Question 1177190: If P(B) > 0,
(a) Show that P(A|B) + P(A^c|B) = 1.
(b) Show that in general the following two statements are false: (i) P(A|B)+ P(A|B^c) = 1,
(ii) P(A|B) + P(A^c|B^c) = 1
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) > 0
**(a) Show that P(A|B) + P(Aᶜ|B) = 1**
1. **Definition of Conditional Probability:**
* P(A|B) = P(A ∩ B) / P(B)
* P(Aᶜ|B) = P(Aᶜ ∩ B) / P(B)
2. **Sum the Conditional Probabilities:**
* P(A|B) + P(Aᶜ|B) = [P(A ∩ B) / P(B)] + [P(Aᶜ ∩ B) / P(B)]
* P(A|B) + P(Aᶜ|B) = [P(A ∩ B) + P(Aᶜ ∩ B)] / P(B)
3. **Recognize the Union:**
* Note that (A ∩ B) and (Aᶜ ∩ B) are mutually exclusive events (they cannot occur simultaneously).
* Also, their union is (A ∩ B) ∪ (Aᶜ ∩ B) = (A ∪ Aᶜ) ∩ B = S ∩ B = B, where S is the sample space.
4. **Simplify the Sum:**
* P(A ∩ B) + P(Aᶜ ∩ B) = P(B)
5. **Substitute and Conclude:**
* P(A|B) + P(Aᶜ|B) = P(B) / P(B) = 1
**(b) Show that in general the following two statements are false:**
**(i) P(A|B) + P(A|Bᶜ) = 1**
1. **Counterexample:**
* Let's consider a simple example.
* Suppose we have a fair coin toss.
* Let A be the event "heads" and B be the event "the coin is fair."
* P(A) = 1/2, P(B) = 1.
* P(A|B) = 1/2 (probability of heads given the coin is fair).
* Bᶜ means "the coin is not fair," which is a rare event.
* Let's say P(Bᶜ) = 0.01.
* Let's assume P(A|Bᶜ) = 0.8 (if the coin is not fair, it's more likely to be heads).
* P(A|B) + P(A|Bᶜ) = 1/2 + 0.8 = 1.3 ≠ 1.
2. **General Explanation:**
* P(A|B) and P(A|Bᶜ) are conditional probabilities based on different events (B and Bᶜ).
* There is no reason to expect that the sum of these probabilities will always be 1.
* They are not complementary events.
**(ii) P(A|B) + P(Aᶜ|Bᶜ) = 1**
1. **Counterexample:**
* Using the same example as above.
* P(A|B) = 1/2
* P(Aᶜ|Bᶜ) = P(tails|unfair)
* Let's assume P(Aᶜ|Bᶜ) = 0.2 (if the coin is unfair, it's less likely to be tails).
* P(A|B) + P(Aᶜ|Bᶜ) = 1/2 + 0.2 = 0.7 ≠ 1.
2. **General Explanation:**
* P(A|B) and P(Aᶜ|Bᶜ) are conditional probabilities involving different events (B and Bᶜ) and different outcomes (A and Aᶜ).
* There is no general relationship that would guarantee their sum to be 1.
* These events are not complementary.
**Conclusion:**
* **(a)** P(A|B) + P(Aᶜ|B) = 1 is always true when P(B) > 0.
* **(b)** The statements P(A|B) + P(A|Bᶜ) = 1 and P(A|B) + P(Aᶜ|Bᶜ) = 1 are generally false.
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