A) \[\frac{61}{72}\]
B) \[\frac{65}{72}\]
C) \[\frac{121}{144}\]
D) \[\frac{121}{216}\]
Correct Answer: B
Solution :
The number of ways of getting the sum 9 is the coefficient of \[{{x}^{9}}\] in the expansion of \[{{(x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{6}})}^{4}}\] i.e., the coefficient of \[{{x}^{5}}\] in \[{{(1+\,x+...+\,{{x}^{5}})}^{4}}\] \[={{(1-{{x}^{6}})}^{4}}\,{{(1-x)}^{-5}}\] \[\therefore \] The number of ways of getting the sum \[\le 9\] is the coefficient of \[{{x}^{5}}\] in \[{{(1-{{x}^{6}})}^{4}}\,{{(1-x)}^{-5}}\] \[=\,(1-4{{x}^{6}}-...)\,(1{{+}^{5}}{{C}_{1}}x{{+}^{6}}{{C}_{2}}{{x}^{2}}+...)\] \[{{=}^{9}}{{C}_{5}}=126\] \[\therefore \] The probability of getting the sum \[\ge 10\] is \[1-\frac{126}{{{6}^{4}}}=1-\frac{7}{72}=\frac{65}{72}\]
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