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(C) Given, $n = 50$. Let $p$ be the probability of success in one trial. Then the parameter of the Poisson distribution is $\lambda = np = 50p$. The p

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$0.04$

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(C) Given, $n = 50$. Let $p$ be the probability of success in one trial. Then the parameter of the Poisson distribution is $\lambda = np = 50p$. The probability mass function for a Poisson variable is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$. Given the equation: $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$. Substituting the formula: $2 \frac{e^{-\lambda} \lambda^1}{1!} = 5 \frac{e^{-\lambda} \lambda^5}{5!} + 2 \frac{e^{-\lambda} \lambda^3}{3!}$. Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} 
eq 0$): $2 \lambda = 5 \frac{\lambda^5}{120} + 2 \frac{\lambda^3}{6}$. $2 \lambda = \frac{\lambda^5}{24} + \frac{\lambda^3}{3}$. Multiplying by $24$: $48 \lambda = \lambda^5 + 8 \lambda^3$. Since $\lambda 
eq 0$, divide by $\lambda$: $\lambda^4 + 8 \lambda^2 - 48 = 0$. Let $u = \lambda^2$, then $u^2 + 8u - 48 = 0$. $(u + 12)(u - 4) = 0$. So, $\lambda^2 = 4$ or $\lambda^2 = -12$. Since $\lambda^2$ must be positive, $\lambda^2 = 4$, which gives $\lambda = 2$. Finally, $p = \frac{\lambda}{n} = \frac{2}{50} = 0.04$.

If $X$ is a Poisson variable representing the 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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