A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{3}\]
Correct Answer: D
Solution :
\[4P(X=4)=P(X=2)\Rightarrow 4.{}^{6}{{C}_{4}}{{p}^{4}}{{q}^{2}}={}^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}}\] \[\Rightarrow 4{{p}^{2}}={{q}^{2}}\Rightarrow 4{{p}^{2}}={{(1-p)}^{2}}\] \[\Rightarrow 3{{p}^{2}}+2p-1=0\Rightarrow p=\frac{1}{3}\].
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