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a box contains 12 transistors, 4 of which are defective. if 4 are selec…

Question

a box contains 12 transistors, 4 of which are defective. if 4 are selected at random, find the probability of the statements below.
a. all are defective
b. none are defective
a. the probability is \\(\square\\).
(type a fraction. simplify your answer.)

Explanation:

Step1: Recall combination formula

The number of ways to choose \( r \) items from \( n \) items is given by the combination formula \( C(n, r)=\frac{n!}{r!(n - r)!} \). We need to find the number of ways to choose 4 defective transistors from 4 defective ones and the number of ways to choose 4 transistors from 12 total.

Step2: Calculate number of favorable outcomes

Number of ways to choose 4 defective transistors from 4 defective ones: \( C(4, 4)=\frac{4!}{4!(4 - 4)!}=\frac{4!}{4!0!}=1 \) (since \( 0!=1 \)).

Step3: Calculate total number of possible outcomes

Number of ways to choose 4 transistors from 12: \( C(12, 4)=\frac{12!}{4!(12 - 4)!}=\frac{12!}{4!8!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495 \).

Step4: Calculate probability

Probability \( P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{C(4, 4)}{C(12, 4)}=\frac{1}{495} \).

Answer:

\(\frac{1}{495}\)