A) \[\frac{1}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{2}{3}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Here, n = 6 According to the question \[{}^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}}=4\cdot {}^{6}{{C}_{4}}{{p}^{4}}{{q}^{2}}\] \[\Rightarrow \] \[{{q}^{2}}=4{{p}^{2}}\] \[\Rightarrow \] \[{{(1-p)}^{2}}=4{{p}^{2}}\] \[\Rightarrow \] \[3{{p}^{2}}+2p-1=0\] \[\Rightarrow \] \[(p+1)\,(3p-1)=0\] \[\Rightarrow \] \[p=\frac{1}{3}\] (\[\because \]p cannot be negative)
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