A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{n}\]
D) \[\frac{n-1}{3n}\]
Correct Answer: A
Solution :
| Total number of cases \[={}^{3n}{{C}_{2}}=\frac{3n(3n-1)}{2}\] |
| Now, \[{{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}-xy+{{y}^{2}})\] |
| \[{{x}^{3}}+{{y}^{3}}\] is divisible by 3 if 3divdes \[x+y\]or divides\[{{x}^{2}}-xy+{{y}^{2}}\] |
| We arrange the \[3n\]numbers in 3 sequences. |
| \[A:\left\{ 1,4,7,.......,3n-2 \right\}\] |
| \[B:\left\{ 2,5,8,.......,3n-1 \right\}\] |
| \[C:\left\{ 3,6,9,.......,3n \right\}\] |
| Clearly we must choose either one number from the first sequence or other number from the second sequence or both numbers from the third sequence only. |
| \[\therefore \]Number of favourable cases \[=n\times n+{}^{n}{{C}_{2}}\]\[={{n}^{2}}+\frac{n(n-1)}{2}=\frac{n}{2}(3n-1)\] |
| \[\therefore \] Required probability =\[\frac{\frac{n}{2}(3n-1)}{\frac{3n(3n-1)}{2}}=\frac{1}{3}\] |
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