A) \[\frac{15}{216}\]
B) \[\frac{5}{54}\]
C) \[\frac{13}{216}\]
D) \[\frac{1}{18}\]
Correct Answer: B
Solution :
Exhaustive number of cases \[={{6}^{3}}=216.\] Obviously, the second number has to be greater than unity. If the second number is i\[(i>1),\] then the first can be chosen in \[i-1\] ways and the third in \[6-i\] ways and hence three numbers can be chosen in \[(i-1)\times 1\times (6-i)\] ways. But the second number can be \[2,\,\,3,\,\,4,\,\,5.\] Thus the favourable number of cases \[=\sum\limits_{i=2}^{5}{(i-1)(6-i)}=1\times 4+2\times 3+3\times 2+4\times 1=20\] Hence the required probability \[=\frac{20}{216}=\frac{5}{54}.\]
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