Probability - constant probability throughout the day The probability of an electronics store selling at least one computer in an 8-hour working day is 0.8. Assuming a constant probability throughout the day, what’s the probability of the store selling at least one computer in any given 2-hour time window?
Hi, i want to understand how to approch this kind of questions.
if any one can explain? and get the answer.
 A: There is no need for Poisson distribution.
Let the probability of not selling a computer in a 2-hour window be $x$. Then the probability of not selling a computer in an 8-hour window is $x^4$: there are four 2-hour windows in which, independently, no computer is sold.
We are given that $x^4 = 1 - 0.8 = 0.2$. So $x = 0.2^{1/4} \approx 0.669$, and the probability we want is $1-x \approx 0.331$.
A: Poisson distribution is suitable for this. I think you may want to read a little about that topic.
$$
0.8=1-\frac{\lambda^{0}e^{-\lambda}}{0!}
$$
From this, you obtain the value of $\lambda$. Then since 2 hours is 0.25 times 8 hours,
$$
1-\frac{(0.25\lambda)^{0}e^{-0.25\lambda}}{0!}
$$
is the answer you are looking for.
A: You have to use a Poisson distribution.
With the given information that you sell one or more computer in 8 hours you can calculate your sell-average:
$$\mathbb{P}[X \geq1]=1-\mathbb{P}[0]=1-e^{-\theta}=0.8 \rightarrow \theta \approx 1.6$$
This is your parameter of the Poisson Law.
Thus in 2 hours windows, the sales can be represented by a poisson distribution with parameter $$\lambda =\frac{\theta}{4}\approx 0.4$$
Concluding:
The probability of selling at least one computer in 2 hours window is
$$\mathbb{P}[X \geq1]=1-\mathbb{P}[0]=1-e^{0.402359}=0.331260 \approx \frac{1}{3}$$
