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Tree diagrams and conditional probability

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)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Question 3
Given that a bag does NOT contain a forbidden item, what is the probability that is does NOT trigger the alarm?
?2=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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)=P(FA)P(A)
Question 4
Find the probability that a randomly selected bag contains a forbidden item AND triggers the alarm.
P(FA)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Question 5
Find the probability that a randomly selected bag triggers the alarm.
P(A)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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+)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Step 2
Find the probability that a random patient tests positive.
P(+)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Step 3
If a random patient tests positive, what is the probability that they have the disease?
Round to three decimal places.
P(D|+)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Want to join the conversation?

  • blobby green style avatar for user Prashant.Kuls
    What is the difference between P(E1 and E2) and P(E1|E2)?
    (13 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      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).
      (32 votes)
  • blobby green style avatar for user jthabet10
    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?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper happy style avatar for user Adam Simpson
      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.
      (6 votes)
  • blobby green style avatar for user Nicolle West
    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?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • cacteye blue style avatar for user Jerry Nilsson
      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
      (4 votes)
  • blobby green style avatar for user joao.boverio
    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?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user daniella
      You're correct that in this scenario, the events are not independent. The probability of a bag containing a forbidden item (F) triggering the alarm (A) is indeed different from the probability of a bag containing a forbidden item (F) overall. However, the reason why we can calculate P(F ∩ A) as P(F) × P(A) in this case is because of the given structure of the problem. The conditional probability formula, P(A ∣ B) = P(A ∩ B) / P(B), can still be used here, but because we have the direct probabilities for P(F ∩ A) and P(A), we can simply multiply P(F) and P(A) to find P(F ∩ A) due to the structure of the problem.
      (1 vote)
  • blobby green style avatar for user 120738
    The walls are bleeding
    (2 votes)
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  • blobby green style avatar for user yumusana
    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
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user daniella
      Yes, you're correct. To find the probability that a bag containing a forbidden item does NOT trigger the alarm, you multiply the probabilities together: P(F) × P(¬A) = 0.05 × 0.02 = 0.001. This calculates the probability of the bag containing a forbidden item (0.05) and not triggering the alarm (1 − 0.98 = 0.02).
      (1 vote)
  • leaf green style avatar for user Matevž Zorec
    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?
    (2 votes)
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    • piceratops ultimate style avatar for user ANB
      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%!
      (1 vote)
  • piceratops seedling style avatar for user Tyler Johnson
    how do you know when to put it in the D(positive)/D(positive)+N(positive)
    (0 votes)
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    • purple pi pink style avatar for user Constanca Sousa
      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!
      (7 votes)
  • duskpin sapling style avatar for user JJ
    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?
    (1 vote)
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    • leaf green style avatar for user cossine
      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).
      (2 votes)
  • blobby green style avatar for user alfred grainger
    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?
    (2 votes)
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