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)=1−b−a1−b

Question

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)=1−b−a1−b

Florence Evans · Accepted Answer

Given information:  1. A and B are independent events  Therefore  P(A∩B)=P(A)P(B)  2. P(B)=b  3. P(A∪B―)=1−P(A∪B)=a  Now, we have  a=1−P(A∪B)  =1−(P(A)+P(B)−P(A∩B))  =1−P(A)−P(B)+P(A)P(B)  =1−P(A)−b+P(A)⋅b  Hence,  (1−b)P(A)=1−b−a  P(A)=1−b−a1−b

fudzisako · Accepted Answer

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)=1−b−a1−b.Since A and B are independent, the probability that both A and B occur is the product of their individual probabilities: P(A∩B)=P(A)·P(B).Given that the probability that neither A nor B occurs is a, we can express this as P(A′∩B′)=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(A′∩B′)=1−P(A∪B), where A∪B 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(A∪B)=P(A)+P(B)−P(A∩B).Substituting this into P(A′∩B′)=1−P(A∪B), we have 1−P(A∪B)=1−(P(A)+P(B)−P(A∩B)).Given that P(A′∩B′)=a, we can rewrite the above equation as 1−(P(A)+P(B)−P(A∩B))=a.Simplifying further, we get 1−P(A)−P(B)+P(A∩B)=a.Since A and B are independent, P(A∩B)=P(A)·P(B). Substituting this in, we have 1−P(A)−P(B)+P(A)·P(B)=a.Rearranging the terms, we get P(A)·(1−P(B))=1−P(B)−a.Dividing both sides by (1−P(B)), we obtain P(A)=1−P(B)−a1−P(B).Since P(B)=b, we can substitute this back into the equation to get P(A)=1−b−a1−b.Thus, we have shown that P(A)=1−b−a1−b.

Jazz Frenia · Accepted Answer

Answer:P(A)=1−b−a1−bExplanation: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(A∩B)=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 1−b.Substituting these values into the equation, we get:a=P(¬A)+(1−b).Now, let&#039;s solve for P(¬A) by rearranging the equation:P(¬A)=a−(1−b).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−(1−b)=1.Simplifying the equation, we get:P(A)=1−a+(1−b).Combining like terms, we have:P(A)=1−a−b+1.Finally, simplifying further, we obtain the desired result:P(A)=1−b−a1−b.Thus, we have shown that P(A)=1−b−a1−b.

Andre BalkonE · Accepted Answer

Step 1:Let&#039;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(A∪B)=1−a.Since events A and B are independent, we can express the probability of their union as the sum of their individual probabilities: P(A∪B)=P(A)+P(B)−P(A∩B).Since we know that the probability of event B is b, and events A and B are independent, the probability of their intersection, P(A∩B), can be expressed as the product of their individual probabilities: P(A∩B)=P(A)·P(B).Substituting these expressions into the equation for P(A∪B), we have:1−a=P(A)+b−P(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)=b−a.Factoring out P(A), we have:P(A)·(P(B)−1)=b−a.Finally, we can solve for P(A) by dividing both sides of the equation by (P(B)−1):P(A)=b−aP(B)−1.Since P(B) is given as b, we can substitute it into the equation:P(A)=b−ab−1.Therefore, we have shown that P(A)=1−b−a1−b.

Suppose that A and B are independent events such that

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