A) \[\frac{1}{4}\]
B) \[\frac{3}{4}\]
C) \[\frac{4}{5}\]
D) \[\frac{5}{6}\]
Correct Answer: B
Solution :
| total number of combinations of numbers on the cube and the tetrahedron \[=6\times 4=24\]favourable number of ways of getting a sum not less than 5 = sum of coefficients of \[{{x}^{6}},{{x}^{7}},....{{x}^{10}}\] in the product |
| \[=(x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+{{x}^{5}}+{{x}^{6}})(x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}})\] |
| \[({{x}^{2}}+2{{x}^{3}}+3{{x}^{4}}+4{{x}^{5}}+4{{x}^{6}}+4{{x}^{7}}+3{{x}^{8}}+2{{x}^{9}}+{{x}^{10}})\]\[=4+4+4+3+2+1=18\] |
| \[\therefore \]Required probability\[=\frac{18}{6\times 4}=\frac{3}{4}\] |
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