Solution Found!
(Opt.) Before a DVD leaves the factory, it is given a
Chapter 5, Problem 24RE(choose chapter or problem)
Before a DVD leaves the factory, it is given a quality control check. The probabilities that a DVD contains 0, 1, or 2 defects are 0.90, 0.06, and 0.04, respectively. In a sample of 12 recorders, find the probability that 8 have 0 defects, 3 have 1 defect, and 1 has 2 defects.
Questions & Answers
QUESTION:
Before a DVD leaves the factory, it is given a quality control check. The probabilities that a DVD contains 0, 1, or 2 defects are 0.90, 0.06, and 0.04, respectively. In a sample of 12 recorders, find the probability that 8 have 0 defects, 3 have 1 defect, and 1 has 2 defects.
ANSWER:Step 1 of 2
Given the probabilities that a DVD contains 0, 1, or 2 defects are 0.90, 0.06, and 0.04.
The total record is 12.
So, n=12.
Our goal is :
We need to find the probability that 8 have 0 defects, 3 have 1 defect, and 1 has 2 defects.
Then the multinomial distribution formula is
\(P=\frac{n !}{x_{1} ! \times x_{2} ! \times x_{3} !, \ldots, x_{n} !} \times p_{1}^{x_{1} \times p_{2}^{x_{2}} \times p_{3}^{x_{3}} \times, \ldots, p_{n}^{x_{n}}}\)
Where, \(p_{1}=0.90, p_{2}=0.06, \text { and } p_{3}=0.04\)
8 have 0 defects \(x_{1}=8\)
3 have 1 defect \(x_{2}=3\) and
1 has 2 defects \(x_{3}=1\).