The number of cars produced in a factory follow a Poisson distribution. The probability of not producing a car in an hour is 0.11. What's the probability of producing 3 cars in an hour, knowingly that at least one car was produced in that time period?
I guess I'm a little confused and my problem relies in the translation of the conditional. I believe that $\lambda = 1 - 0.11 = .89$ is the probability of producing a car in an hour, and being X = "number of cars produced in an hour", then I'm looking for $P(X = 3 | X \geq 1)$. But how does it correctly translate to an expression? My attempt was:
$$P(X=3 | X \geq 1) = P(X=3)\times P(X\geq 1) = \frac{.89^3e^{-.89}}{3!}\times.89$$
But I'm really unsure I got it right. How am I supposed to go for this problem?
You can use the definition of conditional expression, $P(A | B) = P(A\wedge B)/P(B)$.
In this case that would be:
$$P(X = 3 | X \geq 1) = \frac{P(X = 3 \wedge X \geq 1)}{P(X \geq 1)}$$