# 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 $$P(Y < 30)\times(1 - P(Y < 30)),$$ 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!

• Welcome to MSE. Please use MathJax. – José Carlos Santos Sep 16 '17 at 8:00
• I think that you must also take into account that this event can only occur if Jackson manages to find answers for the first 6 questions within 2 hours. – drhab Sep 16 '17 at 9:40

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: $$P\left(X_{1}+X_{2}+X_{3}>30\wedge X_{4}+X_{5}+X_{6}<30\mid X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6}\leq120\right)\tag1$$

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

Then $Y$ and $Z$ are iid with $P\left(Y>t\right)=P\left(Z>t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}\right]$ 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 $P\left(Y+Z>t\right)=e^{-\lambda}\left[1+\lambda t+\frac{1}{2}\lambda^{2}t^{2}+\frac{1}{6}\lambda^{3}t^{3}+\frac{1}{24}\lambda^{4}t^{4}+\frac{1}{120}\lambda^{5}t^{5}\right]$.

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

We can rewrite $(1)$ as:

$$P\left(Y>30\wedge Z<30\mid Y+Z\leq120\right)=\frac{P\left(Y>30\wedge Z<30\wedge Y+Z\leq120\right)}{P\left(Y+Z\leq120\right)}\tag2$$

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

\begin{aligned} & \int P\left(Y>30\wedge Z<30\wedge Y+Z\leq120\mid Y=t\right)f_{Y}(t)dt\\ & =\int_{30}^{120}P\left(Z<30\wedge t+Z\leq120\right)f_{Y}(t)dt\\ & =\int_{30}^{90}P\left(Z<30\right)f_{Y}\left(t\right)dt+\int_{90}^{120}P\left(Z\leq120-t\right)f_{Y}\left(t\right)dt\\ & =P\left(Z<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\\ & =P\left(Z<30\right)P\left(30<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.

• Hmm, @drhab, thank you for the Poisson Process methodology. However, may I check what is wrong with my working above? Can I not just convert the lambda in exponential distribution by dividing 1 after getting the new mean of 3 questions? $$\lambda = 1/(12.5 x 3)] = 2/75$$ – fauxpas Sep 16 '17 at 10:28
• It seems to me (am I correct in this?) that you think that $Y=X_1+X_2+X_3$ has exponential distribution again. This is not the case. Maybe you are mixing up with $\min(X_1,X_2,X_3)$. See here. Secondly in your answer you don't take into account that the test only lasts $2$ hours. – drhab Sep 16 '17 at 10:36