I have a question regarding probability and statistics...
A certain Friday evening 1% of the car drivers are intoxicated. a) At a road check 50 drivers are tested. Compute the probability that at least one intoxicated driver is caught. b) At a road check drivers are tested. Let X be the number of tested drivers when the first intoxicated driver is caught. Compute the expected number E(X) and the standard deviation SD(X).
I wanted to try and use the following method:
X binomial random variable: X ~ Binomial (n, p) where n is the number of trials (50) and p = P(success)
So for task a) we have p = 1/100, n = 50, P(X>=1)=1-P(X=0)=1-binomial(50,0) * (1/100)^0 * (1-1/100)^50
How can I calculate this? It seems a bit confusing to me so I'd really appreciate the help
For A), we have
$1.0 - (0.99)^{50}$, because that is the same thing as one minus the probability that no drivers are caught.
For B), we are dealing with a geometric distribution.
$E(X) = \frac{1-p}{p}$
Where p=0.01.
The variance is $E(X) = \frac{1-p}{p^2}$, so the square root of that is the SD.