I am given:
- I get 11 text messages per hour according to a Poisson process.
- The probability that a given text message is from my mother is $0.62$.
I then have to find the probability that I receive exactly two text messages from my mother in a 40 minute period.
My thought process is:
- Find the expected number, $E(X)$, of text messages in a 40 minute period.
- Use this value as the number of trials, $n$, in a binomial distribution to calculate $P(Y = 2)$ where $Y =$ "the number of text messages I receive from my mother in a 40 minute period".
However, $E(X) = 11 \cdot \frac{40}{60} = \frac{22}{3}$ meaning my idea won't work without rounding the number of trials - giving two wildly different answers (as expected). Am I meant to round up or down, or am I doing this completely wrong? I did try averaging the probabilities of the two answers I obtained, but I'm not sure if this is correct either.
If the probability that a given text message is from my mother is independent from other text messages, you can split the poisson process: you can say that the stream of messages from my mother is a poisson process too with rate $0.62 \cdot 11 = 6.82$. This is known as 'splitting'.
The number of text messages in a 40 minute interval follows a poisson distribution with parameter $6.82 \cdot 40/60 = 341/75$. The probability that I receive exactly two text messages from my mother in a 40 minute period is: $$\frac{e^{-341/75} (341/72)^2}{2!} \approx 0.12.$$