| X: | 1 | 2 | 3 | 4 | 5 |
| P(X): | \[{{K}^{2}}\] | \[2K\] | \[K\] | \[2K\] | \[5{{K}^{2}}\] |
| Then \[P(X>2)\]is equal to | |||||
A) \[\frac{7}{12}\]
B) \[\frac{23}{36}\]
C) \[\frac{1}{36}\]
D) \[\frac{1}{6}\]
Correct Answer: B
Solution :
\[P(x>2)=P(x=3)+P(x=4)+P(x=5)\] \[=k+2k+5{{k}^{2}}\] \[=5{{k}^{2}}+3k\] Now \[\Sigma {{p}_{i}}={{k}^{2}}+2k+k+2k+5{{k}^{2}}\] \[=6{{k}^{2}}+5k\] as \[\Sigma {{p}_{i}}=1\Rightarrow 6{{k}^{2}}+5k=1\Rightarrow (k+1)(6k-1)=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,k=\frac{1}{6}\] \[p(x>2)=\frac{5}{36}+\frac{3}{6}=\frac{23}{36}\]
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