What is the difference between P(E1 and E2) and P(E1|E2)?

P(E1 and E2) means the probability that E1 and E2 both occur, given no information about what has already occurred. P(E1|E2) means the probability that E1 occurs, given that E2 has already occurred. Example: Suppose a box contains 3 white balls and 5 black balls, and two balls are drawn one at a time without replacement. If E2 is the event that the first ball is white and E1 is the event that the second ball is white, P(E1 and E2) = 3/8 * 2/7 = 3/28, but P(E1|E2) = 2/7. In general, note that since P(E1|E2) = P(E1 and E2)/P(E2), P(E1|E2) is always greater than or equal to P(E1 and E2).

I'm in grade 8 and my teacher is testing us on these probability trees. For question 2, why are the chances of the test testing falsely to be 0.099 and 0.001?

10% of the population have the disease. Of that 10% of the population 99% test positive. So we are calculating 99% of 10% which is 0.10*0.99=0.099. This is the true positive rate (test positive and actually have the disease). Of the 10% of the population that have the disease 1% will have a negative test result. (test negative but actually have the disease). 1% of 10% is 0.10*0.01=0.001.

In question 6, shouldn't you divide the total of all alarms (49+76) by the total number of forbidden objects, which is 50 and not 49?

It's true that out of 1,000 bags we can assume that 50 of them (5%) contain forbidden items – but only 49 of them (98% of 50) will trigger the alarm. Likewise, there are 950 legal bags out of which 76 (8%) will also trigger the alarm. So, out of the thousand bags, 49 + 76 = 125 will trigger the alarm, and out of those 125 bags, 49 will contain forbidden items (the 50th forbidden bag will just pass through security without being noticed). So, the probability that a bag that triggers the alarm also contains a forbidden item is 49∕125 = 0.392

First example in problem4, I was taught that P(F ∩ A) is =P(F|A) . P(A) if its dependent, but its not the case here. Independent event formula P(F ∩ A)= P(F). P(A) is what its used here, so do we just assume "contain forbidden item(F)" and " triggering alarm (A)" are both independent events?

So the definition of conditional probability is P(A|B) = P(A and B)/P(B). This definition will hold for all scenarios including dependent/independent events. By multiplying by P(B) we get P(A|B)*P(B) = P(A and B). Definition: Two events are independent if P(A|B) = P(A). If this equation does not hold then the two events are said to be not independent. Note it can be proven if P(A|B)= P(A) then P(B|A) = P(B). I will leave the proof as an exercise. This comes from the fact if two events are independent then P(A and B) = P(A)*P(B).

how do you know when to put it in the D(positive)/D(positive)+N(positive)

You use the conditional probability formula which is P(A/B) = P(A and B)/P(B). P(A/B) translates to "the probability of A occuring given that B has occured". In the above question they ask "If a random patient tests positive, what is the probability that they have the disease?". In other words, "*given* that a random patient tests positive, what is the probability....." so usually there will be the command words, "if... what/then" or "given" and that's when you will know that you have to use the conditional probability formula. Hope it helps!

Main content

AP®︎/College Statistics

Course: AP®︎/College Statistics > Unit 7

Lesson 3: Conditional probability

Tree diagrams and conditional probability

Google Classroom

Example: Bags at an airport

An airport screens bags for forbidden items, and an alarm is supposed to be triggered when a forbidden item is detected.

Suppose that $5 %$ ‍ of bags contain forbidden items.
If a bag contains a forbidden item, there is a $98 %$ ‍ chance that it triggers the alarm.
If a bag doesn't contain a forbidden item, there is an $8 %$ ‍ chance that it triggers the alarm.

Given a randomly chosen bag triggers the alarm, what is the probability that it contains a forbidden item?

Let's break up this problem into smaller parts and solve it step-by-step.

Starting a tree diagram

The chance that the alarm is triggered depends on whether or not the bag contains a forbidden item, so we should first distinguish between bags that contain a forbidden item and those that don't.

"Suppose that $5 %$ ‍ of bags contain forbidden items."

Question 1

What is the probability that a randomly chosen bag does NOT contain a forbidden item?

P (not forbidden) =

Filling in the tree diagram

"If a bag contains a forbidden item, there is a $98 %$ ‍ chance that it triggers the alarm."

"If a bag doesn't contain a forbidden item, there is an $8 %$ ‍ chance that it triggers the alarm."

