # question_answer Three numbers are chosen at random without replacement from 1, 2, 3, ..., 10. The probability that the minimum of the chosen numbers is 4 or their maximum is 8 is A) $\frac{11}{40}$                        B) $\frac{3}{10}$ C) $\frac{1}{40}$              D)  none of these.

[a] : Let us define the events in the following way: A : 4 being the minimum number B : 8 being the maximum number $A\cap B:1$ being the minimum number and 8 being the maximum number Therefore $P(A)=\frac{^{6}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{15}{120}$ $P(B)=\frac{^{7}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{21}{120}$and $P(A\cap B)=\frac{^{3}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{3}{120}$ $\therefore$The required probability, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $=\frac{15}{120}+\frac{21}{120}-\frac{3}{120}=\frac{33}{120}=\frac{11}{40}$