Question

A package contains 11 resistors, 2 of which are defective. If 4 are selected find the probability of getting the following results. P(0defectice)

Answer 1

non-defectectiveThe Solution.

Step 1:

We shall determine the number non defectective resistors.

$\begin{gathered} \text{Non}-\text{defective resistors = total resistors - defective resistors} \\ Non-defective\text{ resistors =11-2 = 9} \end{gathered}$

Step 2:

We shall find the probability of defective and non-defective resistors.

$\begin{gathered} \text{Prob(defective resistor)}=\frac{n\text{umber of required outcomes}}{n\text{umber of total outcomes}} \\ \text{Where number of required outcomes =2 resistors} \\ n\text{umber of total outcomes = 11} \end{gathered}$$\begin{gathered} \text{prob(defective resistor) =}\frac{2}{11} \\ \\ \text{Prob(non}-\text{defective resistor) =1-prob(defective resistor)} \\ \text{Prob(non}-\text{defective resistor) =1-}\frac{2}{11}=\frac{9}{11} \end{gathered}$

Step 3:

We shall find the probability of getting 0 defective when 4 resistors are selected.

Note: Getting 0 defective resistor after 4 selections means that the 4 resistors selected were non-defective resistors. Note also that since the question is silent about replacement, we shall treat it as probability without replacement.

So,

$\text{Prob(selecting 4 non-defective resistors)=pr(N}_1N_2N_3N_4)$$\begin{gathered} \text{where pr(N}_1)=\frac{9}{11} \\ pr(N_2)=\frac{8}{10} \\ pr(N_3)=\frac{7}{9} \\ pr(N_4)=\frac{6}{8} \end{gathered}$

Step 4:

We shall substitute these probabilities and then simplify.

Prob(0 defective resistors) = prob(4 non-defective resistors) =

$\begin{gathered} pr(N_1)\times pr(N_2)\times pr(N_3)\times pr(N_4) \\ =\frac{9}{11}\times\frac{8}{10}\times\frac{7}{9}\times\frac{6}{8}=\frac{42}{110} \\ =\text{ 0.38181818 }\approx\text{ 0.381818} \end{gathered}$

Step 5:

Presentation of the Answer.

So, the probability of getting a 0 defective resistor when 4 resistors are selected is 0.381818

Thus, the correct answer is 0.381818