A) \[\frac{1}{12}\]
B) \[\frac{1}{15}\]
C) \[\frac{1}{40}\]
D) None of these
Correct Answer: C
Solution :
[c] \[n(S){{=}^{10}}{{C}_{3}}\] and \[n(E){{=}^{3}}{{C}_{1}},\] becomes on selecting \[3,7\] and we have to select one from 4, 5 and 6. \[\therefore P(E)=\frac{^{3}{{C}_{1}}}{^{10}{{C}_{3}}}=\frac{1}{40}.\]
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