A) \[\frac{41}{81}\]
B) \[\frac{39}{81}\]
C) \[\frac{40}{81}\]
D) None of these
Correct Answer: A
Solution :
The total number of ways of selecting 4 tickets \[={{3}^{4}}=81\]. The favourable number of ways = sum of coefficients of \[{{x}^{2}},\,{{x}^{4}},\,.......\] in \[{{(x+{{x}^{2}}+{{x}^{3}})}^{4}}\] = sum of coefficients of \[{{x}^{2}},\,{{x}^{4}},\,......\] in \[{{x}^{4}}{{(1+x+{{x}^{2}})}^{4}}.\] Let \[{{(1+x+{{x}^{2}})}^{4}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{8}}{{x}^{8}}.\] Then \[{{3}^{4}}=1+{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{8}}\], (On putting \[x=1)\] and \[1=1-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+.....+{{a}_{8}}\], (On putting \[x=-1)\] \[\therefore \,\,\,{{3}^{4}}+1=2(1+{{a}_{2}}+{{a}_{4}}+{{a}_{6}}+{{a}_{8}})\] \[\Rightarrow {{a}_{2}}+{{a}_{4}}+{{a}_{6}}+{{a}_{8}}=41\] Thus sum of the coefficients of \[{{x}^{2}},\,{{x}^{4}},\,......=41\] Hence the required probaility \[=\frac{41}{81}.\]
You need to login to perform this action.
You will be redirected in
3 sec
