I had assumed that the probability of a defect without the manager there was approximately 8.7%, since to me the question seemed to imply that 5% was the average defect rate over all states. Guess I was overthinking it. 😅
I thought the same thing. The way the problem was stated, it seems like the overall P(Defect) = .05, which would mean P(defect|no manager) would have to be .087 if P(defect|manager)= .03 and P(manager)=.65
@@zafnas5222 because probability of (A|B) = P(A) but he didn’t explicitly say. Independence doesn’t mean the probability don’t affect one another like this it means the two events are uncorrelated eg, the more the manager being on time doesn’t effect whether it’s defective or not but it does change the probability
@alexn4255 increasing the frequency of the manager being on time has no efffect defectivness probability as independence implies uncorrelated but the manager being on duty changing the probability at a specific moment can still be an independent event even though it changes the probability
P(D|~M)=P(D)=3% , P(~D|~M)=P(~D)=97%. It's easy to proof that events are independents with Venn Diagram. Construct a Venn Diagram with D and ~M events or trace a Venn D that show ~D and ~M events. Then professor have an error tree diagram.
35% of 5% defect and 65% of 3% defect, In a average day the probablity of defect would be 5x0,35+3x0,65 = 1,75%+1,95% =3,7% defect on average day. Calced this before watching. Lets say there'll be 5 defects 35% of the time and 3 defects 65% of the time when manager is there. The defect chance on manager time is: 3/2(3(65%)+5(35%)) =3/6,4 =1 in 2,1333 of the time. =~46,875% 2 is for 8 = 3(100%) + 5(100%) 8 = 2(3(50%) + 5(50%)) Probably 2 because there are 2 probabilities makes sense. 47,875% of the time it defect the manager would be there. Cmiiw. Nothing, just had spare time.
Hi everyone, Eddie is a fantastic, charismatic and enthusiastic math teacher and I'm sure he is must be correct in his video above. However, there was something he said that challenged what I thought I knew about the importance of Independence in Probability. For multiplication of probabilities to be valid, the two events MUST be Independent i.e. like flipping a coin multiple time. In this example, the relationship between the Manager and the Failure rate is clearly dependant - for whatever reason, increased Manager time results in less failures! Whatever this manager does (or doesn't do) somehow affects the failure rate of bulbs. The person above/below talks about "correlation" but as far as I can understand, that's irrelevant because independent events have a correlation of zero i.e. they are unrelated? Really hope someone can explain please otherwise my understanding is all screwed up!! Many thanks, Pete
In a tree you multiply probabilities just because at each branch you write a conditional probability. If you see the branch for having the manager and then having a defect you multiply 0.65 times 0.03 because its P(Manager)P(defect|manager)
I had assumed that the probability of a defect without the manager there was approximately 8.7%, since to me the question seemed to imply that 5% was the average defect rate over all states.
Guess I was overthinking it. 😅
Yeah same here. It should have been worded to mean specifically that the manager is not there imo
Especially since he wrote P(defective) and not P(defective | no manager)
I thought the same thing. The way the problem was stated, it seems like the overall P(Defect) = .05, which would mean P(defect|no manager) would have to be .087 if P(defect|manager)= .03 and P(manager)=.65
@@AdamLaMore I was thinking exactly this!
it's always tricky 😭
Good day. Could you make a video on linear programming and parabola
U assumed they were independent events though, didn’t prove that they were
How are they independent events? The manager affects the defect
@@zafnas5222 because probability of (A|B) = P(A) but he didn’t explicitly say. Independence doesn’t mean the probability don’t affect one another like this it means the two events are uncorrelated eg, the more the manager being on time doesn’t effect whether it’s defective or not but it does change the probability
@@hamzazahid2775 That's right. Then P(D|~M)=P(D)=3% , P(~D|~M)=P(~D)=97%.
Then professor have an error tree diagram. Please verify.
@alexn4255 increasing the frequency of the manager being on time has no efffect defectivness probability as independence implies uncorrelated but the manager being on duty changing the probability at a specific moment can still be an independent event even though it changes the probability
P(D|~M)=P(D)=3% , P(~D|~M)=P(~D)=97%. It's easy to proof that events are independents with Venn Diagram. Construct a Venn Diagram with D and ~M events or trace a Venn D that show ~D and ~M events.
Then professor have an error tree diagram.
#. ALL
EYES
ON
RAFAH
🇵🇸❤
Good day. Can you help make vodeo on "proof of universally quantified statements"
I wish the students payed more attention
35% of 5% defect and 65% of 3% defect,
In a average day the probablity of defect would be
5x0,35+3x0,65 = 1,75%+1,95%
=3,7% defect on average day.
Calced this before watching.
Lets say there'll be 5 defects 35% of the time and 3 defects 65% of the time when manager is there. The defect chance on manager time is:
3/2(3(65%)+5(35%))
=3/6,4
=1 in 2,1333 of the time.
=~46,875%
2 is for 8 = 3(100%) + 5(100%)
8 = 2(3(50%) + 5(50%))
Probably 2 because there are 2 probabilities makes sense.
47,875% of the time it defect the manager would be there. Cmiiw.
Nothing, just had spare time.
Eddie come our school
Hi everyone, Eddie is a fantastic, charismatic and enthusiastic math teacher and I'm sure he is must be correct in his video above. However, there was something he said that challenged what I thought I knew about the importance of Independence in Probability. For multiplication of probabilities to be valid, the two events MUST be Independent i.e. like flipping a coin multiple time. In this example, the relationship between the Manager and the Failure rate is clearly dependant - for whatever reason, increased Manager time results in less failures! Whatever this manager does (or doesn't do) somehow affects the failure rate of bulbs. The person above/below talks about "correlation" but as far as I can understand, that's irrelevant because independent events have a correlation of zero i.e. they are unrelated? Really hope someone can explain please otherwise my understanding is all screwed up!! Many thanks, Pete
In a tree you multiply probabilities just because at each branch you write a conditional probability. If you see the branch for having the manager and then having a defect you multiply 0.65 times 0.03 because its P(Manager)P(defect|manager)