Understanding Probabilities: From Rain Predictions to Surgical Success

Welcome back!

 In this blog, we’ll dive into a series of intriguing probability problems, from assessing the reliability of weather forecasts to evaluating the success of surgical procedures. We’ll explore how to apply Bayes' Theorem to understand the likelihood of rain on an important day and use statistical tools to gauge the effectiveness of a new surgical method. 

A. Based on Table 1, let's calculate the probabilities:

The total of all the numbers in the table is:

$$10 + 20 + 20 + 40 = 90$$

• A1. Event A (probability of A happening):

$$P(A) = \frac{\text{Total for row A}}{\text{Grand total}} = \frac{10 + 20}{90} = \frac{30}{90} = 0.333$$

• A2. Event B (probability of B happening):

$$P(B) = \frac{\text{Total for column B}}{\text{Grand total}} = \frac{10 + 20}{90} = \frac{30}{90} = 0.333$$

• A3. Event A or B (probability of A or B happening): We need to find $P(A \cup B)$, the union of A and B. This is calculated by:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Where $P(A \cap B)$ is the intersection, i.e., the probability of both A and B occurring, which is given by the value in cell $A \cap B$, which is 10.

$$P(A \cup B) = \frac{30}{90} + \frac{30}{90} - \frac{10}{90} = \frac{30 + 30 - 10}{90} = \frac{50}{90} = 0.556$$

• A4. $P(A \cup B) = P(A) + P(B)$: This equation is false because the correct formula is:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

We already calculated $P(A \cup B) = 0.556$, and the sum of $P(A) + P(B) = 0.333 + 0.333 = 0.666$, which is not the same as 0.556. Therefore, you need to subtract the intersection.

B1. True

B2. Explanation:

This result is derived correctly using Bayes' Theorem, which allows us to compute the probability of an event given the occurrence of another related event.

In this case, we want to find the probability that it will rain on the day of Jane’s wedding given that the weatherman has predicted rain, i.e., $P(A1 | B)$.

Using Bayes' Theorem:

$$P(A1 | B) = \frac{P(A1) \cdot P(B | A1)}{P(A1) \cdot P(B | A1) + P(A2) \cdot P(B | A2)}$$

Where:

• $P(A1) = 0.0136985$ (the probability it rains),
• $P(A2) = 0.9863014$ (the probability it doesn't rain),
• $P(B | A1) = 0.9$ (the probability the weatherman predicts rain when it rains),
• $P(B | A2) = 0.1$ (the probability the weatherman predicts rain when it doesn't rain).

Substituting these values into the equation:

$$P(A1 | B) = \frac{(0.0136985 \cdot 0.9)}{(0.0136985 \cdot 0.9) + (0.9863014 \cdot 0.1)}$$

This simplifies to:

$$P(A1 | B) = \frac{0.01232865}{0.01232865 + 0.09863014} = \frac{0.01232865}{0.11095879} \approx 0.111$$

This means that, given the weatherman predicts rain, there is about an 11% chance it will actually rain. This is lower than one might expect given the weatherman's 90% accuracy when it does rain, but this low result is due to the fact that it rarely rains in the desert (only 5 days out of the year).

C Explanation:

To find the probability of having 10 successful operations (i.e., no complications) using the traditional method, you can use the dbinom function from R.

Here’s how you set it up:

• size: The number of trials, which is 10.
• prob: The probability of success on each trial. For the traditional method with a 20% complication rate, the success rate is $1 - 0.20 = 0.80$.

So you want the probability of having 0 complications out of 10 trials, which is:

INPUT:

The result of this calculation is approximately $0.1074$

OUTPUT: