Question

The probability of experiencing a significant personnel injury on the shop floor of a
manufacturing plant in any given work day is 1%. What is the probability of having zero significant
personnel injuries during a period of 30 work days?
a) 0.7000 b) 0.6690 c) 0.7397 d) 0.6050 e) 0.9900
What is the probability of having 1 or more significant personnel injuries in that time?
a) 0.3310 b) 0.3950 c) 0.9900 d) 0.5052 e) 0.2603

Answer 1

Binomial formula should work here.

Answer 2

\[P(zero\ injuries)=\left(\begin{matrix}30 \\ 0\end{matrix}\right)0.01^{0}(0.99)^{30}=?\]
\[P(1\ injury)=\left(\begin{matrix}30 \\ 1\end{matrix}\right)0.01^{1}(0.99)^{29}\]
\[P(2\ injuries)=\left(\begin{matrix}30 \\ 2\end{matrix}\right)0.01^{2}(0.99)^{28}\]
\[P(3\ injuries)=\left(\begin{matrix}30 \\ 3\end{matrix}\right)0.01^{3}(0.99)^{27}\]
\[P(4\ injuries)=\left(\begin{matrix}30 \\ 4\end{matrix}\right)0.01^{4}(0.99)^{26}\]
To find the probability of 1 or more injuries in 30 days, sum the probabilities of 1, 2, 3 and 4 injuries.