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Use conditional probability to see if events are independent or not.
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- How does the logic work when we conclude that P(delayed) is independent to P(delayed | snowy) when they are approximately the same?(8 votes)
- When they are approx the same, it means that the probability of a delay is the same whether or not it snows. In other words, the probability of delay has nothing to do with whether or not it snows; the event Delay is independent from the event Snow. Not sure if that helps/answers your question! :)(52 votes)
- why do we do (delayed|snowy) but not (snowy|delayed)? How do we know which one to use for future questions that are similar.(8 votes)
- I do not understand how can we understand that is it independent or dependent the problem after solving it.(1 vote)
- Intuitively, in other case whenever the value of P(delayed) = P(delayed | snowy), we could see that with or without the snowy condition, the probability of the flight delay stays the same. So we can conclude that they're independent to each other. But in this case, their values are different. So the existence snowy condition affects the flight delay probability and they're dependent.(14 votes)
- I don't understand how to find if it is independent or dependent. What makes something independent or dependent?(3 votes)
- When something is independent, then the probability of A given B is the same as the probability of just A. In other words, both the probability of A and probability of B do not affect one another.(5 votes)
- At1:23,Sal talks about the existence of a theoretical probability value. I do not quite understand this. Does this mean that there is an exact real number assigned to every probability we try to calculate and even though it is magically hidden from us, it somehow exists in some place?(Where,actually?) I understand that the more sample we have, the more confident we can be about our calculated probability but this notion of absolute theoretical probability is bugging me.(3 votes)
- I have this problem in my book and I am very frustrated.
Four trumpet players' instruments are mixed up and the trumpets are given to the players just before concert. What is the probability that no one gets his or her trumpet back. The answer in the book is 3/8(2 votes)
- Interesting problem!
We can use the inclusion-exclusion principle to find the probability that at least one player gets his/her trumpet back. Then we can subtract this from 1.
Let A, B, C, D represent the events that the first, second, third, and fourth player, respectively, individually gets his/her trumpet back.
Note that the probability that a specific collection of k of these events all occur is 1/(4Pk).
The probability that at least one player gets his/her trumpet back is
= [P(A) + P(B) + P(C) + P(D)] - [P(A n B) + P(A n C) + P(A n D) + P(B n C) + P(B n D) + P(C n D)] + [P(A n B n C) + P(A n B n D) + P(A n C n D) + P(B n C n D)] - P(A n B n C n D)
= (4C1)/(4P1) - (4C2)/(4P2) + (4C3)/(4P3) - (4C4)/(4P4)
= 1/1! - 1/2! + 1/3! - 1/4!
= 1 - 1/2 + 1/6 - 1/24
So the probability that no player gets his/her trumpet back is 1-5/8 = 3/8.
We can extend this to n players. A similar calculation gives a final answer of
1/2! - 1/3! + 1/4! - ... + (-1)^n / n!.
This implies that as n becomes large, the final answer approaches 1/e.
Have a blessed, wonderful day!(3 votes)
- how do we know, wen should we use additive law and multiplicative and conditional probability.(1 vote)
- Roughly, “or” means “add”; “and” means multiply, but sometimes with modifications.
If A and B are mutually exclusive (that is, they can’t both occur), then
P(A or B) = P(A) + P(B).
Without the assumption of mutual exclusivity, we have to modify this rule by subtracting the probability of overlap:
P(A or B) = P(A)+P(B)-P(A and B).
If A and B are independent (that is, the occurrence of a specific one of these two events does not influence the probability of the other event), then
P(A and B) = P(A)P(B).
Without the assumption of independence, we have to modify this rule by replacing one of the individual probabilities by a conditional probability:
P(A and B) = P(A given B)P(B) assuming P(B) is nonzero;
P(A and B) = P(A)P(B given A) assuming P(A) is nonzero.
Have a blessed, wonderful day!(4 votes)
- It looks like the audio on this video is gone. All other videos have perfect audio, and I've tried loading this in other browsers but no audio in all scenarios. Hope it's fixed soon because I'm interested in learning about this(2 votes)
- [Instructor] James is interested in weather conditions and whether the downtown train he sometimes takes runs on time. For a year, James records whether each day is sunny, cloudy, rainy or snowy, as well as whether this train arrives on time or is delayed. His results are displayed in the table below. Alright, this is interesting. These columns, on time, delayed and the total, so for example, when it was sunny, there's a total of 170 sunny days that year, 167 of which the train was on time, three of which the train was delayed, and we can look at that by the different types of weather conditions, and then they say for these days, are the events delayed and snowy independent? So to think about this, and remember, we're only going to be able to figure out experimental probabilities, and you should always view experimental probabilities as somewhat suspect. The more experiments you're able to take, the more likely it is to approximate the true theoretical probability, but there's always some chance that they might be different or even quite different. Let's use this data to try to calculate the experimental probability. So the key question here is what is the probability that the train is delayed? And then we wanna think about what is the probability that the train is delayed given that it is snowy? If we knew the theoretical probabilities and if they were exactly the same, if the probability of being delayed was exactly the same as the probability of being delayed given snowy, then being delayed or being snowy would be independent, but if we knew the theoretical probabilities and the probability of being delayed given snowy were different than the probability of being delayed, then we would not say that these are independent variables. Now, we don't know the theoretical probabilities. We're just going to calculate the experimental probabilities and we do have a good number of experiments here, so if these are quite different, I would feel confident saying that they are dependent. If they are pretty close with the experimental probability, I would say that it would be hard to make the statement that they are dependent, and that you would probably lean towards independence, but let's calculate this. What is the probability that the train is just delayed? Pause this video and try to figure that out. Well, let's see. If we just think in general, we have a total of 365 trials, or 365 experiments, and of them, the train was delayed 35 times. Now, what's the probability that the train is delayed given that it is snowy? Pause the video and try to figure that out. Well, let's see. We have a total of 20 snowy days and we are delayed 12 of those 20 snowy days, and so this is going to be a probability, 12/20 is the same thing as, if we multiply both the numerator and the denominator by five, this is a 60% probability, or I could say a 0.6 probability of being delayed when it is snowy. This is, of course, an experimental probability, which is much higher than this. This is less than 10% right over here. This right over here is less than 0.1. I could get a calculator to calculate it exactly. It'll be nine point something percent or zero point nine something, but clearly, this, you are much more likely, at least from the experimental data, it seems like you have a much higher proportion of your snowy days are delayed than just general days in general, than just general days, and so based on this data, because the experimental probability of being delayed given snowy is so much higher than the experimental probability of just being delayed, I would make the statement that these are not independent, so for these days, are the events delayed and snowy independent? No.