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If $X$ is a Poísion variable representing number of successes in 50 trials such that $2 P(X=1)=5 P(X=5)+2 P(X=3)$, then the probability of getting success in one trial is
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The correct answer is:
0.04
Given, $n=50$
$$
\begin{aligned}
& p=\text { ? } \\
& \lambda=n p=50 p \\
& 2 P(x=1)=5 P(x=5)+2 P(x=3) \\
& 2 e^{-\lambda} \lambda=\frac{5 e^{-\lambda} \lambda^5}{5 !}+\frac{2 e^{-\lambda} \lambda^3}{3 !} \\
& \Rightarrow \quad 2 \lambda=\frac{\lambda^5}{24}+\frac{\lambda^3}{3} \\
& \Rightarrow \quad \lambda^4+8 \lambda^2-48=0 \\
& \left(\lambda^2+12\right)\left(\lambda^2-4\right)=0 \\
& \lambda^2=4, \lambda^2 \neq-12 \\
& \therefore \quad \lambda=2 \\
& p=\frac{\lambda}{n}=\frac{2}{50}=\frac{1}{25}=0.04 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& p=\text { ? } \\
& \lambda=n p=50 p \\
& 2 P(x=1)=5 P(x=5)+2 P(x=3) \\
& 2 e^{-\lambda} \lambda=\frac{5 e^{-\lambda} \lambda^5}{5 !}+\frac{2 e^{-\lambda} \lambda^3}{3 !} \\
& \Rightarrow \quad 2 \lambda=\frac{\lambda^5}{24}+\frac{\lambda^3}{3} \\
& \Rightarrow \quad \lambda^4+8 \lambda^2-48=0 \\
& \left(\lambda^2+12\right)\left(\lambda^2-4\right)=0 \\
& \lambda^2=4, \lambda^2 \neq-12 \\
& \therefore \quad \lambda=2 \\
& p=\frac{\lambda}{n}=\frac{2}{50}=\frac{1}{25}=0.04 \\
&
\end{aligned}
$$
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