# A 2nd year stat/probability question about P(A|B) I'm sooo close

• laura_a
In summary, the conversation is about a 2nd year stat/probability question regarding the conditional probability P(A|B). The goal is to prove that P(A|B) is greater than or equal to 1 - [(1-x)/y], given that P(A)=x, P(B)=y, and x+y>1. The solution involves manipulating the formula P(A|B) = P(A n B) / P(B) by using the fact that P(not A) = 1-x and P(not B) = 1-y, and the equation P(A|B) = P(A n B) / P(B). The final result is P(A|B) >=
laura_a
A 2nd year stat/probability question about P(A|B) I'm sooo close!

## Homework Statement

Consider 2 events A and B such that P(A)=x and P(B)=y, given x,y >0 and x + y > 1

Prove that
P(A|B) >= 1 - [(1-x)/y]

## Homework Equations

P(A|B) = P(A n B) / P(B)

## The Attempt at a Solution

Now I know that P(A)= x so P(not a) = 1-x and P(B) = y and P(Not B) = 1-y

so the answer that I have to prove is actually the same as 1 - P(not a)/P(b)

So all I have to work out is how to go from

P(A|B) = P(A n B) / P(B)

to

1 - P(not a)/P(b)

And I'm going off the part where x + y > 1

What step am I missing? Am I using the right formula? There must be a trick or formula I can manipulate as this is a question at the start of a chapter so should be easyish..?

laura_a said:

## Homework Statement

Consider 2 events A and B such that P(A)=x and P(B)=y, given x,y >0 and x + y > 1

Prove that
P(A|B) >= 1 - [(1-x)/y]

## Homework Equations

P(A|B) = P(A n B) / P(B)

## The Attempt at a Solution

Now I know that P(A)= x so P(not a) = 1-x and P(B) = y and P(Not B) = 1-y

so the answer that I have to prove is actually the same as 1 - P(not a)/P(b)

So all I have to work out is how to go from

P(A|B) = P(A n B) / P(B)

to

1 - P(not a)/P(b)

And I'm going off the part where x + y > 1

What step am I missing? Am I using the right formula? There must be a trick or formula I can manipulate as this is a question at the start of a chapter so should be easyish..?

$$1\geq P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$1\geq x+y-P(A\cap B)$$

$$P(A\cap B)\geq x+y-1$$

$$\frac{P(A\cap B)}{P(B)}\geq \frac{x+y-1}{y}$$

$$P(A|B)\geq \frac{y-(1-x)}{y}$$

$$P(A|B)\geq 1-\frac{1-x}{y}$$

## 1. What is P(A|B)?

P(A|B) refers to the conditional probability of event A occurring given that event B has already occurred. It is calculated by dividing the probability of both A and B occurring by the probability of B occurring.

## 2. How is P(A|B) calculated?

P(A|B) is calculated by dividing the probability of both A and B occurring by the probability of B occurring. This can be represented by the formula P(A|B) = P(A∩B) / P(B).

## 3. What does P(A|B) represent?

P(A|B) represents the probability of event A occurring, taking into account the occurrence of event B. It is a way to measure the likelihood of an event happening, given that another event has already occurred.

## 4. How is P(A|B) different from P(A) or P(B)?

P(A|B) is different from P(A) and P(B) because it takes into account the occurrence of event B. P(A) represents the probability of event A occurring without considering any other events, while P(B) represents the probability of event B occurring without considering any other events.

## 5. Can P(A|B) be greater than 1?

No, P(A|B) cannot be greater than 1. This is because the probability of an event occurring cannot be greater than the probability of both events occurring together. If P(A|B) is greater than 1, it means that the occurrence of event A is more likely than the occurrence of both A and B together, which is not possible.

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