A road has 2 "states" each day - a busy state and a free state. Both states follows a Poisson distribution.
- Busy day - 120 cars drive the road in a busy day.
- Free day - 80 cars drive the road in a free day.
The probability that tomorrow will have at least 40 cars is 0.8.
The probability for a busy day is p and the probability for a free day is 1-p.
Using that information, how can I find the value of p?
I've tried to calculate the probability that each day will have 120 and 80 cars but couldn't do it with the given data, and doesn't seem like the right direction.
Any help would be much appreciated,
Thanks
Hint: Let's denote $A$ as the event that tomorrow is a busy day. Then $\overline A$ is the event that tomorrow is a free day. Additionally $X$ is poisson distributed as $X\sim Poi(\lambda)$. Now apply the total law of probability. The equation becomes
$$P(X\geq 40|A)\cdot P(A)+P(X\geq 40|\overline A)\cdot P(\overline A)=0.8,$$
where $P(\overline A)=1-P(A)$