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 \[^{6}{{C}_{2}}{{p}^{2}}{{q}^{4}}=4{{\cdot }^{6}}{{C}_{3}}{{p}^{4}}{{q}^{2}}\] \[\Rightarrow \] \[{{q}^{2}}=4{{p}^{2}}\] \[\Rightarrow \] \[{{(1-p)}^{2}}=4{{p}^{2}}\] \[\Rightarrow \] \[3{{p}^{2}}+2p-1=0\] \[\Rightarrow \] \[(p+1)(3p-1)=0\] \[\Rightarrow \] \[p=\frac{1}{3}\] (\[\therefore \,\,p\]cannot be negative)You need to login to perform this action.
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