Why P(A and B) != P(A)*P(B). Because those events dependent?

When A and B are independent, P(A and B) = P(A) * P(B); but when A and B are dependent, things get a little complicated, and the formula (also known as Bayes Rule) is P(A and B) = P(A | B) * P(B). The intuition here is that the probability of B being True times probability of A being True given B is True (since A depends on B) is the probability of both A and B are True.

I'm a bit confused by this video. Basically, the video says that P(A) [Eating a bagel] is dependent on P(B) [Eating a pizza for lunch]. But what I don't understand is why is P(B) dependent on P(A)? Nothing in the question indicates that having a bagel for breakfast affects having a pizza for lunch. Why, therefore, is P(B) dependent on P(A)?

There does not have to be a causative relationship between A and B, just a correlation. Suppose I have 4 friends: Amy is on the basketball team and has red hair. Bill is a cheerleader for the basketball team and has black hair. Chad is also on the basketball team and has blond hair. Diego hates basketball and has brown hair. Only 25% of my friends have red hair (Amy). But if I tell you that I'm going to watch one of my friends play basketball, it has to be Amy (red hair) or Chad (blond hair), so there is a 50% chance that I am referring to a friend with red hair. Amy's red hair did not make her more likely to play basketball, it is just an accident of the situation.

I don't get it, HOW can the likelihood of what you have for breakfast be influenced what you have for lunch! I can't even begin to try to draw this out via a tree diagram.

Don't infer cause from dependence. If we look at the likelihood of what you eat for breakfast, there will be a frequency of each choice. The dependence merely observes that the frequency of choosing a bagel happens to be higher than usual on the days you also choose to eat pizza for lunch. There is no consideration of how, or even a claim that there is such a how. We simply notice the likelihood is higher and treat as dependent.

Can you explain how to work out Pr(A'|B')?

When we say S ∩ Q , we mean that both S and Q should happen isn't it. So we can express that in 2 ways. Why 2 ways?, Because S ∩ Q = Q ∩ S (Commutative law is satisfied) (Eg. Bagel and Pizza <=> Pizza and Bagel) 1) Event S happening and Event Q happening given that S has already happened (or) 2)Event Q happening and Event S happening given that Q has already happened We can express 1) as: P(S ∩ Q) = P(S) * P(Q | S)---------------(alpha) and we can express 2) as: P(Q ∩ S) = P(Q) * P(S | Q)--------------(beta) Since intersection is commutative , (alpha) and (beta) both are same. so using the values you mentioned, you can substitute them and solve it. :)

I'm confused with this. Wouldn't P(A and B) be equal to A * B, or 0.6 * 0.5? To find P(A and B), Sal is multiplying P(A | B) by P(B), or 0.7 * 0.5, which gives another answer. Can anyone explain which method is correct?

Multiplying P(A) and P(B) only works when A and B are independent. When they are not independent, then you need to use the conditional probability.

So P(A) = 0.6 and P(A&B) = 0.7 and they are dependent. But would the P(A&B) = 0.6 if they were independent? If they would be why??

P(A&B) can't be greater than P(A), I assume what you meant to say is P(A|B) which is the probability of A given that you know B has occurred. In that case, yes if A and B are independent then P(A) = P(A|B) because this is the definition of independence, the outcome of B has no bearing on the outcome of A.

I know this sounds really basic, but how would you go about answering an example like this? A survey of college students finds that 20% like country music, 15% like gospel music and 10% like both country and gospel music. The conditional probability that a student likes country music given that they like gospel music is.. ??

Conditional probability P(A | B) = P(AnB) / P(B) So you're looking for the probability of both, divided by the probability of the thing that is the given that. 10 / 15. 66.7%

Main content

Statistics and probability

Course: Statistics and probability > Unit 7

Lesson 9: Conditional probability and independence

Calculating conditional probability

Name: Calculating conditional probability
Uploaded: 2014-04-09T17:59:57Z
Description: Conditional probabilities are written like P(A|B), which can be read to mean, "the probability that A happens GIVEN b has happened." If we know probabilities like P(A), P(B), and P(A|B), we can solve for other probabilties like P(B|A).

