Math Problem Statement

Consider two events A and B.   You know that the probability of B given A is 0.29, and the probability of B given that A has not occured is 0.27.    Also it is known that the probability of A occuring is 0.26.    Calculate to four decimal places the probility that A has occured given that B occurs. 

Solution

To calculate the probability that event AA has occurred given that event BB occurs, denoted as P(AB)P(A \mid B), we can use Bayes' Theorem.

Bayes' Theorem is formulated as:

P(AB)=P(BA)P(A)P(B)P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}

where:

  • P(AB)P(A \mid B) is the probability of AA given BB,
  • P(BA)P(B \mid A) is the probability of BB given AA,
  • P(A)P(A) is the probability of AA,
  • P(B)P(B) is the probability of BB.

Given Information

  • P(BA)=0.29P(B \mid A) = 0.29: the probability of BB given AA.
  • P(BAc)=0.27P(B \mid A^c) = 0.27: the probability of BB given AA has not occurred (where AcA^c denotes the complement of AA).
  • P(A)=0.26P(A) = 0.26: the probability of AA.

We need to find P(AB)P(A \mid B).

Step-by-Step Solution

  1. Find P(B)P(B) using the Law of Total Probability:

    The Law of Total Probability states that if AA and AcA^c are mutually exclusive and exhaustive events, then:

    P(B)=P(BA)P(A)+P(BAc)P(Ac)P(B) = P(B \mid A) \cdot P(A) + P(B \mid A^c) \cdot P(A^c)

    Here, P(Ac)=1P(A)P(A^c) = 1 - P(A).

    So, we calculate P(B)P(B) as:

    P(B)=(0.29)(0.26)+(0.27)(10.26)P(B) = (0.29) \cdot (0.26) + (0.27) \cdot (1 - 0.26)

    Let's calculate P(B)P(B):

P(B)=(0.29×0.26)+(0.27×0.74)P(B) = (0.29 \times 0.26) + (0.27 \times 0.74)

  1. Compute P(B)P(B):

P(B)=(0.0754)+(0.1998)P(B) = (0.0754) + (0.1998)

P(B)=0.2752P(B) = 0.2752

  1. Apply Bayes' Theorem to find P(AB)P(A \mid B):

P(AB)=P(BA)P(A)P(B)P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}

Substitute the values:

P(AB)=0.290.260.2752P(A \mid B) = \frac{0.29 \cdot 0.26}{0.2752}

Now calculate:

P(AB)=0.07540.2752P(A \mid B) = \frac{0.0754}{0.2752}

P(AB)0.2741P(A \mid B) \approx 0.2741

Final Answer

The probability that AA has occurred given that BB occurs is 0.2741 to four decimal places.

Would you like more details, or do you have any questions?


Here are 5 related questions for further exploration:

  1. What is the probability that event AcA^c occurs given that event BB occurs?
  2. How would the probability P(AB)P(A \mid B) change if P(BA)P(B \mid A) was 0.5 instead of 0.29?
  3. How can you use the complement rule to verify your calculations?
  4. What would the calculation look like if events AA and BB were independent?
  5. How does Bayes' Theorem apply to continuous probability distributions?

Tip: Remember that Bayes' Theorem is very powerful for updating probabilities when new information is available, particularly in real-world applications like medical testing and machine learning.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Theory
Bayes' Theorem
Conditional Probability

Formulas

Bayes' Theorem

Theorems

Law of Total Probability

Suitable Grade Level

Undergraduate