Three numbers are chosen at random without replacement from $\{1,2, \ldots \ldots . . . . . . . . .

Question

Three numbers are chosen at random without replacement from $\{1,2, \ldots \ldots . . . . . . . . . . ., 10\}$. The probability that the minimum of the chosen numbers is 3 , or their maximum is 7 , is m/n in the lowest form. report m+n

Professor PC Thomas Classes · Answer

Total number of ways of selecting 3 numbers from $1,2, \ldots, 10$ is $n(S)={ }^{10} C_{3}$\ Number of choosing a minimum of 3 and other is from 4 to 10 is $n(A)={ }^{7} C_{2}$\ Number of choosing a maximum of 7 and other is from 1 to 6 is $n(B)={ }^{6} C_{2}$\ $n(A \cap B)={ }^{3} C_{2}$\ Required probability $=P(A \cup B)$\ $=P(A)+P(B)-P(A \cap B)$\ $=\frac{{ }^{7} C_{2}}{{ }^{10} C_{3}}+\frac{{ }^{6} C_{2}}{{ }^{10} C_{3}}-\frac{{ }^{3} C_{2}}{{ }^{10} C_{3}}$\ $=\frac{21+15-3}{120}=\frac{33}{120}=\frac{11}{40}$.\