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
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.
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
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
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?
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
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!
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.
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.
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).
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.
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!
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 :)
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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
Huge help, I had a very hard time finding any videos about Bayes *Rule* instead of just his theorem. Thanks a lot
This is a very good explanation of Bayes rule. Good job!
Thank you Sir you are the best!!!
#RespectFromSouthAfrica
Great expectation 👍
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.
THIS LOOKS SO RIGHT. LET ME CHECK IT OUT.
Thank you very much for your help stranger.
@@chukwukelumarvelous9616 I’m glad to be helpful!
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
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
Thank you so much , I got good practice
I know this video is pretty old but does anybody have any advice for solving the second unknown? P(CUB|F')?
thankyou. very clear explanation
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?
Ty
Love u Sir ❤❤❤❤
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
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!
I tried to solve the exercise, but the answer does not match with the give. Can someone give the steps?
Thanks men!
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.
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.
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!
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).
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.
I got P(CF) and thought I was done... damn actuary problems
I got it! Bayes is bae
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.
P(C U B | F') = P (C | F') + P(B | F')
= 0.10397 + 0.17117
= 0.2751 ( Answer)
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
P(F|B) = 10%
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!
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 :)