A) \[\frac{16}{15}\]
B) \[\frac{15}{16}\]
C) \[\frac{31}{16}\]
D) None of these
Correct Answer: D
Solution :
Given, \[f(x)=\frac{k}{2},\] Where \[x=0,1,2,3,4,\] We know that, Sum of probability distribution \[=1\] \[\frac{k}{{{2}^{0}}}+\frac{k}{{{2}^{1}}}+\frac{k}{{{2}^{2}}}+\frac{k}{{{2}^{3}}}+\frac{k}{{{2}^{4}}}=1\] \[\Rightarrow \] \[\frac{1}{k}=1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+\frac{1}{{{2}^{4}}}\] \[\Rightarrow \] \[\frac{1}{k}=\frac{1.\left( 1-{{\left( \frac{1}{2} \right)}^{5}} \right)}{1-\frac{1}{2}}=\frac{1-\frac{1}{32}}{\frac{1}{2}}=\frac{31}{16}\] \[\Rightarrow \] \[k=\frac{16}{31}\]
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