SOLUTION: 4. Given that P(B) = 0.3, P (B/A) = 0.3, P(A∩B) = 0.02, find the following probabilities: a. P(A) b. P(A/B) c. P (A U B) d. P ((A∩B)/(A U B)) e. P (A/(A U B))

Algebra ->  Probability-and-statistics -> SOLUTION: 4. Given that P(B) = 0.3, P (B/A) = 0.3, P(A∩B) = 0.02, find the following probabilities: a. P(A) b. P(A/B) c. P (A U B) d. P ((A∩B)/(A U B)) e. P (A/(A U B))       Log On


   

Question 1192080: 4. Given that P(B) = 0.3, P (B/A) = 0.3, P(A∩B) = 0.02, find the following probabilities:
a. P(A)
b. P(A/B)
c. P (A U B)
d. P ((A∩B)/(A U B))
e. P (A/(A U B))
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Certainly, let's calculate the probabilities step-by-step.
**Given:**
* P(B) = 0.3
* P(B|A) = 0.3
* P(A ∩ B) = 0.02
**a. P(A)**
We can use the formula:
P(B|A) = P(A ∩ B) / P(A)
Rearranging:
P(A) = P(A ∩ B) / P(B|A)
P(A) = 0.02 / 0.3
P(A) = 2/30
P(A) ≈ 0.0667
**b. P(A|B)**
We can use Bayes' Theorem:
P(A|B) = P(B|A) * P(A) / P(B)
P(A|B) = 0.3 * (2/30) / 0.3
P(A|B) = 2/30
P(A|B) ≈ 0.0667
**c. P(A U B)**
We can use the formula:
P(A U B) = P(A) + P(B) - P(A ∩ B)
P(A U B) = (2/30) + 0.3 - 0.02
P(A U B) = 0.3 - 0.02 + (2/30)
P(A U B) ≈ 0.3467
**d. P((A ∩ B) / (A U B))**
This represents the probability of the intersection of A and B given that either A or B occurs.
P((A ∩ B) / (A U B)) = P(A ∩ B) / P(A U B)
P((A ∩ B) / (A U B)) = 0.02 / (0.3 - 0.02 + (2/30))
P((A ∩ B) / (A U B)) ≈ 0.0577
**e. P(A / (A U B))**
This represents the probability of A given that either A or B occurs.
P(A / (A U B)) = P(A) / P(A U B)
P(A / (A U B)) = (2/30) / (0.3 - 0.02 + (2/30))
P(A / (A U B)) ≈ 0.1923
**Summary:**
* P(A) ≈ 0.0667
* P(A|B) ≈ 0.0667
* P(A U B) ≈ 0.3467
* P((A ∩ B) / (A U B)) ≈ 0.0577
* P(A / (A U B)) ≈ 0.1923