The Law of Total Probability

[@sudhakarreddy1599]

Perfectly explained. Couldn't be better than this. Thank you so much and please continue the work.

[@jaretwilliams2564]

I have to take an engineering statistics course this summer and as someone who has not liked statistics much in the past, your channel and videos are a lifesaver!

[@Saeed.n1]

Mate this is the best of all probability youtube channel. Thank you so much I learned a lot

[@chajataal5033]

Thank you, finally someone who explains something difficult in an easy way

[@williamtownsend3395]

Thank you for being clear and precise with worked out examples. This video greatly helped me with some proofs for advanced econometrics.

[@Colors2000]

I don't know if I or my Professor should feel ashamed that I didn't understand a thing about this. Sir You made everything looks so much easier.

[@kennethmcdoogle3860]

This is the best video on this topic I have seen. Really really good work mate.

[@ChocooWafeer]

This vid is sooo good! Everything is crystal clear. Thank you so much for sharing this!

[@sergeijegorov1279]

watched a ton of material on this, but understood only after this video. Thanks a lot

[@gloystar]

Thank you! Now I totally understand this concept. Amazing illustrations!

[@amintahiri3053]

Thans you sooooo mUch Im in love with ur explanation its like im getting prived lessons from my last professor of probability theory. Thanks alot 😊😊

[@shashinidulakshi7318]

simple,, but explained perfectly,,,thank you very much again and again

[@thedramatiks6221]

This was so helpful. Many thanks, wish you immense growth!

[@thomasoffenbecher2196]

Very excellent video. You explained the law very clearly and with good examples. Helped a lot!

[@augustinejunior3361]

Thank you. Let me just subscribe I never knew that there is the best channel for probability

[@azombieee]

Thannnk you. This formula and sorting the data was throwin me into space lol - you explained it so well!

[@jbstatistics]

I'm glad to be of help!

[@navidahmed7922]

the best explanation i have ever heard

[@sivaarun4351]

Best underrated video

[@ianeyre7162]

Brilliant. The best I’ve seen.

[@saketkumar4972]

if u put all the balls in one urn then P(blue)=0.3823. BUT WHY IS IT DIFFERENT FROM THE FIRST ANSWER, IT SHOULD BE SAME, ISN'T IT?

[@Bridgelessalex]

No, because there are two events: first, pick an urn, second, pick a ball

[@TheLegendaryMudkipz]

The process of grouping the balls into the urn is what creates the difference.

[@Syed-wj4pj]

Imagine if the blue balls were distributed such that they ended up in only one of the 4 urns

[@juliocardenas4485]

Again. Wonderfully explained

[@amit4learners]

Nicely explained . Example shown is perfect

[@hamzanasir1590]

Best explanation ever 👍👍👍

[@romaion4024]

awesome job. clear and to the point. helped me a lot. thank you!!

[@Bridgelessalex]

Hi Jeremy. Thank you so much for the video! I have a quick question regarding the sample spaces of the partition and the even A. For example, in the second exercise (randomly selecting a ball from a cup), the sample space of the first event (selecting a cup) is {cup 1, cup 2, cup 3, cup 4}, however, the sample space of the second event (selecting a ball) is {blue, red}. Since the sample spaces are different, could you please elaborate on why would be the law of total probability works in the case? Thanks in advance.

[@jbstatistics]

The probability experiment is randomly selecting an urn, then randomly selecting a ball from that urn. There are different ways we could define the sample space, but one is S = {1B, 1R, 2B, 2R, 3B, 3R, 4B, 4R}. It's the same idea as if we have a probability experiment where we toss a coin twice and observe whether heads or tails comes up on each toss. There are 4 possibilities, S = {HH, HT, TH, TT} (where HT represents getting heads on the first toss and tails on the second). A: The first toss is heads, and B: the second toss is heads, are defined on the same sample space.

[@Bridgelessalex]

@@jbstatistics Thank you so much for the detailed explanation Jeremy!

[@petercourt]

For the final problem (where all the balls are mixed), I got P(Ball = Blue) = 0.382, and I can't figure out why the probability would be higher if all the balls are mixed up, compared with when they're separated into four urns?

[@pavanchaudhari5245]

What I think is that since drawing a ball from the urn's is weighted, meaning that each urn has a different composition of blue balls, so drawing a blue ball in each scenario would be different, still it is 37.7% likely to draw a blue ball. But, mixing all the balls into one huge bag breaks the idea of weights as everything is in one single place, therefore drawing a ball from a huge tank of balls would make the probability 13/34. Anybody can add in or correct me if I am wrong. Thanks.

