Bayes rule Example

[@skooba7202]

Huge help, I had a very hard time finding any videos about Bayes *Rule* instead of just his theorem. Thanks a lot

[@ugururesin8405]

Here is the solution for the exercise: P(CUB | F') = [ P(C)*P(F' | C) + P(B)*P(F' | B) ] / [ P(A)*P(F' | A) +P(B)*P(F' | B) +P(C)*P(F' | C) ] P(CUB | F') = [ 0.12*(1-0.18) + 0.18*(1-0.10) ] / [ 0.70*(1-0.02) +0.18*(1-0.10) +0.12*(1-0.18) ] P(CUB | F') = 0.2751 . You can also solve this problem by using Venn diagrams or a table: Let's say there are 10,000 products in total, Q . Total . Fail . Fail' A . 7000 . 140 . 6860 B . 1800 . 180 . 1620 C . 1200 . 216 . 984 Based on the table above, P(CUB | F') = P(C & B Products when FAIL') / P(ALL FAILS') P(CUB | F') = (1620+984) / (6860+1620+984) = 0.2751

[@aliceiceberg581]

I got the value equal to 0.275. Here is my logic: P(C U B | F') = (P(B)xP(F') + P(C)xP(F'))/(P(A)xP(F') + P(B)xP(F') + P(C)xP(F')) where P(F') = 1- P(F) (the value of P(F') is different for A, B and C) Correct me if I am wrong.

[@chukwukelumarvelous9616]

THIS LOOKS SO RIGHT. LET ME CHECK IT OUT.

[@chukwukelumarvelous9616]

Thank you very much for your help stranger.

[@aliceiceberg581]

@@chukwukelumarvelous9616 I’m glad to be helpful!

[@firstkaransingh]

Great expectation 👍

[@reubenwilliammpembe667]

Thank you Sir you are the best!!! #RespectFromSouthAfrica

[@Borzacchinni]

answer to last question: p(b)*p(not fail B) + p(c)*p(not fail C) / p(not fail A) * p(A) + p(not fail B) * p(b) + p(not fail C) * p(c) with numbers: 0.18*0.9 + 0.12*0.82 / 0.7*0.98 + 0.18*0.9 + 0.12*0.82

[@Oncus2]

This is a very good explanation of Bayes rule. Good job!

[@DavidWittler]

I found the same answer to the last question through a more convoluted way. P(CUB/F')=P(CUB)P(F'/CUB)/Law of total probability for F' well I figured C and B were disjoint so P(CUB)= P(C)+P(B), but the hard part was finding P(F'/CUB). P(F'/CUB) is the probability that the component didn't fail out of the class C or B components. given earlier P(CUB)= P(C)+P(B)= .12+.18= .3 If .9 of Components B worked and .82 of components C worked (given by P(F'/B)=1-P(F/B) and P(F'/C)=1-P(F/B)) and they are disjoint one could take the average of the working components in class B and C. Class C and B take up .3 of the sample space, and class B takes up .18 of that .3, while C takes up .12 of that .3. So class B is .6 from sample BUC (18/30) and class C is .4 of BUC (12/30). Taking a weighted average of (.9x.6)+(.82x.4) finds the average probability that Components didn't fail out of class C or B. P(F'/CUB)= (.9x.6)+(.82x.4) = .868. Plugging that back in to the original equation finds 0.2751

[@alyaqistina597]

thankyou. very clear explanation

[@oluremiojo4551]

I understand the numerator is the intersection of B and F. I naturally want to multiply P(B) by P(F) but I see you instead you multiplied P(B) by P(B|F). Is this because P(B) and the P(F) are dependent events? If so, what should I pay attention to in the problem to also know they are dependent events?

[@homeslice86]

I know this video is pretty old but does anybody have any advice for solving the second unknown? P(CUB|F')?

[@AntiPeach]

I see my mistake. I was trying to add the two in the sample space, as if they were not mutually exclusive.e.g. P(C U B|F`)=(P(C)(F`|C)+P(B)P(F`|B))/(P(A)P(F`|A)+P(B)P(F`|B)+P(C)P(F`|C)). Even though I have that wrong do I have the formula right for if they were mutually inclusive

[@rupamajmudar8966]

Thank you so much , I got good practice

[@RideSmoothTV]

I tried your last problem but I keep getting a value greater than one. Do you mind sharing your value of P(F'), P(C ^ F') & P(B ^ F')?!!? Just looking for clarification with the values in my formula. Anyways great videos. Keep em coming! God Bless!

[@shumailsajjad2967]

Love u Sir ❤❤❤❤

[@KnowledgeGuide859]

I tried to solve the exercise, but the answer does not match with the give. Can someone give the steps?

[@RicharD-fi3ho]

Ty

[@darkan101]

For some reason I got the last problem wrong. I did : (P(C∪B) ∗ P(F'|C∪B))/((P(A)*P(F'|A) + P(B)*P(F'|B) + P(C)*P(F'|C)) = ((0.3)(1.72))/((0.7)(0.98)+(0.18)(0.9)+(0.12)(0.82) and I got 0.545. If you can, I'd like to know what I did wrong for this problem. Thank you so much for your videos! They are a great help.

[@StatCourses]

The key solving the problem is to realize that the events A, B and C are mutually exclusive (i.e., disjoint). With that P(C U B |F') = P(C|F') + P(B|F') . P(C|F') = 0.103973, and P(B|F') = 0.171175. Adding the two you get the answer 0.2751.

[@darkan101]

Actuarial Path AH! Thank you so much! I was having some problems with this. =] Your videos help a lot, like I said earlier, so thank you so much!

[@BoringMathTutor]

Also having trouble. I used: P(F`) = 1 - P(F) = 1 - denominator P(C U B) = .3 I plugged in .3358 for the answer and solved to find P(F` | [C U B]) but got 1.06 (which makes no sense as it's greater than 1).

[@StatCourses]

The key solving the problem is to realize that the events A, B and C are mutually exclusive (i.e., disjoint). With that P(C U B |F') = P(C|F') + P(B|F') . P(C|F') = 0.103973, and P(B|F') = 0.171175. Adding the two you get the answer 0.2751.

[@seoexperimentations6933]

I got P(CF) and thought I was done... damn actuary problems

[@MrAlaalaMo]

I got it! Bayes is bae

[@alaaaltuhl3750]

If we knew that A,B and C is disjoint , then shouldn't we go with P(C U B |F') = P(C) + P(B|F') ? that gave me 0.291.

[@openfor..823]

P(C U B | F') = P (C | F') + P(B | F') = 0.10397 + 0.17117 = 0.2751 ( Answer)

[@omegatechhd8332]

Thanks men!

[@darrenisaacs1423]

i think there's a mistake with regards to the values substituted into the denominator for P(B/F) shouldn't it be 0.12*0.1 +0.18*0.18 +0.7*0.02

[@KnowledgeGuide859]

P(F|B) = 10%

[@whittakerdanielj]

Wow! Another video that doesn't show the order of operations and what that stupid dot stands for!! It obviously doesn't meant multiplication! My math book doesn't show diddly squat either!

[@teaadvice4996]

Daniel Whittaker I think you are talking about the compliment and that basically means the opposite, and if you rewatch all of his videos you will understand more, repetition is key to understanding :)