Poisson or Exponential Distribution? There are $10$ questions in a two-hour test. Suppose that the time taken for Jackson to answer each equations is independent of each other and has exponential distribution with $\lambda = 0.08$ questions per min. 

What is the probability that he will take more than $30$ min to answer the first $3$ questions and less than $30$ min to answer the next $3$ questions?

My initial thought would be to use exponential distribution of $\lambda =\frac2{75}$ to solve the question, by finding $$P(Y &lt; 30)\times(1 - P(Y &lt; 30)),$$ where Y is the time taken for Jackson to finish 3 questions. 

However, I have contesting views on using Poisson distribution as well. Perhaps the community could enlighten me. Cheers!
\begin{align}
&amp; \Pr( \,\overbrace{\text{time to answer the first three}}^{\begin{smallmatrix} \text{the sum of three exponentially} \\ \text{distributed random variables} \end{smallmatrix} }{} &gt; 30 \text{ minutes}\, ) \\[8pt]
= {} &amp; \Pr( \, \underbrace{ \text{number of questions answered in 30 minutes}}_\text{a Poisson-distributed random variable} {} &lt;3 )
\end{align}
Numerically, finding the second of these two numbers is quick and easy, so my way of finding the first one would be to find the second one and rely on your understanding of why they must be equal.
For $i=1,2,\dots,10$ let $X_{i}$ denote the time needed to get an
answer on question $i$. 

Then to be found is: $$P\left(X_{1}+X_{2}+X_{3}&gt;30\wedge X_{4}+X_{5}+X_{6}&lt;30\mid X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}\leq120\right)\tag1$$

Define $Y:=X_{1}+X_{2}+X_{3}$ and $Z:=X_{4}+X_{5}+X_{6}$. 

Then $Y$ and $Z$ are iid with $P\left(Y&gt;t\right)=P\left(Z&gt;t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}\right]$ with $\lambda=0.08$.

Note that this expression corresponds with $P\left(N\left(t\right)\leq2\right)$
where $N\left(t\right)$ denotes a Poisson counting process with rate $\lambda$.

Likewise we have $P\left(Y+Z&gt;t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}+\frac{1}{6}\lambda^{3}t^{3}+\frac{1}{24}\lambda^{4}t^{4}+\frac{1}{120}\lambda^{5}t^{5}\right]$.

This expression corresponds with $P\left(N\left(t\right)\leq5\right)$

We can rewrite $(1)$ as:

$$P\left(Y&gt;30\wedge Z&lt;30\mid Y+Z\leq120\right)=\frac{P\left(Y&gt;30\wedge Z&lt;30\wedge Y+Z\leq120\right)}{P\left(Y+Z\leq120\right)}\tag2$$

The denominator is straightforward and the numerator can be calculated as: 

$$\begin{aligned} &amp; \int P\left(Y&gt;30\wedge Z&lt;30\wedge Y+Z\leq120\mid Y=t\right)f_{Y}(t)dt\\
 &amp; =\int_{30}^{120}P\left(Z&lt;30\wedge t+Z\leq120\right)f_{Y}(t)dt\\
 &amp; =\int_{30}^{90}P\left(Z&lt;30\right)f_{Y}\left(t\right)dt+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt\\
 &amp; =P\left(Z&lt;30\right)\int_{30}^{90}f_{Y}\left(t\right)dt+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt\\
 &amp; =P\left(Z&lt;30\right)P\left(30&lt;Y\leq90\right)+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt
\end{aligned}
$$
Here in the first equality it is used that $Y$ and $Z$ are independent and $f_{Y}$ denotes the PDF of $Y$ which can be found as derivative of $P\left(Y\leq t\right)$.

The working out is left to you.