[@JoaoVitorBRgomes]

@@pavanchaudhari5245 I also think it is because of weighting. I think it is like comparing average vs weighted average. If you do the same in the machine exercise, outcomes will be very different I think. If instead of 5/9 in the last urn it would be 5/50, calculating unweighted would skew the probability.

[@fasthand2000]

I think if you put all the balls together, then each blue ball will have equal chance to be picked. But if you put the balls in urns, then choose a ball is first depends on which urn do you pick. Use this extreme example, if I rearrange the balls. I put all the red balls in one urn, and put 4, 4 and 5 blue balls in the other 3 urns. there are 3/4 chance to select the urns with only blue balls, and the probability to pick a blue ball is 0.75 now. 1/4 x 1 + 1/4 x 1 + 1/4 x 1 + 1/4 x 0 = 3/4 = 0.75

[@Syed-wj4pj]

Imagine if the blue balls were distributed such that they ended up in only one of the 4 urns

[@mukamafrancois2776]

Perfect and understandable..

[@camerongridley9065]

Terrific video. Very helpful!

[@saugatbhattarai327]

Nice explained. Thank you for such videos. Please create some more videos

[@sandeshdeshkar1424]

nice video. superb explanation. Do you also have on Bayes theorem?

[@benfield1866]

very well done explanation, thank you

[@shirshak6738]

fav teacher

[@oyunaeagle34]

Great video! Thank you.

[@olivershao3983]

Great Video. Thank you so much!

[@omkarkachi-h4r]

Hi Jeremy. Thank you so much for the video! Can you please tell me that how identify when we have to use this particular "Law of Total probability" for which type of question

[@karthiksukumaran85]

You explained it good, thank you.

[@RaviRahulKumarShah]

Thanks a lot !! You explained very clearly !!! you gained a subscriber

[@ateequllah7670]

nicely explained. love it

[@oak6677]

Thanks soo much :-)) this video is really helpful

[@ratnamuthusudhakar1575]

Thank u sir wonderful explanation

[@Nazgenyanly]

You're really good

[@jbstatistics]

Thanks!

[@zhenminliu]

The explanation is clear, using a heuristic approach. For a more mathematical approach, you have to search elsewhere.

[@desertezz]

Great video, do you have a video on Bayes theorem?

[@che5738]

Crystal clear...thanks a lot

[@aymanadam7825]

best explanation!!! Thanks a lot!

[@duckymomo7935]

Would this, in measure theory, be sigma additivity?

[@constanzamiguel9241]

Amazing video, super clear and easy to follow. Thank you!

[@jbstatistics]

Glad to be of help!

[@aquilagaming6996]

looking for a Baye's theorem video, thanks for the contents

[@bensmyth450]

Excellent stuff, did you ever make the video on Baye's theorem?

[@aniketrane]

Thanks for the nice video. Please give us some examples where events are not mutually exclusive and non exhaustive. Will be looking forward to your video on practical applications with Baye's theorem.

[@kgpianlaskar]

Sir I have a qsn. We are using conditional probability because the events are dependent. But in the case of independent events I think the law of total probability will be like multiplication of individual events instead of conditional probability for 2nd event. Am I right???

[@hammadullahshaikh4735]

If the events are independent then the formula would be P(A)=£P(A)P(B)...plz guide how it will work?

[@99BeastMaker]

Nice and simple.

[@jeweljoya]

Why do you do 1/4 ? At 8:32

[@jbstatistics]

"One of these urns is randomly selected, in such a fashion as each urn is equally like to be chosen." There are 4 urns, and they all have the same probability of being selected.

[@AidenIndeed]

Thanks so much!

[@husseinsaad6185]

Thank you so much sir.

[@maxchen9919]

This is so good

[@RagnarLothbrok-bk7zi]

best exolenation so far

[@malds129]

Hi, can you please make a video explaining Bayes. Thank you.

[@vishallohare2984]

is it the same as Bayes Theorum @jbstatistics

[@curtismoxam5382]

How can A exist if B1,B2 etc cannot overlap? Do disjoint events suddenly gain the ability to intersect?

[@jbstatistics]

I don't understand what you're asking . B_1 through B_k are mutually exclusive events that cover the sample space. A is another event in that sample space. A is going to intersect with at least one of the Bs. Why is the existence of A in question?

