Answer:
The probability is 0.9824
Step-by-step explanation:
If a variable x follows a poisson distribution, the probability P is given by:
[tex]P(x)=\frac{e^{-λ}*λ^{x}}{x!}[/tex]
Where x counts the number of events that happens in a specific time and λ is the mean number of events in that specific time.
So, if the probability that a machine will become inoperative during a given day is 0.002 and there are 100 drill presses, then the value of λ or the mean is calculated as:
λ = 100*0.002 = 0.2
Then, the probability that fewer than two machines will be inoperative during a particular day is:
P(x<2) = P(x=0) + P(x=1)
[tex]P(x<2)=\frac{e^{-0.2}*0.2^{0}}{0!}+\frac{e^{-0.2}*0.2^{1}}{1!}[/tex]
P(x<2) = 0.8187 + 0.1637
P(x<2) = 0.9824
Finally, the probability that fewer than two machines will be inoperative during a particular day is 0.9824
