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Let $\lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K$ and $X$ be the random variable representing 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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Verified Answer
The correct answer is:
$\frac{k-1}{k}$
We have,
$$
\begin{aligned}
& \lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K \\
& \Rightarrow \quad K=e^{\lim _{t \rightarrow 0} \frac{5 t}{t}}=e^5 \\
& \text { Here, } n=100, \quad p=0.05 \\
& \\
& \quad \lambda=n p=5
\end{aligned}
$$
$$
\begin{aligned}
& P(X \geq 1)=1-P(X=0)=1-e^{-5} \\
& P(X \geq 1)=\frac{e^5-1}{e^5}=\frac{K-1}{K}
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K \\
& \Rightarrow \quad K=e^{\lim _{t \rightarrow 0} \frac{5 t}{t}}=e^5 \\
& \text { Here, } n=100, \quad p=0.05 \\
& \\
& \quad \lambda=n p=5
\end{aligned}
$$
$$
\begin{aligned}
& P(X \geq 1)=1-P(X=0)=1-e^{-5} \\
& P(X \geq 1)=\frac{e^5-1}{e^5}=\frac{K-1}{K}
\end{aligned}
$$
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