Conditional Probability Example Problems
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- Опубликовано: 31 янв 2018
- 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)?
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.
By far, this is the best explanation I have seen of this topic. Congratulations!
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
YAY!!! Prof Balka is back! So exicited. Happy New Year Prof!
Thanks! I'm glad to be back!
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 😀
one of the best videos i watched 2020 so far as per my course is concerned
Good job on the visualisations of the conditional probabilities with the Venn Diagrams!
Thanks!
I love your video. It helped a lot. Keep up the good work.
thank you sir you had cleared my doubts it was exceptionally a good video
This is the Best explaination I ever came across!!!
Thanks!
Simply incredible vids.
Thanks!
Amazing explanations!
Thank you for the clarification!
The video is really helpful. Can you please upload a video on Random variables in statistics?
Finally. Found the video ive been looking for
Thank You!! Great video!!
Thanks for the compliment! I'm glad to be of help!
helped my intuition quite a bit. thnx ;^)
I WOULD USE THE REDUCED SAMPLE SPACE IF MY PROF DIDNT WHAT US TO “shOw tHe WoRk”
Great videos and looking at the “why we do things” helps a lot. Not just here’s the formula remember it and apply it. Thank you very much
Thanks for the kind words. If I had to just teach what the formula was and how to use it, then I'd be inclined to quit and go off and start a trucking company. (And I don't know anything about that!) I definitely try to help with the "why" of things.
I understand what you're saying about your prof, and the use of the formula, and if that's the way they require you to do it then you should go with that. But the reduced sample space argument is 100% legitimate, and still fits with the notion of showing your work.
Great video!!
Thanks!
Great help
Good explanation
I believe you have saved me from failing :)
What's up with the party part at the end? Mmm. I must be out of touch with the modern era?
your video makes me successful, thank you very much!!
This is really helpful nd was taught this yesterday by prof.Thron
I'm glad to hear it!
Brilliant.
I'm so into this👌🏼
Me too :)
@@jbstatistics doesn't in the last example, P(A | B) = P(A) / P(B) ?, since P(A and B) = P(A) if A is a subset of B. I thought you were hinting at that idea with your 'what can be said of these formulas'
@@alexgabriel5877 i agree with you but don't know if this is correct.
Thanks!!!
You are welcome!
Perfect you are genius
Very sweet
Superb
Great!!
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.
There is no easy way to get the probability of the intersection of B complement and A U C. The best way to visualize it is with the Venn Diagram. If you wanted to do it formulaically, we could say that intersections distribute over unions, so B^c n (A U C) = (B^c n A) U (B^c n C) and keep going down that path. But that's not especially helpful, as at some point we're simply going to have to figure out what region we need to find and find the probability of that.
First of all, I appreciate you updating your question with an answer for those who had a similar question. But, shouldn't the P(A and B and C) be added instead of subtracting it twice again as you already have removed it both times. Also, doing this is easy only because of the diagram. It would be hard to solve if something else was asked in some random question x.
Great video! Demystifies everything except the last comment. :-). Pray tell: what was the party comment about?
please, I'm so utterly confused at what he meant by that lol
I think the joke is that no one at a party would want to hear a math joke, so he IS "really out of touch with the modern era."
@@hellmuth26 I think so too xPP I just imagine someone going around a party trying to start a conversation about probability theory and not getting anywhere xPPPP
thanks your explication it could be helpful while the course is online during covid19 pandemic
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?
Why you did not want to write/say at 16:12 that P(A|B)=P(A)/P(B) ?
How do I know when to apply the probability formula and when I want to apply reduced sample space method?
Um, its personal preference (you can use either method, it just depends which one your brain sees first, but best to PRACTICE both, you dnt necessarily have to USE both). The real question is when to use probability rules and when not to use probability rules, which he beautiful explains at 1:38.
How do you prove the part P(A∪B|C) using conditional probability formulae?
scroll to the name The Establishment. He explains it there.
@@thehamsterarmy2380 "/pjilummArijuana of us, and u l please do you y2k last p Lilly ppl l
Ppl puLp: ppl injuryp0ppppppppppppppppppp0 imP the yyyy6
@@thehamsterarmy2380 silly question but where? I can't find it
@@Sk8erMorris you can just calculate it by yourself, given all those little areas in the Venn diagram. A union B intersect C, the numerator, is going to be 0.19, and C is going to be 0.3. then you do division.
@@reyrey4993 that's not helping, we are asking for the formula, not what goes on top and bottom
Hi how did you get the 0.30 in 9:00?
Add some videos on inclusion probability in sampling
There is something off at 11:11 why not B is 0.41 instead of 0.71 (1-P(B)=0.71), i probably didn't get the notation.
Thanks for the video.