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Conditional probabilities are written like P(A|B), which can be read to mean, "the probability that A happens GIVEN b has happened." If we know probabilities like P(A), P(B), and P(A|B), we can solve for other probabilties like P(B|A). Created by Sal Khan.

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Viacheslav Kondratiuk
Posted 9 years ago. Direct link to Viacheslav Kondratiuk's post “Why P(A and B) != P(A)*P(...”
Why P(A and B) != P(A)*P(B). Because those events dependent?
Button navigates to signup pageComment on Viacheslav Kondratiuk's post “Why P(A and B) != P(A)*P(...”
(58 votes)
Answer
- Shuai Wang
  Posted 9 years ago. Direct link to Shuai Wang's post “When A and B are independ...”
  When A and B are independent, P(A and B) = P(A) * P(B); but when A and B are dependent, things get a little complicated, and the formula (also known as Bayes Rule) is P(A and B) = P(A | B) * P(B). The intuition here is that the probability of B being True times probability of A being True given B is True (since A depends on B) is the probability of both A and B are True.
  Comment on Shuai Wang's post “When A and B are independ...”
  (102 votes)
bbrelin
Posted 8 years ago. Direct link to bbrelin's post “I'm a bit confused by thi...”
I'm a bit confused by this video. Basically, the video says that P(A) [Eating a bagel] is dependent
on P(B) [Eating a pizza for lunch]. But what I don't understand is why is P(B) dependent on P(A)? Nothing in the question indicates that having a bagel for breakfast affects having a pizza for lunch. Why, therefore, is P(B) dependent on P(A)?
Button navigates to signup pageComment on bbrelin's post “I'm a bit confused by thi...”
(36 votes)
Answer
- JaniceHolz
  Posted 8 years ago. Direct link to JaniceHolz's post “There does not have to be...”
  There does not have to be a causative relationship between A and B, just a correlation.
  Suppose I have 4 friends:
  Amy is on the basketball team and has red hair.
  Bill is a cheerleader for the basketball team and has black hair.
  Chad is also on the basketball team and has blond hair.
  Diego hates basketball and has brown hair.
  
  Only 25% of my friends have red hair (Amy).
  But if I tell you that I'm going to watch one of my friends play basketball, it has to be Amy (red hair) or Chad (blond hair), so there is a 50% chance that I am referring to a friend with red hair. Amy's red hair did not make her more likely to play basketball, it is just an accident of the situation.
  Comment on JaniceHolz's post “There does not have to be...”
  (51 votes)
mangohayley
Posted 9 years ago. Direct link to mangohayley's post “Can you explain how to wo...”
Can you explain how to work out Pr(A'|B')?
Button navigates to signup pageComment on mangohayley's post “Can you explain how to wo...”
(7 votes)
Answer
- be.cse.deepak
  Posted 6 years ago. Direct link to be.cse.deepak's post “When we say S ∩ Q , we me...”
  When we say S ∩ Q , we mean that both S and Q should happen isn't it.
  So we can express that in 2 ways.
  Why 2 ways?, Because S ∩ Q = Q ∩ S (Commutative law is satisfied) (Eg. Bagel and Pizza <=> Pizza and Bagel)
  1) Event S happening and Event Q happening given that S has already happened
  (or)
  2)Event Q happening and Event S happening given that Q has already happened
  
  We can express 1) as: P(S ∩ Q) = P(S) * P(Q | S)---------------(alpha)
  and we can express 2) as: P(Q ∩ S) = P(Q) * P(S | Q)--------------(beta)
  
  Since intersection is commutative , (alpha) and (beta) both are same.
  so using the values you mentioned, you can substitute them and solve it.
  :)
  Button navigates to signup page
  (2 votes)
Ahmad Badawy
Posted 7 years ago. Direct link to Ahmad Badawy's post “Is it possible that a dep...”
Is it possible that a dependency is valid only in one-direction? In other words, is it possible that B affects A, but A does not affect B. I believe there are real life examples of one-way dependencies.

