Module #4 Assignment - Probability Theory

 Part A:

1. The probability of Event A occurring is calculated with P(A) = 30/90, which equals approximately a 33% chance of occurring. 

2. The probability of Event B occurring is calculated with P(B) = 30/90, which equals approximately a 33% chance of occurring as well. 

3.  In this case, both A and B occur for 10 outcomes. So, P(A ∩ B) = 10/90 = 0.11. Therefore, P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.33 + 0.33 - 0.11 = 0.56.

4. This formula is used to calculate the probability of either event A or event B occurring. However, this formula assumes that events A and B are mutually exclusive, which means they cannot occur at the same time. If events A and B can occur at the same time, then we need to subtract the probability of both events occurring from the sum of their individual probabilities to avoid double-counting.

Part B

B1. The answer is True.

B2. The calculation is based on Bayes’ theorem, which is a fundamental concept in probability theory and statistics that describes how to update the probabilities of hypotheses when given evidence. It’s used to revise existing predictions or theories (update probabilities) given new or additional evidence. In this case, the evidence is the weatherman’s forecast.

The calculation correctly applies Bayes’ theorem as follows:

P(A1 | B) = P(A1) P(B | A1) / [P(A1) P(B | A1) + P(A2) P(B | A2)]

Substituting in the given probabilities:

P(A1 | B) = (0.014)(0.9) / [(0.014)(0.9) + (0.986)(0.1)] = 0.111

This result means that, given the weatherman has predicted rain, there is an 11% chance that it will actually rain on Jane’s wedding day. This might seem counterintuitive because even though the weatherman predicted rain, the overall probability of rain on any given day is quite low (5 days out of the year), and the weatherman has a 10% rate of false positives where he predicts rain but it doesn’t actually occur. These factors combine to make the actual probability of rain, given the forecast, lower than one might intuitively expect.

Part C 

In this case, the parameters for the dbinom function would be:

x = 10: This is the number of successful operations, which is 10 in this case.

size = 10: This is the total number of operations, which is also 10.

prob = 0.8: This is the probability of a successful operation with no complications. Given that the postoperative complication frequency is 20%, the probability of a successful operation is 1 - 0.20 = 0.80.

The dbinmon function would look like:

dbinom(10, size=10, prob=0.8)