Suppose that A and B are independent events such that

ahgan3j

ahgan3j

Answered question

2021-11-15

Suppose that A and B are independent events such that the probability that neither occurs is a and the probability of B is b. Show that
P(A)=1ba1b

Answer & Explanation

Florence Evans

Florence Evans

Beginner2021-11-16Added 16 answers

Given information:
1. A and B are independent events
Therefore
P(AB)=P(A)P(B)
2. P(B)=b
3. P(AB)=1P(AB)=a
Now, we have
a=1P(AB)
=1(P(A)+P(B)P(AB))
=1P(A)P(B)+P(A)P(B)
=1P(A)b+P(A)b
Hence,
(1b)P(A)=1ba
P(A)=1ba1b
fudzisako

fudzisako

Skilled2023-06-14Added 105 answers

Suppose that A and B are independent events such that the probability that neither occurs is a, and the probability of B is b. We want to show that P(A)=1ba1b.
Since A and B are independent, the probability that both A and B occur is the product of their individual probabilities: P(AB)=P(A)·P(B).
Given that the probability that neither A nor B occurs is a, we can express this as P(AB)=a, where A and B represent the complements of A and B, respectively.
The complement of an event A is the event that A does not occur. So, A represents the event that A does not occur, and B represents the event that B does not occur.
Using the complement rule, we can write P(AB)=1P(AB), where AB represents the union of events A and B.
Since A and B are independent, the probability of their union is the sum of their individual probabilities minus the product of their probabilities: P(AB)=P(A)+P(B)P(AB).
Substituting this into P(AB)=1P(AB), we have 1P(AB)=1(P(A)+P(B)P(AB)).
Given that P(AB)=a, we can rewrite the above equation as 1(P(A)+P(B)P(AB))=a.
Simplifying further, we get 1P(A)P(B)+P(AB)=a.
Since A and B are independent, P(AB)=P(A)·P(B). Substituting this in, we have 1P(A)P(B)+P(A)·P(B)=a.
Rearranging the terms, we get P(A)·(1P(B))=1P(B)a.
Dividing both sides by (1P(B)), we obtain P(A)=1P(B)a1P(B).
Since P(B)=b, we can substitute this back into the equation to get P(A)=1ba1b.
Thus, we have shown that P(A)=1ba1b.
Jazz Frenia

Jazz Frenia

Skilled2023-06-14Added 106 answers

Answer:
P(A)=1ba1b
Explanation:
We are given that the probability that neither event A nor event B occurs is a, and the probability of event B is b. We need to find the probability of event A, denoted as P(A).
Since A and B are independent events, we know that the probability of both events occurring is the product of their individual probabilities, i.e., P(AB)=P(A)·P(B).
The probability of neither event A nor event B occurring, denoted as P(¬A¬B), is equal to a. Since A and B are mutually exclusive (if one occurs, the other cannot occur), we can write this probability as the sum of the probabilities of the complementary events:
P(¬A¬B)=P(¬A)+P(¬B).
We also know that the probability of event B occurring is b. Therefore, the probability of the complementary event, event B not occurring (P(¬B)), is 1b.
Substituting these values into the equation, we get:
a=P(¬A)+(1b).
Now, let's solve for P(¬A) by rearranging the equation:
P(¬A)=a(1b).
Since A and ¬A are complementary events (either one occurs or the other occurs), we have P(A)+P(¬A)=1. Therefore, we can substitute P(¬A) in terms of P(A):
P(A)+a(1b)=1.
Simplifying the equation, we get:
P(A)=1a+(1b).
Combining like terms, we have:
P(A)=1ab+1.
Finally, simplifying further, we obtain the desired result:
P(A)=1ba1b.
Thus, we have shown that P(A)=1ba1b.
Andre BalkonE

Andre BalkonE

Skilled2023-06-14Added 110 answers

Step 1:
Let's start by defining the probability of event A as P(A). We are given that events A and B are independent, with the probability that neither occurs being a and the probability of event B being b.
Now, the probability that either event A or event B occurs (or both) can be represented as the complement of the probability that neither occurs. In other words, P(AB)=1a.
Since events A and B are independent, we can express the probability of their union as the sum of their individual probabilities: P(AB)=P(A)+P(B)P(AB).
Since we know that the probability of event B is b, and events A and B are independent, the probability of their intersection, P(AB), can be expressed as the product of their individual probabilities: P(AB)=P(A)·P(B).
Substituting these expressions into the equation for P(AB), we have:
1a=P(A)+bP(A)·P(B).
Step 2:
Now, we can solve for P(A) in terms of a and b. Rearranging the equation, we get:
P(A)·P(B)P(A)=ba.
Factoring out P(A), we have:
P(A)·(P(B)1)=ba.
Finally, we can solve for P(A) by dividing both sides of the equation by (P(B)1):
P(A)=baP(B)1.
Since P(B) is given as b, we can substitute it into the equation:
P(A)=bab1.
Therefore, we have shown that P(A)=1ba1b.

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