If that is the case, then
P(A & B) = P(A given B) . P(B) = P(B given A) . P(A) could be rewritten as follows:
P(A & B) = P(A given B) . P(B) = P(B) . P(A) and if that is true, then
P(A given B) must be equal to P(A) which indicates that A & B are independent? I appreciate your feedback. Thank you.
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
jamie_chu78
Posted 8 years ago. Direct link to jamie_chu78's post “I don't get it, HOW can t...”
I don't get it, HOW can the likelihood of what you have for breakfast be influenced what you have for lunch! I can't even begin to try to draw this out via a tree diagram.
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- jdowling
  Posted 4 years ago. Direct link to jdowling's post “Don't infer cause from de...”
  Don't infer cause from dependence. If we look at the likelihood of what you eat for breakfast, there will be a frequency of each choice. The dependence merely observes that the frequency of choosing a bagel happens to be higher than usual on the days you also choose to eat pizza for lunch. There is no consideration of how, or even a claim that there is such a how. We simply notice the likelihood is higher and treat as dependent.
  Button navigates to signup page
  (7 votes)
Academic Ninja
Posted 7 years ago. Direct link to Academic Ninja's post “I'm confused with this. W...”
I'm confused with this. Wouldn't P(A and B) be equal to A * B, or 0.6 * 0.5? To find P(A and B), Sal is multiplying P(A | B) by P(B), or 0.7 * 0.5, which gives another answer. Can anyone explain which method is correct?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Dr C
  Posted 7 years ago. Direct link to Dr C's post “Multiplying P(A) and P(B)...”
  Multiplying P(A) and P(B) only works when A and B are independent. When they are not independent, then you need to use the conditional probability.
  Comment on Dr C's post “Multiplying P(A) and P(B)...”
  (5 votes)
Reina K
Posted 8 years ago. Direct link to Reina K's post “I know this sounds really...”
I know this sounds really basic, but how would you go about answering an example like this?

A survey of college students finds that 20% like country music, 15% like gospel music and 10% like both country and gospel music.

The conditional probability that a student likes country music given that they like gospel music is.. ??
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Tom Giles
  Posted 8 years ago. Direct link to Tom Giles's post “Conditional probability P...”
  Conditional probability P(A | B) = P(AnB) / P(B)
  
  So you're looking for the probability of both, divided by the probability of the thing that is the given that. 10 / 15. 66.7%
  Button navigates to signup page
  (9 votes)
Ken Florek
Posted 9 years ago. Direct link to Ken Florek's post “At 6:26, we get the same ...”
At
6:26
, we get the same answer by treating the question as ratios. That is, 0.6:0.7 as 0.5:x. Is the reasoning sound?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Maahi Ameer
  Posted 6 years ago. Direct link to Maahi Ameer's post “Yup, because ratios are j...”
  Yup, because ratios are just fractions, so in the end everything will work out; your reasoning is sound.
  Button navigates to signup page
  (1 vote)
Niklas.Vesterinen
Posted 8 years ago. Direct link to Niklas.Vesterinen's post “When i draw this out on a...”
When i draw this out on a tree i see that there are two possible ways to make A & B happend. Either A then B or B then A. Then why is not P(A&B) = the sum of P(A)*P(A|B)+P(B)*P(AB)?
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(3 votes)
Answer
- Thu Trang Nguyen
  Posted 8 years ago. Direct link to Thu Trang Nguyen's post “I was wondering the same ...”
  I was wondering the same thing! Anybody have an answer?
  Button navigates to signup page
  (1 vote)
minihardison001
Posted 8 years ago. Direct link to minihardison001's post “So P(A) = 0.6 and P(A&B) ...”
So P(A) = 0.6 and P(A&B) = 0.7 and they are dependent. But would the P(A&B) = 0.6 if they were independent? If they would be why??
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Thomas B
  Posted 8 years ago. Direct link to Thomas B's post “P(A&B) can't be greater t...”
  P(A&B) can't be greater than P(A), I assume what you meant to say is P(A|B) which is the probability of A given that you know B has occurred. In that case, yes if A and B are independent then P(A) = P(A|B) because this is the definition of independence, the outcome of B has no bearing on the outcome of A.
  Button navigates to signup page
  (2 votes)

