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### Course: AP®︎/College Statistics > Unit 7

Lesson 3: Conditional probability- Conditional probability and independence
- Conditional probability with Bayes' Theorem
- Conditional probability using two-way tables
- Calculate conditional probability
- Conditional probability and independence
- Conditional probability tree diagram example
- Tree diagrams and conditional probability

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

**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 thatof bags contain forbidden items." $5\mathrm{\%}$

## Filling in the tree diagram

"If a bag contains a forbidden item, there is achance that it triggers the alarm." $98\mathrm{\%}$

"If a bag doesn't contain a forbidden item, there is anchance that it triggers the alarm." $8\mathrm{\%}$

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

## 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:

## 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.

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

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

## Want to join the conversation?

- What is the difference between P(E1 and E2) and P(E1|E2)?(13 votes)
- 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).(34 votes)

- 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)
- 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.(7 votes)

- The walls are bleeding(3 votes)
- yes they are, time has come(3 votes)

- 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)
- 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)

- 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)
- 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)

- 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)- 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)

- 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)
- 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)

- how do you know when to put it in the D(positive)/D(positive)+N(positive)(0 votes)
- 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)

- 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)
- 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)

- 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)
- 25%. P(W) = 5white/20 balls(1 vote)