We can use these facts to fill in the next branches in the tree diagram like this:

Question 2

Given that a bag contains a forbidden item, what is the probability that it does NOT trigger the alarm?

?_{1} =

Question 3

Given that a bag does NOT contain a forbidden item, what is the probability that is does NOT trigger the alarm?

?_{2} =

Completing the tree diagram

We multiply the probabilities along the branches to complete the tree diagram.

Here's the completed diagram:

Solving the original problem

"Given a randomly chosen bag triggers the alarm, what is the probability that it contains a forbidden item?"

Use the probabilities from the tree diagram and the conditional probability formula:

P (forbidden | alarm) = \frac{P (F \cap A)}{P (A)}

[What do those symbols mean?]

Question 4

Find the probability that a randomly selected bag contains a forbidden item AND triggers the alarm.

P (F \cap A) =

Question 5

Find the probability that a randomly selected bag triggers the alarm.

P (A) =

Question 6

Given a randomly chosen bag triggers the alarm, what is the probability that it contains a forbidden item?
Use three decimal places in your answer.

P (F | A) =

[Wait, why is that probability so low?]

Try one on your own!

A hospital is testing patients for a certain disease. If a patient has the disease, the test is designed to return a "positive" result. If a patient does not have the disease, the test should return a "negative" result. No test is perfect though.

$99 %$ ‍ of patients who have the disease will test positive.
$5 %$ ‍ of patients who don't have the disease will also test positive.
$10 %$ ‍ of the population in question has the disease.

If a random patient tests positive, what is the probability that they have the disease?

Step 1

Find the probability that a randomly selected patient has the disease AND tests positive.

P (D \cap +) =

Step 2

Find the probability that a random patient tests positive.

P (+) =

Step 3

If a random patient tests positive, what is the probability that they have the disease?
Round to three decimal places.

P (D | +) =

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Prashant.Kuls
Posted 6 years ago. Direct link to Prashant.Kuls's post “What is the difference be...”
What is the difference between P(E1 and E2) and P(E1|E2)?
Button navigates to signup pageComment on Prashant.Kuls's post “What is the difference be...”
(12 votes)
Answer
- Ian Pulizzotto
  Posted 6 years ago. Direct link to Ian Pulizzotto's post “P(E1 and E2) means the pr...”
  P(E1 and E2) means the probability that E1 and E2 both occur, given no information about what has already occurred.
  P(E1|E2) means the probability that E1 occurs, given that E2 has already occurred.
  Example: Suppose a box contains 3 white balls and 5 black balls, and two balls are drawn one at a time without replacement. If E2 is the event that the first ball is white and E1 is the event that the second ball is white, P(E1 and E2) = 3/8 * 2/7 = 3/28, but P(E1|E2) = 2/7.
  
  In general, note that since P(E1|E2) = P(E1 and E2)/P(E2), P(E1|E2) is always greater than or equal to P(E1 and E2).
  Comment on Ian Pulizzotto's post “P(E1 and E2) means the pr...”
  (30 votes)
jthabet10
Posted 5 years ago. Direct link to jthabet10's post “I'm in grade 8 and my tea...”
I'm in grade 8 and my teacher is testing us on these probability trees. For question 2, why are the chances of the test testing falsely to be 0.099 and 0.001?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Adam Simpson
  Posted 4 years ago. Direct link to Adam Simpson's post “10% of the population hav...”
  10% of the population have the disease. Of that 10% of the population 99% test positive. So we are calculating 99% of 10% which is 0.10*0.99=0.099. This is the true positive rate (test positive and actually have the disease).
  Of the 10% of the population that have the disease 1% will have a negative test result. (test negative but actually have the disease). 1% of 10% is 0.10*0.01=0.001.
  Button navigates to signup page
  (6 votes)
joao.boverio
Posted 4 years ago. Direct link to joao.boverio's post “So they are not independe...”
So they are not independent right? P(F∣A)=0.392 != P(F) = 0.05 ? But why we an calculate P(F&A) as P(F)*P(A) ? Isn't this case only for independent events?
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(3 votes)
Answer
Nicolle West
Posted 5 years ago. Direct link to Nicolle West's post “In question 6, shouldn't ...”
In question 6, shouldn't you divide the total of all alarms (49+76) by the total number of forbidden objects, which is 50 and not 49?
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(2 votes)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “It's true that out of 1,0...”
  It's true that out of 1,000 bags we can assume that 50 of them (5%) contain forbidden items – but only 49 of them (98% of 50) will trigger the alarm.
  
