If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
0.42, 0.99 respectively
Understanding Probability Concepts: Conditional Probability and Union
This problem involves calculating two important probabilities: conditional probability, specifically P(A/B), and the probability of the union of two events, P(A ∪ B). We are given the individual probabilities of events A and B, and the conditional probability of event B given event A has occurred, P(B/A).
Given Information:
- Probability of event A, P(A) = 0.7
- Probability of event B, P(B) = 0.5
- Conditional probability of event B given A, P(B/A) = 0.3
Step-by-Step Calculation of P(A ∩ B)
To find P(A/B) and P(A ∪ B), we first need to calculate the probability of the intersection of events A and B, denoted as P(A ∩ B). This is the probability that both A and B occur.
The formula for conditional probability P(B/A) relates the intersection probability to the individual probability of A:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
We can rearrange this formula to find P(A ∩ B):
\( P(A \cap B) = P(B/A) \times P(A) \)
Substitute the given values:
\( P(A \cap B) = 0.3 \times 0.7 \)
\( P(A \cap B) = 0.21 \)
So, the probability that both A and B occur is 0.21.
Calculating P(A/B): Conditional Probability of A given B
Now we can calculate the conditional probability of event A given event B has occurred, P(A/B). The formula for P(A/B) relates the intersection probability to the individual probability of B:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
We have calculated P(A ∩ B) as 0.21, and we are given P(B) as 0.5. Substitute these values:
\( P(A/B) = \frac{0.21}{0.5} \)
\( P(A/B) = 0.42 \)
Thus, the conditional probability of A given B is 0.42.
Calculating P(A ∪ B): Probability of the Union of A and B
Next, we calculate the probability of the union of events A and B, denoted as P(A ∪ B). This is the probability that event A occurs, or event B occurs, or both occur.
The formula for the probability of the union of two events is:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
We are given P(A) = 0.7 and P(B) = 0.5, and we calculated P(A ∩ B) = 0.21. Substitute these values into the formula:
\( P(A \cup B) = 0.7 + 0.5 - 0.21 \)
\( P(A \cup B) = 1.2 - 0.21 \)
\( P(A \cup B) = 0.99 \)
Therefore, the probability of the union of A and B is 0.99.
Summary of Results
Based on the calculations:
- (i) P(A/B) = 0.42
- (ii) P(A ∪ B) = 0.99
The values are 0.42 and 0.99 respectively.
Revision Table: Key Probability Formulas
| Concept | Formula |
|---|---|
| Conditional Probability (A given B) | \( P(A/B) = \frac{P(A \cap B)}{P(B)} \) |
| Conditional Probability (B given A) | \( P(B/A) = \frac{P(A \cap B)}{P(A)} \) |
| Intersection of A and B | \( P(A \cap B) = P(B/A) \times P(A) \) or \( P(A \cap B) = P(A/B) \times P(B) \) |
| Union of A and B | \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) |
Additional Information: Understanding Events
In probability, events are outcomes or sets of outcomes from an experiment. Understanding the relationship between events is crucial for applying the correct formulas.
- Intersection (A ∩ B): Represents the event where both A and B occur. This is the 'and' condition.
- Union (A ∪ B): Represents the event where A occurs, or B occurs, or both occur. This is the 'or' condition.
- Conditional Probability (A/B): Represents the probability that event A occurs given that event B has already occurred. The occurrence of B affects the sample space for A.
- Mutually Exclusive Events: Events A and B are mutually exclusive if they cannot occur at the same time, meaning P(A ∩ B) = 0. In this case, the union formula simplifies to P(A ∪ B) = P(A) + P(B). The events in this problem are not mutually exclusive since P(A ∩ B) = 0.21 ≠ 0.
- Independent Events: Events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this means P(A/B) = P(A) and P(B/A) = P(B). Also, P(A ∩ B) = P(A) × P(B). In this problem, P(B/A) = 0.3 while P(B) = 0.5, so the events are not independent.
Important Questions from Conditional Probability
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
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