What would be the answer please?
The mean number of accidents in a shoe factory is 0.10 per day. What is the probability that during a randomly selected day, there will be:
a) no accidents
b) exactly 1 accident
c) at least 1 accident
The Poisson distribution gives the probability of k events in an interval as \(p = \frac{\lambda^ke^{-\lambda}}{k!}\) where \(\lambda\) is the mean.
So:
1. \(p(k=0)=e^{-0.1} \rightarrow 0.905\)
2. \(p(k=1)=0.1e^{-0.1} \rightarrow 0.0905\)
3. Ther probability of at least one accident is 1 - (the probability of exactly no accidents plus the probability of exactly one accident).
The Poisson distribution gives the probability of k events in an interval as \(p = \frac{\lambda^ke^{-\lambda}}{k!}\) where \(\lambda\) is the mean.
So:
1. \(p(k=0)=e^{-0.1} \rightarrow 0.905\)
2. \(p(k=1)=0.1e^{-0.1} \rightarrow 0.0905\)
3. Ther probability of at least one accident is 1 - (the probability of exactly no accidents plus the probability of exactly one accident).