Question:
If a sample of 2 hammer is selected
(a) find the probability that all in the sample are defective.
(b) find the probability that none in the sample are defective.
Answer:
a [tex]Pr = \frac{2}{110}[/tex]
b [tex]Pr = \frac{72}{110}[/tex]
Step-by-step explanation:
Given
[tex]n = 11[/tex] --- hammers
[tex]r = 2[/tex] --- selection
This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities
Solving (a): Probability that both selection are defective.
For two selections, the probability that all are defective is:
[tex]Pr = P(D) * P(D)[/tex]
[tex]Pr = \frac{2}{11} * \frac{2-1}{11-1}[/tex]
[tex]Pr = \frac{2}{11} * \frac{1}{10}[/tex]
[tex]Pr = \frac{2}{110}[/tex]
Solving (b): Probability that none are defective.
The probability that a selection is not defective is:
[tex]P(D') = \frac{9}{11}[/tex]
For two selections, the probability that all are not defective is:
[tex]Pr = P(D') * P(D')[/tex]
[tex]Pr = \frac{9}{11} * \frac{9-1}{11-1}[/tex]
[tex]Pr = \frac{9}{11} * \frac{8}{10}[/tex]
[tex]Pr = \frac{72}{110}[/tex]