Question 1177191: If P(B) = p,P(A^c|B) = q, andP(A^c ∩ B^c) = r, find :
(a)P(A ∩ B^c),
(b)P(A),
(c)P(B|A).
thank you :)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**Given:**
* P(B) = p
* P(Aᶜ|B) = q
* P(Aᶜ ∩ Bᶜ) = r
**Understanding the Notation:**
* Aᶜ: Complement of A (not A)
* Bᶜ: Complement of B (not B)
* A ∩ B: Intersection of A and B (A and B)
* A ∪ B: Union of A and B (A or B or both)
* P(A|B): Conditional probability of A given B
**Solving the Problem:**
**(a) P(A ∩ Bᶜ)**
1. **Use the Complement Rule:**
* P(Bᶜ) = 1 - P(B) = 1 - p
2. **Use De Morgan's Law:**
* (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ
* P((A ∪ B)ᶜ) = P(Aᶜ ∩ Bᶜ) = r
3. **Relate P(A ∪ B) to P(A ∩ Bᶜ):**
* P(A ∪ B) = 1 - P((A ∪ B)ᶜ) = 1 - r
4. **Use the Inclusion-Exclusion Principle:**
* P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
* 1 - r = P(A) + p - P(A ∩ B)
5. **Use the Conditional Probability P(Aᶜ|B) = q:**
* P(Aᶜ|B) = P(Aᶜ ∩ B) / P(B) = q
* P(Aᶜ ∩ B) = qP(B) = qp
6. **Use the Complement Rule for P(B):**
* P(B) = P(A ∩ B) + P(Aᶜ ∩ B)
* p = P(A ∩ B) + qp
* P(A ∩ B) = p - qp = p(1 - q)
7. **Find P(A ∩ Bᶜ):**
* P(A) = P(A ∩ B) + P(A ∩ Bᶜ)
* From step 4, P(A) = 1-r - p + P(A ∩ B)
* P(A) = 1-r-p + p(1-q)
* P(A) = 1-r-pq
* P(A ∩ B^c) = P(A) - P(A ∩ B)
* P(A ∩ B^c) = 1-r-pq - p(1-q)
* P(A ∩ B^c) = 1-r-pq - p + pq
* P(A ∩ B^c) = 1-r-p
**(b) P(A)**
* From step 7 in part (a):
* P(A) = 1 - r - pq
**(c) P(B|A)**
1. **Use the Conditional Probability Formula:**
* P(B|A) = P(A ∩ B) / P(A)
2. **Substitute the values:**
* P(B|A) = p(1 - q) / (1 - r - pq)
**Final Answers:**
* **(a) P(A ∩ Bᶜ) = 1 - r - p**
* **(b) P(A) = 1 - r - pq**
* **(c) P(B|A) = p(1 - q) / (1 - r - pq)**
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