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)?

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 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). 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.

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.

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.

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?

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.

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 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? 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

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!

P($40,000 and over ∣ Uni. B) equals 40/120, but NOT .33

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!

Main content

Course: AP®︎/College Statistics > Unit 7

Lesson 3: Conditional probability

Conditional probability and independence

Google Classroom

In probability, we say two events are independent if knowing one event occurred doesn't change the probability of the other event.

For example, the probability that a fair coin shows "heads" after being flipped is

1 / 2

. What if we knew the day was Tuesday? Does this change the probability of getting "heads?" Of course not. The probability of getting "heads," given that it's a Tuesday, is still

1 / 2

. So the result of a coin flip and the day being Tuesday are independent events; knowing it was a Tuesday didn't change the probability of getting "heads."

Not every situation is this obvious. What about gender and handedness (left handed vs. right handed)? It may seem like a person's gender and whether or not they are left-handed are totally independent events. When we look at probabilities though, we see that about

10 %

of all people are left-handed, but about

12 %

of males are left-handed. So these events are not independent, since knowing a random person is a male increases the probability that they are left-handed.

The big idea is that we check for independence with probabilities.

Two events, A and B, are independent if

P (A | B) = P (A)

and

P (B | A) = P (B)

Example 1: Income and universities

Researchers surveyed recent graduates of two different universities about their annual incomes. The following two-way table displays data for the

300

graduates who responded to the survey.

Annual income	University A	University B	TOTAL
Under $20,000	$36$ ‍	$24$ ‍	$60$ ‍
$20,000 to 39,999	$109$ ‍	$56$ ‍	$165$ ‍
$40,000 and over	$35$ ‍	$40$ ‍	$75$ ‍
TOTAL	$180$ ‍	$120$ ‍	$300$ ‍

Suppose we choose a random graduate from this data.

Are the events "income is $40,000 and over" and "attended University B" independent?

Let's check using conditional probability.

Example 1: Problem A

What is the probability that a randomly selected graduate earns $40,000 and over?

P ($ 40,000 and over) =

Example 1: Problem B

What is the probability that a randomly selected graduate earns $40,000 and over given they are from University B?

P ($ 40,000 and over | Uni. B) =

Example 1: Problem C

Are the events "income is $40,000 and over" and "attended University B" independent?

Example 2: Income and universities (continued)

Here is the same data from the previous example:

Annual income	University A	University B	TOTAL
Under $20,000	$36$ ‍	$24$ ‍	$60$ ‍
$20,000 to 39,999	$109$ ‍	$56$ ‍	$165$ ‍
$40,000 and over	$35$ ‍	$40$ ‍	$75$ ‍
TOTAL	$180$ ‍	$120$ ‍	$300$ ‍

Suppose we choose a random graduate from this data.

Are the events "income under $20,000" and "attended University B" independent?

Let's check using conditional probability.

Example 2: Problem A

What is the probability that a randomly selected graduate earns under $20,000?

P (under $ 20,000) =

Example 2: Problem B

What is the probability that a randomly selected graduate earns under $20,000 given they are from University B?

P (under $ 20,000 | Uni. B) =

Example 2: Problem C

Are the events "income is under $20,000" and "attended University B" independent?

What if the probabilities are close?

When we check for independence in real world data sets, it's rare to get perfectly equal probabilities. Just about all real events that don't involve games of chance are dependent to some degree.

In practice, we often assume that events are independent and test that assumption on sample data. If the probabilities are significantly different, then we conclude the events are not independent. We'll learn more about this process in inferential statistics.

Finally, be careful not to make conclusions about cause and effect unless the data came from a well-designed experiment. For a challenge, can you think of some outside variables — apart from the universities — that may be the cause of the income disparity between the graduates at the two universities in Example 2?

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chris
Posted 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)?
Button navigates to signup pageComment on chris's post “At the top it says two ev...”
(29 votes)
Answer
- Ian Pulizzotto
  Posted 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
  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).
  
  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.
  Button navigates to signup page
  (55 votes)
ytcsplayz2018
Posted 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.
Button navigates to signup pageComment on ytcsplayz2018's post “Hello everybody. I had a ...”
(7 votes)
Answer
- Nabil Daoud
  Posted 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.
  Button navigates to signup page
  (9 votes)
Page Ellsworth
Posted 6 years ago. Direct link to Page Ellsworth's post “P($40,000 and over ∣ Uni....”
P($40,000 and over ∣ Uni. B) equals
40/120, but NOT .33
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Ian Pulizzotto
  Posted 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!
  Button navigates to signup page
  (12 votes)
Moin M
Posted 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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Martin
  Posted 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.
  Button navigates to signup page
  (4 votes)
gurushishya
Posted 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)
'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?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- cossine
  Posted a year ago. Direct link to cossine's post “Theorem: P(A ∣ B) = P(A) ...”
  Theorem:
  P(A ∣ B) = P(A) => P(B|A) = P(B)
  
  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)
  Comment on cossine's post “Theorem: P(A ∣ B) = P(A) ...”
  (1 vote)
Daksh Gargas
Posted 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".
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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Jerry Nilsson
  Posted 5 years ago. Direct link to Jerry Nilsson's post “In this case we have thre...”
  In this case we have three different events:
  A – "earning at least $40,000"
  B – "attended University B"
  C – "earning under $20,000"
  
  In example 1 we find that events A and B are dependent.
  
  This doesn't mean that events B and C are dependent.
  Comment on Jerry Nilsson's post “In this case we have thre...”
  (1 vote)
NB
Posted 4 years ago. Direct link to NB's post “Hi and thank you Sooo muc...”
Hi and thank you Sooo much for these videos Sal.
Is there a way to downnload the lesson summaries in pdf format? I would like to save them for later reviews.
Button navigates to signup pageComment on NB's post “Hi and thank you Sooo muc...”
(2 votes)
Answer
ninolatimer
Posted 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
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?

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
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Sergey Korotkov
  Posted 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!
  Button navigates to signup page
  (3 votes)
jazlyn.trejogonzalez-90533
Posted 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
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
Victor Gutierrez
Posted 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?
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(1 vote)
Answer