Calls arriving at a telephone exchange follow a Poisson distribution with an average of 3 calls every 10 minutes. If the telephone exchange opens at 8:00 am, what is the probability that the 5th call arrives between 8:10 am to 8:20 am?
Here I am creating two events:
Event A: Number of calls between 8:00 am to 8:20 am is $\geq 5$
Event B: Number of calls between 8:00 am to 8:10 am is $\leq 4$
Then the probability that the 5th call will arrive between 8:10 to 8:20 is $P(A \cap B)$.
Now I have found $P(A)=0.71494$ and $P(B)=0.81526$.
How do I find $P(A \cap B)$?
Best Answer
It is easier to handle independent or mutually distinct events so you can multiply or add their probabilities.
Here you have three mutually distinct events, so you can use addition and subtraction
In effect the argument is that $$P(A \cup B)= P(A)+P(B) -P(A \cap B)$$ so $$P(A \cap B)= P(A)+P(B) -P(A \cup B)$$ $$P(A \cap B)= P(A)+P(B) -(1 - P(A^c \cap B^c))$$ and in this case $$P(A^c \cap B^c)=0$$