A manufacturing line produces mechanical parts. It is estimated that $5 \%$ of the parts produced by this line are defective.
An engineer has developed a test to apply to the parts. This test has two possible results: ``positive'' or ``negative''. This test is applied to a part chosen at random from the production of the line. We denote $p ( E )$ the probability of an event $E$. We consider the following events:
- $D$: ``the part is defective'';
- T: ``the part shows a positive test'';
- $\bar { D }$ and $\bar { T }$ denote respectively the complementary events of $D$ and $T$.
Given the characteristics of the test, we know that:
- The probability that a part shows a positive test given that it is defective is equal to 0.98;
- the probability that a part shows a negative test given that it is not defective is equal to 0.97.
PART I
- Represent the situation using a probability tree.
- a. Determine the probability that a part chosen at random from the production line is defective and shows a positive test. b. Prove that: $p ( T ) = 0.0775$.
- The positive predictive value of the test is called the probability that a part is defective given that the test is positive. A test is considered effective if it has a positive predictive value greater than 0.95.
Calculate the positive predictive value of this test and specify whether it is effective.
PART II
A sample of 20 parts is chosen from the production line, treating this choice as a draw with replacement. Let $X$ be the random variable that gives the number of defective parts in this sample. Recall that: $p ( D ) = 0.05$.
- Justify that $X$ follows a binomial distribution and determine the parameters of this distribution.
- Calculate the probability that this sample contains at least one defective part.
Give a result rounded to the nearest hundredth.
3. Calculate the expected value of the random variable $X$ and interpret the result obtained.