Let lim _{t rightarrow 0}(1+5 t)^{frac{1}{t}} | vedclass

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(B) Given that $\lim _{t \rightarrow 0}(1+5t)^{\frac{1}{t}} = K$. Using the standard limit $\lim _{t \rightarrow 0}(1+at)^{\frac{1}{t}} = e^a$, we get

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$\frac{K-1}{K}$

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(B) Given that $\lim _{t ightarrow 0}(1+5t)^{\frac{1}{t}} = K$. Using the standard limit $\lim _{t ightarrow 0}(1+at)^{\frac{1}{t}} = e^a$, we get $K = e^5$. For the random variable $X$ representing the number of successes in $n = 100$ independent trials with probability of success $p = 0.05$, the distribution follows a Binomial distribution $B(n, p)$. Since $n$ is large and $p$ is small, we can approximate this using a Poisson distribution with parameter $\lambda = np = 100 	imes 0.05 = 5$. The probability of getting at least one success is $P(X \geq 1) = 1 - P(X = 0)$. Using the Poisson formula $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, we have $P(X = 0) = e^{-5}$. Therefore, $P(X \geq 1) = 1 - e^{-5} = 1 - \frac{1}{e^5} = 1 - \frac{1}{K} = \frac{K-1}{K}$.

Let $\lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K$ and $X$ be the random variable representing the number of successes in $100$ independent trials. If the probability of success in each trial is $0.05$, then the probability of getting at least one success is

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