  Likewise, there are 950 legal bags out of which 76 (8%) will also trigger the alarm.
  
  So, out of the thousand bags, 49 + 76 = 125 will trigger the alarm, and out of those 125 bags, 49 will contain forbidden items (the 50th forbidden bag will just pass through security without being noticed).
  
  So, the probability that a bag that triggers the alarm also contains a forbidden item is 49∕125 = 0.392
  Button navigates to signup page
  (3 votes)
yumusana
Posted 4 months ago. Direct link to yumusana's post “Given that a bag contains...”
Given that a bag contains a forbidden item, what is the probability that it does NOT trigger the alarm?

I think the solution to this question is = 0.05x0.02= 0.001
Button navigates to signup pageComment on yumusana's post “Given that a bag contains...”
(2 votes)
Answer
Matevž Zorec
Posted 2 years ago. Direct link to Matevž Zorec's post “So if they were to test p...”
So if they were to test positive again... we should just use the probability they were sick in the first place from the first positive test to calculate the overall probability they are sick?
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(2 votes)
Answer
- ANB
  Posted a year ago. Direct link to ANB's post “Good question. If they te...”
  Good question. If they test positive the first time, that means that there is a 68.75% chance that they have the disease. To calculate the probability that they have the disease after testing positive twice, we use .6875 instead of .05 as we used when trying to calculate the probability the first time.
  
  If the person tests positive two times in a row for this disease, the chances that they have the disease is 97.76%!
  Button navigates to signup page
  (1 vote)
Tyler Johnson
Posted 6 years ago. Direct link to Tyler Johnson's post “how do you know when to p...”
how do you know when to put it in the D(positive)/D(positive)+N(positive)
Button navigates to signup pageComment on Tyler Johnson's post “how do you know when to p...”
(0 votes)
Answer
- Constanca Sousa
  Posted 6 years ago. Direct link to Constanca Sousa's post “You use the conditional p...”
  You use the conditional probability formula which is P(A/B) = P(A and B)/P(B). P(A/B) translates to "the probability of A occuring given that B has occured". In the above question they ask "If a random patient tests positive, what is the probability that they have the disease?". In other words, "given that a random patient tests positive, what is the probability....." so usually there will be the command words, "if... what/then" or "given" and that's when you will know that you have to use the conditional probability formula.
  
  Hope it helps!
  Comment on Constanca Sousa's post “You use the conditional p...”
  (7 votes)
JJ
Posted a month ago. Direct link to JJ's post “First example in problem4...”
First example in problem4, I was taught that P(F ∩ A) is =P(F|A) . P(A) if its dependent, but its not the case here. Independent event formula P(F ∩ A)= P(F). P(A) is what its used here, so do we just assume "contain forbidden item(F)" and " triggering alarm (A)" are both independent events?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- cossine
  Posted a month ago. Direct link to cossine's post “So the definition of cond...”
  So the definition of conditional probability is
  
  P(A|B) = P(A and B)/P(B).
  
  This definition will hold for all scenarios including dependent/independent events.
  
  By multiplying by P(B) we get
  
  P(A|B)*P(B) = P(A and B).
  
  Definition: Two events are independent if
  
  P(A|B) = P(A).
  
  If this equation does not hold then the two events are said to be not independent.
  
  Note it can be proven if P(A|B)= P(A) then P(B|A) = P(B). I will leave the proof as an exercise. This comes from the fact if two events are independent then P(A and B) = P(A)*P(B).
  Comment on cossine's post “So the definition of cond...”
  (2 votes)
alfred grainger
Posted 3 years ago. Direct link to alfred grainger's post “10 red balls are in a bag...”
10 red balls are in a bag along with 5 white, 4 blue and 1 green. What is the probability that a white ball is randomly selected from the bag in one pick?
Button navigates to signup pageComment on alfred grainger's post “10 red balls are in a bag...”
(2 votes)
Answer
- rupeshmoosari
  Posted 4 months ago. Direct link to rupeshmoosari's post “25%. P(W) = 5white/20 ba...”
  25%. P(W) = 5white/20 balls
  Button navigates to signup page
  (1 vote)
aveloz
Posted a month ago. Direct link to aveloz's post “why is it that the formul...”
why is it that the formula is P(A and B) OVER P(B) but when we plug in the number it switches places. now the B is the numerator and the 2 numbers are now at the bottom.
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(1 vote)
Answer