This guy is legit.
for the second question there are 5 numbers shared in both AuB so shouldn't AuB = 5, and since two of those numbers are shared with C, shouldn't the P(AuB|C) = 2/5
2/5 is like saying "giving that A U C occured" and not "C occured" ....C is an event of three numbers.... and the sum of the conditional values where AUB and C meet is 2, so 2/3....
can you please show me how to incorporate "c" into our formula aka a question like P(B|A^c)
Nevermind I finished the class with an A🌚
@@ayah7056 But how do you do it??
labai geras
Dėkoju!
Could you please do an example with the conditional probability rule, so I can actually see how I will calculate P(Male "given" (Cerebral OR Respiratory). In all the examples you do the easy way, and just say I can work that out myself, but the problem is I don't understand how to do it.. I have tried for days now.. I almost killed my statistics book because of the frustration this causes :P
Try this:
NB:
S : Sample Space
In a population, a person can either be a male or a female, not both
Male and Female are independent events
We also notice that the probabilities all add up to one.
M or m : "a male", F or f : " a female"
S= {M1, M2, M3,.......M359......F1, F2, F3,.........F164}
S= sum(M) + sum(F) => 523 = 359 + 164
Sm = M = { CAm, Cm, Rm, Om} = { 264, 38, 36, 21}
Sf = F = { CAf, Cf, Rf, Of} = { 89, 27, 29, 19}
S= Sm + Sf = {CAm, Cm, Rm, Om, CAf, Cf, Rf, Of}
P(M|(C u R)) = [P (M n (C u R))] / P(C u R)
C= { Cm, Cf}, R= {Rm, Rf}
C u R = { Cm, Cf, Rm, Rf}
M = {Cm, CAm, Rm, Om}
M n (C u R) = {Cm, Rm} ==> P(M n (C u R)) = (38 + 36) / 523 = 74/523
P(C u R ) = (38 + 27 + 36 +29)/ 523 = 130/523
P(M|(C u R)) = [P (M n (C u R))] / P(C u R)
= (74/523)/(130/523)
P(M|(C u R)) = P(M| C u R) .... I remove the extra pair of bracket
P(M| C u R) = 74/130 = 0.57
Hope it helps!
What was the last part about being out of touch with the modern era? I don't get it.
jbstatistics is the coolest kid on the block and that's why he knows that that last example is a good one for parties. ;)
This is so helpful
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.
I can! A U B = {1,2,3,4,6}. C = {1,3,5}. P(A U B | C) = P((A U B) n C)/P(C). A U B intersects with C at 1 and 3 (i.e. What values are in both A U B and C? 1 and 3). So P((AUB) n C) = 2/6. And thus P(A U B | C) = (2/6)/(3/6) = 2/3.
For the final example can we say that P(A|B) >= P(A)?
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.
The sample space is only reduced if P(B) < 1, but yes, under that restriction you are correct.
1:39 you got me there :)
I see it coming a mile away ;)
@@jbstatistics now, I also see it , just a little bit of concentration :))
12:40 why not 3rd option be the answer. p(a/b)must be smaller than p(a).how come greater ? Please xplain
I explained to the best of my capabilities in the video. Males tend to be taller than females. The proportion of adult Canadian males that are over 6' tall is greater than the corresponding proportion of adult Canadian females. So, if we know the randomly selected person is male, that information increases the probability that they are over 6' tall. Hence, P(A|B) > P(A).
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!
A U B = {1,2,3,4,6}. C = {1,3,5}. P(A U B | C) = P( (AUB) n C)/P(C). Finding P(C) is easy (3/6), so the only tricky bit is finding the numerator. Where does A U B intersect with C? What sample points do they have in common? 1 and 3, so (AUB) n C = {1,3}, P( (AUB) n C) = 2/6, and (AUB) | C) = (2/6)/(3/6) = 2/3.
but why P(C) is 3*(1/6)? why not (1/6)^3 ? the event is rolling the dice three times, and each time getting 1, and then 3, and then 5, (or whatever 3 nums of a die, for that matter), so its the principle of multiplication, no? especially that the events 1,3,5 are independent
@@tubics1 you are not rolling dice thrice. You are rolling it once
do you study that at high school in the united states ???
I need exact conditional probability formula for first questions
I don't know what you're asking. In the first example, I use the conditional probability formula to answer the question.
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?
I don't know what you did to get 0.20. (Perhaps P(A U B) - P(B), which is the probability of A but not B.) P(A n B) = 0.14, for the reasons I outline in the video.
@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 ???
Sure, the addition rule works there, but it's not like we *must* use that formula whenever we want the probability of a union. The union of A and C is the event that A or C or both happen. We can see what regions that comprises in the Venn diagram. The probability of the union of A and C can be found with the addition rule, but it is also equal to the sum of the probabilities of the 6 mutually exclusive regions contained therein.
@@jbstatistics I got it thanks a lot.
