College Algebra, Chapter 10, 10.4, Section 10.4, Problem 30

A machine that is used in manufacturing pricess has four separate components, each of which has a $0.01$ probability of failing on any given day. If any component fails, the entire machine breaks down. Find the probability on a given day the indicated event occurs.

a.) The machine breaks down.

b.) The machine does not break down.

c.) Only one component does not fail.

Recall that the formula for the binomial probability is given by

$C(n,r) p^r q^{n-r}$

In this case, $p =0.01, q = 1-p = 0.99$ and $n =4$.

If the machines break down, then the required probability is the probability that at least one component fails. To solve this in most efficient way, we can apply the component to the probability that none of the components fail. Thus, we have,


$\begin{equation} \begin{aligned} =& 1 - \left[C(4,0)(0.01)^0 (0.99)^{4-0}\right] \\ \\ =& 1-0.9606 \\ \\ =& 0.0394 \end{aligned} \end{equation}$


b.) If the machine does not breakdown, then the required probability is the probability that none of the components fail from part (a), we have

$= C(4,0)(0.01)^0 (0.99)^{4-0}$

$= 0.9606$

c.) If only one component does not fail, then exactly 3 of the components fail. Thus, we have

$= C(4,3)(0.01)^3 (0.99)^{4-3}$

$= 0.00000396$