Question

a box contains 15 transistors, 5 of which are defective. if 5 are selected at random, find the probability of the statements below
a. all are defective
b. none are defective
a. the probability is
(type a fraction. simplify your answer)

Explanation:

Step1: Calculate the total number of ways to select 5 transistors

The total number of ways to select 5 transistors out of 15 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 15\) and \(r=5\).

$$C(15,5)=\frac{15!}{5!(15 - 5)!}=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}=3003$$

Step2: Calculate the number of ways to select 5 defective transistors

The number of ways to select 5 defective transistors out of 5 is \(C(5,5)\)

$$C(5,5)=\frac{5!}{5!(5 - 5)!}=1$$

Step3: Calculate the probability

The probability \(P\) that all 5 are defective is the number of favorable outcomes divided by the total number of outcomes.

$$P=\frac{C(5,5)}{C(15,5)}=\frac{1}{3003}$$

Answer:

\(\frac{1}{3003}\)