Question

1. defective resistors a package contains 12 resistors, 3 of which are defective. if 4 are selected, find the probability of getting a. 0 defective resistors b. 1 defective resistor c. 3 defective resistors

Explanation:

Step1: Calculate the total number of ways to select 4 resistors

We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 12\) and \(r=4\).

$$C(12,4)=\frac{12!}{4!(12 - 4)!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495$$

Step2: Calculate the number of non - defective resistors

The number of non - defective resistors is \(12-3 = 9\)

Part a: 0 defective resistors

We need to select 4 non - defective resistors. Using the combination formula with \(n = 9\) and \(r = 4\)

$$C(9,4)=\frac{9!}{4!(9 - 4)!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$$

The probability \(P(0)=\frac{C(9,4)}{C(12,4)}=\frac{126}{495}=\frac{14}{55}\approx0.2545\)

Part b: 1 defective resistor

We need to select 1 defective resistor (from 3) and 3 non - defective resistors (from 9).

$$C(3,1)=\frac{3!}{1!(3 - 1)!}=3$$

$$C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7}{3\times2\times1}=84$$

By the multiplication principle, the number of ways is \(C(3,1)\times C(9,3)=3\times84 = 252\)
The probability \(P(1)=\frac{252}{495}=\frac{28}{55}\approx0.5091\)

Part c: 3 defective resistors

We need to select 3 defective resistors (from 3) and 1 non - defective resistor (from 9)

$$C(3,3)=\frac{3!}{3!(3 - 3)!}=1$$

$$C(9,1)=\frac{9!}{1!(9 - 1)!}=9$$

By the multiplication principle, the number of ways is \(C(3,3)\times C(9,1)=1\times9 = 9\)
The probability \(P(3)=\frac{9}{495}=\frac{1}{55}\approx0.0182\)

Answer:

a. \(\frac{14}{55}\)
b. \(\frac{28}{55}\)
c. \(\frac{1}{55}\)