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The length of time between arrivals at a hospital clinic follows approximately exponential probability distribution.

Mean time between the arrivals for patients at a clinic is 4 minute.

Given mean time is 4 minute hence, the pdf can be given as

$f\left(t\right)=\frac{1}{4}exp\left(-\frac{t}{4}\right)t鈮�0$

We have to find the probability that a particular interarrival time is less than 1 minute, that is to find $P\left(T<1\right)$

$$\begin{gathered} P\left(T<1\right)={鈭�}_0^{1}f\left(t\right)dt \\ ={鈭�}_0^1\frac{1}{4}exp\left(-\frac{1}{4}t\right)dt \\ =-{\left[exp\left(-\frac{1}{4}t\right)\right]}_0^1 \\ =-exp\left(\frac{-1}{4}\right)-\left(-exp\left(0\right)\right) \\ =-exp\left(\frac{-1}{4}\right)-\left(-1\right) \\ P\left(T<1\right)=1-exp\left(-0.25\right) \\ =0.22 \end{gathered}$$

Hence Probability that a particular interarrival time is less than 1 minute is 0.22.

Exponential distribution has the memoryless property that is the next interarrival time re independent of each other.

Hence the required probability is

$$\begin{gathered} \mathrm{P}\left(T<1\right)={\left(0.22\right)}^{meantime} \\ ={\left(0.22\right)}^4 \\ =0.0023 \end{gathered}$$

Therefore probability that the next four interarrival times are all less than 1 minute is 0.0023

We have to calculate $P\left(T>10\right)$

$$\begin{gathered} P\left(T>10\right)={鈭�}_{10}^{鈭�}f\left(t\right)dt \\ ={鈭�}_{10}^{鈭�}\frac{1}{4}exp\left(-\frac{1}{4}t\right)dt \\ =-{\left[exp\left(-\frac{1}{4}t\right)\right]}_{10}^{鈭�} \end{gathered}$$

$$\begin{gathered} P\left(T>10\right)=exp\left(-25\right) \\ =0.082 \end{gathered}$$