If we take the values of p(A) and p(C) provided in given question then we have to subtract p(A n C). Because we are considering this region twice.
???
@@jbstatistics Thanks a lot for your videos..!!
At @9:04, where did the 0.30 outside of the three circles come from???
The probability of the entire sample space is 1, and so the probability of the region outside the circles is 1-P(A U B U C) = 1- (0.18 + 0.12 + 0.10 + 0.03 + 0.12 + 0.04 + 0.11) = 0.30.
@@jbstatistics Ah, I see. Thanks, mate!
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.
His answers are all correct, where you able to get it right later on?
What did u mean than under independence??
for 2:01 why is the probability of A and B is not 0.70?
I don't know why you think it would be. P(A U B) is given as 0.70, and that's the probability that A or B (or both) occurs.
Hi how did you get the answer .14 in 2:52? Thank you
Solve for P(A n B) in 0.70 = 0.34 + 0.50 - P(A n B). This implies P(A n B) = 0.84 - 0.70.
for the second question I get 5/6 to be the P(A U B) and P(C)= 1/2, so for the final answer using the formula I get 10/6 which is impossible? what am I doing wrong?
The question asks for P(AUB|C). If you feel compelled to use the conditional probability formula here, then you need to find P((AUB) n C)/P(C). A U B intersects C at the numbers 1 and 3, so P((AUB) n C) = 2/6, and P(AUB|C) = (2/6)/(3/6) = 2/3.
do you get it? I see no where 3/6 came from. I see a 5/6 for AUB
@@thehamsterarmy2380 You're probably way past this, but for future readers: the 3/6 = P(C) = {1,3,5}. So where P(AUB) {1,2,3,4,6} intersects P(C) {1,3,5} is {1,3} = 2/6
@@shdyo Thank youuuuu! I still had no clue
How is it more likely that a Canadian adult is tall and male rather than the probability that the Canadian adult is tall?
Tall *and* male is not more likely than tall, of course. The probability the individual is tall *given* they are male is greater than the probability they are tall (which is what I state in the video).
In case anyone wants to use the formula for question 3 :
P(B'| A u C ) = P (B' n (A u C))/ P(A u C) = [P( A n B') u P( C n B') ] / P(A u C)
Numerator:
A n B' = A - (A n B), C n B' = C - (C n B)
(A n B' ) u (C n B' ) = A n B' + C n B' - [ A n B' n C n B'] = A n B' + C n B' - [ A n C n B']
(A n B' ) u (C n B' ) = A - (A n B) + C - (C n B) - [(A n C) - (A n B n C)]
P(A n B' ) u P(C n B' ) = P(A) - P(A n B) + P(C) -P(C n B) - [P(A n C) - P(A n B n C)]
P(A n B' ) u P(C n B' ) = 0.43 - 0.13 + 0.30 - 0.07 - [ 0.15 - 0.03] = 0.41
Denominator:
P(A u C) = P(A) + P(C) - P(A n C)
P(A u C) = 0.43 + 0.30 - 0.15 = 0.58
P(B'| A u C ) = 0.41/0.58 = 0.707
it is longer to do this way than using the vain diagram, but using the formula make us dig more since the formula is not a direct one.
Happy to have any remark on my approach to the answer be it positive or negative.
Thanks.
I forgot to thanks @jbstatistics for the marvellous videos on stats ! thanks a million.
You the best man!!!!!!
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?
Time and Major coin flipping possibilities and then Unoin of Balls( AUB ) &C
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?
I think I can help:
S = { X, Y, ............}
P(Y/X') = P (Y n X') / P(X')
S= X + X'
P(S) = P(X) + P(X') => P(X') = 1 - 0.5 = 0.5
P(Y n X') = ??
Y n X' = Y - (Y n X) ... I advice you to visulaise this on a vein diagram for clear understanding
P(Y n X') =P( Y - (Y n X)) = P(Y) - P(Y n X)
P(Y n X') = 0.4 - O.1 = 0.3
P(Y/X') = 0.3/0.5 = 0.6
P(Y/X') = 0.6
Hope it helps.
it would have helped if you would have said how you got certain things instead of saying "by that logic" .....
you said we can't multiply P(A) * P(B) because you can't assume independence, but you can assume the events are non-mutually exclusive? P(A U B) = P(A) + P(B) if they are mutually exclusive, and P(A U B) = P(A) + P(B) - P(A n B) if they are not mutually exclusive
It is *always* the case that P(A U B) = P(A) + P(B) - P(A n B). If A and B are mutually exclusive, then P(AnB) = 0 and the addition rule reduces to P(A U B) = P(A) + P(B) - 0 = P(A) + P(B).
I had the same doubt, good to see the answer
7:59
please.
5:10
What is the probability I laugh at your last joke given that I have already shoved my pencil through my ear?
You're 1/1000 of God.
Not really a good teacher imo..