5  Probability by Counting for Equally Likely Outcomes

5.1 Probability for Equally Likely Outcomes

Let \(E\) be an event from a finite sample space \(\Omega\) in which all outcomes are equally likely. The probability of \(E\), denoted \(P(E)\) is defined by \[P(E) = \frac{|E|}{|\Omega|}\] Example: A company has 200 employees. 50 of these employees are smokers. One employee is selected at random. What is the probability that the selected employee is a smoker (S)?

Solution: \[P(S) = \frac{|S|}{|\Omega|} = \frac{50}{200} = .25\]

5.2 Probability Rules for Compound Events

Proposition: For equally likely outcomes, the following rules hold: Negation Rule: \[P(E^c) = 1 - P(E)\] Disjunction Rule: \[P(A\cup B) = P(A) + P(B) - P(A\cap B)\] Addition Rule for Mutually Exclusive Events: \[\text{If } A\cap B = \emptyset, \text{ then } P(A\cup B) = P(A) + P(B)\]

Example: A manufacturer has received a shipment of 50 parts. Unfortunately, 20 of the parts are defective. The manufacturer is going to test a sample of 5 parts chosen at random from the shipment. What is the probability that the sample contains - 3 defective parts and 2 good parts? - All defective parts? - No defective parts?

Solution: - \(\frac{{20 \choose 3}{30 \choose 2}}{50 \choose 5} = \frac{495,900}{2,119,760} \approx .234\) - \(\frac{{20 \choose 5}{30 \choose 0}}{50 \choose 5} = \frac{15,504}{2,119,760} \approx .007\) - \(\frac{{20 \choose 0}{30 \choose 5}}{50 \choose 5} = \frac{142,506}{2,119,760} \approx .067\)

Example: What is the probability that at least two out of forty people have the same birthday? Solution: The probability that at least two have the same birthday is: \[1 - \frac{P(365,40)}{365^{40}} \approx .891\]