Question

# A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged ident

Upper level probability
A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random. What is the probability that:
a) All four units are good?
b) Exactly two units are good?
c) At least two units are good?

2020-12-28

Given that there are 12 microwave ovens i.e., n=12
in that 3 are deffective
then number of good units =9
The probability of good unit is $$\displaystyle={\frac{{{9}}}{{{12}}}}$$
$$\displaystyle={0.75}$$
A vending company has ordered four ofthese units then the probability that: all four units are good is
$$P[X=4]=\frac{\left(\begin{array}{c}9\\ 4\end{array}\right)}{\left(\begin{array}{c}12\\ 4\end{array}\right)}$$
Probability of exactly two are good is
$$P[X=2]=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)}{\left(\begin{array}{c}12\\ 4\end{array}\right)}$$
Probability of at least two are good is
$$\displaystyle{P}{\left[{x}\geq{2}\right]}={P}{\left[{X}={2}\right]}+{P}{\left[{X}={3}\right]}+{P}{\left[{X}={4}\right]}$$
$$=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)}{\left(\begin{array}{c}12\\ 4\end{array}\right)}+\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}12\\ 4\end{array}\right)}+\frac{\left(\begin{array}{c}9\\ 4\end{array}\right)}{\left(\begin{array}{c}12\\ 4\end{array}\right)}$$
Solve the above we get the required answer