Discussion Forum

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

Answer:

Explanation:

We have

  $\lim_{t \rightarrow 0}(1+5t)^{1/t}=K$

 $\Rightarrow$     $K= e^{\lim_{t \rightarrow 0}\frac{5t}{t}}=e^{5}$

 Here, n=100 , p=0.05

 $\lambda$=np=5

$P(X\geq1)=1-P(X=0)=1-e^{-5}$

$P(X\geq1)=\frac{e^{5}-1}{e^{5}}=\frac{K-1}{K}$