Video transcript

Voiceover:Rahul's two favorite foods are bagels and pizza. Let A represent the event that he eats a bagel for breakfast and let B represent the event that he eats pizza for lunch. Fair enough. On a randomly selected day, the probability that Rahul will eat a bagel for breakfast, probability of A, is .6. Let me write that down. So the probability that he eats a bagel for breakfast is 0.6. The probability that he will eat a pizza for lunch, probability of event B ... So the probability of ... Let me do that in that red color. The probability of event B, that he eats a pizza for lunch, is 0.5. And the conditional probability, that he eats a bagel for breakfast given that he eats a pizza for lunch, so probability of event A happening, that he eats a bagel for breakfast, given that he's had a pizza for lunch is equal to 0.7, which is interesting. So let me write this down. The probability of A given, given that B is true. Given B, is not 0.6, it's equal to 0.7. And because these two things are not the same, because the probability of A by itself is different than the probability of A given that B is true, this tells us that these two events are not independent. That we're dealing with dependent probability. This shows us. The fact that B being true has changed the probability of A being true, this tells us that A and B are dependent. Dependent. And so when we start thinking ... Well actually let's just, before I start going on my little soapbox about dependent events, let's just think about what they actually want us to figure out. So the probability, the probability of A given B is equal to 0.7, that's what we wrote right over here. Based on this information, what is the probability of B given A? So they want us to figure out the probability of B given ... Probability of B given A. That's what they want us to figure out. The conditional probability that Rahul eats pizza for lunch, given that he eats a bagel for breakfast, rounded to the nearest hundredth. So how would we think about this? And I encourage you to pause this video before I work through it. So I'm assuming you've given a go at it. So the best way to tackle this is to just think about, well, what's the probability that both A and B are going to happen? Well, the probably of A and B happening ... And let me do this in a new color. The probability of A and B happening. A and B. I want to stay true to the colors. Is equal to ... There's a couple of ways you could write this out. This is equivalent to, this is equivalent to the probability, probability of A given B. Given B, times the probability of B. And hopefully that makes sense. The probability that B happens and that given that B had happened, the probability that A happens. And that would also be equal to ... So obviously this is A and B is happening, or you could do it the other way around. You could view it as the probability that B, the probability that B given A happens. Given A happens, times the probability of A. Times the probability of A. This also makes sense. What's the probability that A happened? And that, given A happened, what's the probability of B? You multiply those together, you get the probability that both happened. So why is this helpful for us? Well, we know a lot of this information. We know the probability of A given B is 0.7. So let me write that, I'll scroll down a little bit. This is 0.7. We know that the probability of B is 0.5. So this is 0.5. So we know that the probability of A and B is the product of these two things. That's going to be 0.35. Seven times five is 35 or, I guess you could say, half of .7 is 0.35. .5 of .7. And that is going to be equal to what we need to figure out. Probability of B given A times probability of A. But we know probability of A. We know that that is 0.6. We know that this is 0.6. So just like that, we've set up a situation, an equation, where we can solve for the probability of B given A. The probability of B given A. Notice, let me just rewrite it right over here. Actually, I'll write this part first. 0.6, 0.6 times the probability of B given A. Times that, right over there. And I'll just copy and paste it so I don't have to keep changing colors. That, over there, is equal to 0.35. Is equal to 0.35. And so to solve for the probability of B given A, we can just divide both sides by 0.6. 0.6, 0.6 and we get the probability of B given A is equal to ... Let me get our calculator out. So 0.35 divided by, divided by 0.6 and we deserve a little bit of a drum roll here, is .5833 ... It keeps going. They tell us to round to the nearest hundredth. So it's 0.58. Is approximately, is approximately 0.58. So notice, this is equal to 0 ... or I'll say approximately equal to 0.58. Once again, verifying that these are dependent. The probability that B happens given A is true, is higher than just the probability that B by itself, or without knowing anything else. Just the probability of B is lower than the probability of B given that you know, given that you know A has happened, or event A is true. And we're done.