A) \[{{(3/4)}^{n}}\cdot \,n\]
B) \[n\,\,\cdot \,\,{{(3/4)}^{n-1}}\]
C) \[(n-1)\cdot {{(3/4)}^{n}}\]
D) None of these
Correct Answer: B
Solution :
[b] Let \[{{x}_{i}}\] be any element of set P, we have following possibilities (i) \[{{x}_{i}}\in A,{{x}_{i}}\in B;\] (ii) \[{{x}_{i}}\in A,{{x}_{i}}\notin B;\] (iii) \[{{x}_{i}}\notin A,{{x}_{i}}\in B;\] (iv) \[{{x}_{i}}\notin A,{{x}_{i}}\notin B;\] Clearly, the element \[{{x}_{i}}\in A\cap B\] if it belongs to A and B both. Thus out of these 4 ways only first way is favourable. Now the element that we want to be in the intersection can be chosen in ?n? different ways, Hence required probability is \[n\,\cdot \,{{(3/4)}^{n-1.}}\].
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