[@nakshathra5149]

"if we put all the balls in one urn, mix them up and drew one ball at random, would the probability of getting a blue ball be the same?".

[@achyutagoura3267]

How can we say all events are mutually exclusive in above examples

[@jbstatistics]

I'm not sure at what level you're asking this question. Events are mutually exclusive if they share no portion of the sample space. In the Venn (Euler, actually) diagram examples, they were mutually exclusive if they didn't overlap (didn't share any sample points -- any portion of the sample space). I also showed a situation in which events shared common ground, and said they were not ME. In the example with the machines, each part was made by one and only one machine.

[@MwilimaIan]

But P(Urn) is 1/4 , why for the first example was the P(machine) not equal to 1/3 for all of them, instead he paid attention the quantity that each machine was making, if this is the case should P(Urn) be equal to number of balls in that particular urn/ total number of balls?...anyone please

[@whateveruwant6440]

absolute GOAT

[@Saad_LH]

thank you sir

[@kodeater]

thankyou sir🙏

[@Pixel_AI_guy]

Thank you so much......

[@KeyserTheRedBeard]

cool content jbstatistics. I broke that thumbs up on your video. Continue to keep up the exceptional work.

[@davidotniel2643]

underated!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@jbstatistics]

Indeed!

[@wahabfiles6260]

Is this also called marginalization?

[@sunilmalla4768]

why does The probality of 🔵 when all are in one urn is not equal to the total probability

[@jbstatistics]

Imagine an extreme scenario: 1000 balls, with 1 blue and 999 red. Put them all in a single urn, and randomly pick a ball. The probability you pick a blue ball is 0.001. Now put the 999 red in an urn, the 1 blue in another urn, and choose between the urns with probability 0.5 then pick a ball from that urn. What's the probability you get a blue ball? 0.5. If each of the two urns contained 500 balls (the balls were evenly split), then the probability of getting a blue ball would be 0.001. The different number of balls in each urn messes with this.

[@karlmagsino3126]

yup they're the same

[@cocoarecords]

THE STA HERCULES IS BACK

[@hinarehman6231]

👍👍👍👍👍👍

[@saraalenezi7057]

Hi u r the bestttttrttttttttt

[@vanshikhapankhuri-dancinga8057]

Complete description of Total Probability Theorem in Hindi Language- kzbin.info/www/bejne/naXPgZxnbrl1r5Y

[@amitkumarsharma5]

Why is 1/4 multiplied with number of balls in each urn? Shouldn't it be 1/(number of blue balls in each urs)

[@jbstatistics]

We're asked for the probability we draw a blue ball, if we select one of the 4 urns at random and then draw a ball. P(Blue) = sum P(urn_i)P(blue | urn_i) = sum (1/4)*proportion of blue balls in that urn. I don't see how 1/number of blue balls could come into play.

[@studytime-art]

Thank u

[@kwame7003]

Great

[@wesamelbaz7811]

Where's the Bayes Theorem?

[@38zech]

The questions at 9:29. Why do you get a different answer when all the balls are in the same urn?

[@yulinliu850]

Much appreciated!

[@lekishaadriaanse3274]

i love you so much.

[@cococnk388]

Great video! my answer for the quizz : B = "Picking a blue ball" NB : One of the urn has 13 blue balls and 21 red balls and the other urns are empty... P(B)= 1/4 * 13/34 = 0,096 do we all agree? if not, tell me why in the comment section? thanks!

[@Loona_r_]

If the empty urns are removed, then P(blue) = 13/34 Otherwise you're right, I think

[@cococnk388]

@@Loona_r_ yeah correct if they are removed, but they are present

[@cococnk388]

@@Loona_r_ B = B n U1 + B n U2 + B n U3 + B n U3 P(B) = P(B n U1) + 0 + 0 + 0 P (B) = P(U1) P(B/U1) P(B) = 1/4 * 13/34 It is possible because the URNs are mutually exclusive and exhaustive events...same as the balls

[@franciscoromogaray3076]

te rifaste papito, taba bien facil. mi profe es re malo

[@yayabean32]

sooo, what's the answer to the question if all the balls were in one urn???

[@dereknigten8757]

good

[@dhakshinamurthy51]

Tq

[@mirrorball1]

you speak so eloquently

[@aabens]

I got blue balls with probability of 1

[@lokman175200]

Thank you so much

[@singcovers8479]

thank you so much