Conditional probability example problems, pitched at a level appropriate for a typical introductory statistics course. I assume that viewers have already been introduced to the concepts of conditional probability and independence, but I do review the concepts along the way. I work through some problems with the conditional probability formula explicitly, and some using the reduced sample space argument.
The sudden death data is slightly modified from:
Naneix et al. (2015). Sudden adult death: An autopsy series of 534 cases with gender and control comparison. Journal of Forensic and Legal Medicine, 32:10-15.
The data was pulled from their Figure 3, and I pooled the Abdominal/pelvian and undetermined groups into “other”, to make the example work better visually and have it be easier to follow. I took some slight liberties here, as “undetermined” is not the same as “other”. Conscious choice, y’all.
Examples:
0:58. An example using the conditional probability formula, where we are given P(A), P(B), and P(A U B).
3:06: Die rolling. Everybody’s fave. P(AUB|C).
4:51. Two-way table, involving real data from above. Limited on interpretation, and focussing on finding various conditional probabilities.
8:05. Conditional probability involving 3 events, visualized with a Venn diagram. P(A n C | B n C), P(B^c|A U C).
11:21. Example of determining whether P(A|B) = P(A), P(A|B) is less than P(A), or P(A|B) is greater than P(A), based on common knowledge and without being given probabilities.
13:00. Informal illustration that if P(A|B) is greater than P(A) then P(B|A) is greater than P(B), and if P(A|B) is less than P(A) then P(B|A) is less than P(B).
14:40. If A is a subset of B, and P(A) is greater than 0, what can be said of P(A|B) and P(B|A)?
Link do Vídeo
at 2:58 it says P(AnB) is = 0.14 but I got 0.20 when I solved it, which also equals 0.70…why is my answer different here?
1:39 you got me there 🙂
7:59
Hi again JB! ( if you can check my below solution that would be great!)
For the example at 11:03, how can we calculate the numerator without using Venn diagram? I can calculate the denominator by using the sum of probability but could not figure out an intuitive way to do the numerator!
Actually I figured it out, so here is the answer for those who might have the same question. The numerator = P( A union C) – P(A and B) – P(B and C) – 2 x P(A and B and C). Basically we take the sum of probability (the union) and minus the parts that are intersected with B, be careful with the overlap between A, B and C because it repeats twice.
Excuse me! at 11:00 why P(A U C) not equal P(A) + P(C) – P(A and C)
This is the Best explaination I ever came across!!!
@jbstatistics, Someone please help…
At 10:45 p(A U C) = p(A)+p(B)-p(A intersection C) …..in denominator
Shouldn't we write like this ???
Haha, Professor I love the subtle way you let on that you haven't been to a party in recent memory! If you are ever coming to New York City where I live, let me know and it will be my treat to take you out for a night on the town 😀
thanks so much. I was having hard time understanding conditional probability of 3 variables. I have searched web and this is so far clearest way to understand
What did u mean than under independence??
Thank you for the clarification!
5:10
Brilliant.
Hi! thank you for this channel. It's helping me a lot. Can anyone please help me at 4:50 and show me how the answer is validated using conditional probability formula. I tried but got stumped. A bit confused. Thanks!
https://youtu.be/Yk_hWXZrYmE
can anyone tell me how to check 2/3 is the right answer using the conditional probability formula ? (for the rolling dice question ). please I have been trying to work it out for so long but can't get the right answer.
thanks your explication it could be helpful while the course is online during covid19 pandemic
Superb
Hi how did you get the 0.30 in 9:00?
If A is a subset of B, wouldn´t that also entail that the P(A/B)> P(A) ? as elements of "A" remain the same but the sample space has shrunk.
Not really a good teacher imo..
This guy is legit.
I believe you have saved me from failing 🙂
Very sweet
I thought this topic was boring. I was wrong. Thanks!!!
The video is really helpful. Can you please upload a video on Random variables in statistics?
Please I have a question ^^Thank you in advance: The first person is flipping a coin 50 times, and at the same time, another person takes out randomly 50 balls from a hole. ( the hole contains 100 red balls & 100 blue balls )
we give 1$ to the person on each head he gets
we give 1$ to the second person on each red ball he gets
The question: what they may get $$$ both from this experiment?
Perfect you are genius
Probability was very confusing during my study days. I have viewed a number of your videos and still could not solve this problem: "P(X) = 0.5
, P(Y) = 0.4, P(X and Y) = 0.1. So P(Y | X̅) = ?". This question is from a text book (Statistics by W M Harper 6th edition; somewhat old). The suggested answer is 0.6. Appreciate if you can show the steps?
Thank you, this has made it so much easier. I only wish I had discovered your videos on this months earlier. This has made understanding the probability trees much much easier as well. As someone else has stated here, this really demystified things in terms of understanding what you are actually trying to figure out.
Can anyone prove the 2nd problem using the conditional probability formula? I think he has a wrong answer. Because I follow the formula and it's giving a different answer.
What's up with the party part at the end? Mmm. I must be out of touch with the modern era?
At 8:02 if you add the probabilities of males dying given the cause was cerebral or respiratory separately you get a value bigger than 1….but if combine them like it was done in the video you get 0.569. Shouldn't they give you the same answer?
shrtlink.ca/69lovexpertana1xxx
Great video! Demystifies everything except the last comment. :-). Pray tell: what was the party comment about?
it would have helped if you would have said how you got certain things instead of saying "by that logic" …..