Three numbers are chosen at random with out replacement from the set of integers {1, 2, 3, ……, 10}. The probability that the minimum of the chosen numbers is 3 or the maximum of the chosen numbers is 7, is equal to

$\frac{11}{40}$

Let A : Minimum of the chosen number is 3

B : Maximum of the chosen number is 7

$P(A)=\frac{{ }^7 C_2}{{ }^{10} C_3}=\frac{21}{120}$

$P(B)=\frac{{ }^6 C_2}{{ }^{10} C_3}=\frac{15}{120}$

$P(A \cap B)=\frac{{ }^3 C_1}{{ }^{10} C_3}=\frac{3}{120}$

Thus, required probability

$=P(A)+P(B)-P(A \cap B)$

$=\frac{11}{40}$