Log in chris 6 years agoPosted 6 years ago. Direct link to chris's post “At the top it says two ev...” At the top it says two events, A and B, are independent if P(A|B) = P(A) and P(B|A) = P(B). But in the last exercise we are only asked to find P(A|B) = P(A) and judge independence solely on that. Why are we not asked to find P(B|A) = P(B)? • (29 votes) Ian Pulizzotto 6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “Assuming that A and B are...” Assuming that A and B are events with nonzero probabilities, P(A|B) = P(A) is actually mathematically equivalent to P(B|A) = P(B). We can see this because So, assuming that P(A) and P(B) are nonzero, it's enough to test just one of P(A|B) = P(A), P(B|A) = P(B) to determine if A and B are independent. (58 votes) ytcsplayz2018 6 years agoPosted 6 years ago. Direct link to ytcsplayz2018's post “Hello everybody. I had a ...” Hello everybody. I had a very challenging question in class today. There are two parts to this question. Hope this does not bug anybody. Based on an online poll, 35% of motorists routinely use their cell phone while driving. Tree people are chosen at random from a group of 100. a) What is the probability of at least two people of the three people use their cell phone while driving? b) What is the probability that no more than one person of the three people use their cell phone while driving? Thank you so much to everybody for reading this and helping me solve this problem. • (7 votes) Nabil Daoud 6 years agoPosted 6 years ago. Direct link to Nabil Daoud's post “For this question I notic...” For this question I notice that we are given the probability that a motorist routinely uses their cell phone while driving. Then I'm given a finite number of independent trials with each classified as a success or failure. I'm am immediately thinking about binomial variables. Head over to https://www.khanacademy.org/math/statistics-probability/random-variables-stats-library/binomial-random-variables/v/binomial-distribution and you should be able to figure it out for yourself from there. (10 votes) Page Ellsworth 6 years agoPosted 6 years ago. Direct link to Page Ellsworth's post “P($40,000 and over ∣ Uni....” P($40,000 and over ∣ Uni. B) equals • (0 votes) Ian Pulizzotto 6 years agoPosted 6 years ago. Direct link to Ian Pulizzotto's post “Note that the correct ans...” Note that the correct answer is 40/120 = 1/3, but 1/3 is the repeating decimal 0.333... which is not exactly the same as 0.33. The question asks for a fraction or an exact decimal, so this is why 0.33 is marked wrong. Had the question said you could round to two decimal places, then 0.33 would have been acceptable. Have a blessed, wonderful day! (12 votes) Moin M 5 years agoPosted 5 years ago. Direct link to Moin M's post “Since 10% of all people a...” Since 10% of all people are left-handed and 12% of all males are left-handed. This means 8% of all females are left-handed. Am I right? • (2 votes) Martin 5 years agoPosted 5 years ago. Direct link to Martin's post “Assuming an even distribu...” Assuming an even distribution of men and women, yes. So for example you have 100 people of which 50 are men and 50 are women, an 10% are left handed, then you 10 left handed people. 12% of those men are left handed. So that's 6 men and 10 - 6 is 4, so you have 4 left handed women, which is 8% of 50. In uneven distribution of men in women that wouldn't work like that though. (5 votes) gurushishya a year agoPosted a year ago. Direct link to gurushishya's post “This page says that event...” This page says that events are independent if: P(A ∣ B) = P(A) • (3 votes) cossine a year agoPosted a year ago. Direct link to cossine's post “Theorem:P(A ∣ B) = P(A) ...” Theorem: Proving the theorem is straight forward just apply definition of conditional probability (hopefully you know the definition) then make P(A and B) the subject. You will find P(A and B) = p(A)p(B). Using this you will find P(B|A) = P(B) (1 vote) Daksh Gargas 5 years agoPosted 5 years ago. Direct link to Daksh Gargas's post “As per my understanding, ...” As per my understanding, "two events A and B are independent if the probability of occurrence of an event A is not affected by the happening of event B and vice-versa". • (2 votes) Jerry Nilsson 5 years agoPosted 5 years ago. Direct link to Jerry Nilsson's post “In this case we have thre...” In this case we have three different events: In example 1 we find that events A and B are dependent. This doesn't mean that events B and C are dependent. (1 vote) NB 5 years agoPosted 5 years ago. Direct link to NB's post “Hi and thank you Sooo muc...” Hi and thank you Sooo much for these videos Sal. • (2 votes) ninolatimer a year agoPosted a year ago. Direct link to ninolatimer's post “I don't get the p(A) and ...” I don't get the p(A) and p(B|A) part of [p(A|B)= p(A) and p(B|A) = p(B)] Let's say Example 2: I'm assuming A=income under 20k and B=attended university B So the probability of A would equal 60/300=.2 But instead we just do p(A)=p(A|B)? so p that income under 20k=income under 20 and from uni B? Edit: Ohhh! By p(A) and p(B|A), did they mean to multiply p(A) and p(B|A) or to add them? Cause when you multiply them you get the right answer but I assumed "and" meant to add them together • (1 vote) Sergey Korotkov a year agoPosted a year ago. Direct link to Sergey Korotkov's post “In such questions "and" u...” In such questions "and" usually means multiplication (one event AND another happening at the same time, you may also see sign ⋂ "intersection"), while "or" means addition (one happening OR another happening, you may also see sign ⋃ "union"). Hope it helps! (3 votes) jazlyn.trejogonzalez-90533 a year agoPosted a year ago. Direct link to jazlyn.trejogonzalez-90533's post “confusing but soon i thin...” confusing but soon i think ill get the hang of it • (1 vote) Victor Gutierrez 4 years agoPosted 4 years ago. Direct link to Victor Gutierrez's post “Is there a relation betwe...” Is there a relation between dependence-independence and asociation between 2 variables?? I mean, if 2 events are independent, the correlation coeficient will be close to zero right? • (1 vote)Want to join the conversation?
P(A|B) = P(A) means P(A and B)/P(B) = P(A) from definition of conditional probability,
P(B|A) = P(B) means P(A and B)/P(A) = P(B) from definition of conditional probability, and
P(A and B)/P(A) = P(B) is obtained from P(A and B)/P(B) = P(A) by multiplying both sides by the well-defined, nonzero quantity P(B)/P(A).
40/120, but NOT .33
'and' (B ∣ A) = P(B). But on Exercise 2, only one equation is found to be true, then it declares the events to be independent without doing the 'and' and checking the other direction. Which one is correct?
P(A ∣ B) = P(A) => P(B|A) = P(B)
Which is proved in Example 1
but in Example 2, as per the mathematical formula we have proven that the two events are independent but this is counter intuitive taking Example 1 into consideration because both the cases are similar, aren't they?
A – "earning at least $40,000"
B – "attended University B"
C – "earning under $20,000"
Is there a way to downnload the lesson summaries in pdf format? I would like to save them for later reviews.
Probability of B = 120/300 = .4
(B|A) is they attend uni b given they make under 20k which would be 24/60 which equals .4?
and so p(A) and p(B|A) is .2+.4 so .6 while p(B) = .4 so they are not equal?
