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Let the random variable $\mathrm{X}$ follow $\mathrm{B}(6, \mathrm{p})$. If $16 \mathrm{P}(\mathrm{X}=4)=$ $\mathrm{P}(\mathrm{X}=2)$, then what is the value of $\mathrm{p}$ ?
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The correct answer is:
$\frac{1}{5}$
$\begin{aligned} & \text { X follows } B(6, p)=16 P(x=4)=P(x=2) \\ & \Rightarrow 16{ }^{6} C_{4} \cdot p^{4}(1-p)^{6-4}={ }^{6} C_{2} \cdot p^{2}(1-p)^{6-2} \\ & \Rightarrow 16 \times \frac{6 !}{4 ! 2 !} p^{4}(1-p)^{2}=\frac{6 !}{2 ! 4 !} p^{2}(1-p)^{4} \\ & \Rightarrow 16 p^{2}=(1-p)^{2} \Rightarrow 16 p^{2}=1+p^{2}-2 p \\ & \Rightarrow 15 p^{2}+2 p-1=0 \Rightarrow 15 p^{2}+5 p-3 p-1=0 \\ & \Rightarrow p=\frac{1}{5},-\frac{1}{3} \\ & \text { As } P>0 \Rightarrow P=\frac{1}{5} \end{aligned}$
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