Understanding Conditional Probability BasicsHelp understanding conditional probabilityConditional probability...
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Understanding Conditional Probability Basics
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In many online sources I've read a statement similar to:
Probability of B happening given A is equal to the probability of A and B both happening divided by B happening
or
$p(A | B) = p(A cap B) / p(B)$
But what is the difference between "B happening given A" and "A and B happening at the same time" ? If B happens given A, didn't A and B happen at the same time ? Why is that chance smaller than the intersection ?
probability bayesian
asked Jul 4 '15 at 6:31
pwnedpwned
34229
$endgroup$
add a comment |
$begingroup$
In many online sources I've read a statement similar to:
Probability of B happening given A is equal to the probability of A and B both happening divided by B happening
or
$p(A | B) = p(A cap B) / p(B)$
But what is the difference between "B happening given A" and "A and B happening at the same time" ? If B happens given A, didn't A and B happen at the same time ? Why is that chance smaller than the intersection ?
probability bayesian
asked Jul 4 '15 at 6:31
pwnedpwned
34229
$endgroup$
$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
1
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
$begingroup$
I understood your questions. I only make comments on you false statements. I hope it is clear now.
$endgroup$
– callculus
Jul 4 '15 at 7:27
$begingroup$
@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
$endgroup$
– Jared
Jul 4 '15 at 7:37
add a comment |
$begingroup$
In many online sources I've read a statement similar to:
Probability of B happening given A is equal to the probability of A and B both happening divided by B happening
or
$p(A | B) = p(A cap B) / p(B)$
But what is the difference between "B happening given A" and "A and B happening at the same time" ? If B happens given A, didn't A and B happen at the same time ? Why is that chance smaller than the intersection ?
probability bayesian
asked Jul 4 '15 at 6:31
pwnedpwned
34229
$endgroup$
In many online sources I've read a statement similar to:
Probability of B happening given A is equal to the probability of A and B both happening divided by B happening
or
$p(A | B) = p(A cap B) / p(B)$
But what is the difference between "B happening given A" and "A and B happening at the same time" ? If B happens given A, didn't A and B happen at the same time ? Why is that chance smaller than the intersection ?
probability bayesian
probability bayesian
asked Jul 4 '15 at 6:31
pwnedpwned
34229
asked Jul 4 '15 at 6:31
pwnedpwned
34229
asked Jul 4 '15 at 6:31
pwnedpwned
34229
asked Jul 4 '15 at 6:31
pwnedpwned
34229
asked Jul 4 '15 at 6:31
pwnedpwned
34229
34229
$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
1
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
$begingroup$
I understood your questions. I only make comments on you false statements. I hope it is clear now.
$endgroup$
– callculus
Jul 4 '15 at 7:27
$begingroup$
@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
$endgroup$
– Jared
Jul 4 '15 at 7:37
add a comment |
$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
1
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
$begingroup$
I understood your questions. I only make comments on you false statements. I hope it is clear now.
$endgroup$
– callculus
Jul 4 '15 at 7:27
$begingroup$
@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
$endgroup$
– Jared
Jul 4 '15 at 7:37
$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
1
1
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
$begingroup$
I understood your questions. I only make comments on you false statements. I hope it is clear now.
$endgroup$
– callculus
Jul 4 '15 at 7:27
$begingroup$
I understood your questions. I only make comments on you false statements. I hope it is clear now.
$endgroup$
– callculus
Jul 4 '15 at 7:27
$begingroup$
@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
$endgroup$
– Jared
Jul 4 '15 at 7:37
$begingroup$
@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
$endgroup$
– Jared
Jul 4 '15 at 7:37
add a comment |
3 Answers
3
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oldest
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$begingroup$
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $frac{2}{3}$ and the probability of A and B occurring is $frac{1}{3}$ therefore the probability of A given B is: $frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y neq x) = frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have ${HH, TT, HT, TH}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $frac{2}{4} = frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $frac{1}{4}$, divided by the probability of getting heads the first time, $frac{2}{4}$: $frac{frac{1}{4}}{frac{2}{4}} = frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
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The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
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In summary, the essence of conditional probability is in every probability calculation, find the Whole (i.e. universal set), find the Part (intersection between the event you are trying to calculate the probability and the universal set) and divide the Part by the Whole to find the probability of given event (i.e the comparative relation between Part and Whole).
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory $ A cap B = A $ when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set (U).
Why do we need to divide?
Proportions: Finding the likelihood or probability of some event can be interpreted as a proportion problem, which implies the existence of a Part and a Whole. The way that you find how much a Part represent from the Whole is by dividing the Part by the Whole: $ frac{Part}{Whole} $ (if you are still uncomfortable with the concept and for some reason can't visualize why division would provide you with such information, just take it as a rule for now, at some point the concept will click). So whenever you are thinking about the probability or likelihood of an event always think about the Part and the Whole, which leads to the next question: How do we find the Part and the Whole? We will need to understand how we represent events to answer that question.
Sets, Subsets and Cardinalities: An event is a set of outcomes, the cardinality of a set is the measure of the size of that same set. Now, for the sake of our probabilistic theory discussion, you will see that a Part is always a subset of the Whole (you may be thinking of some edge cases that do not fit this rule, keep reading and you will see that this idea indeed holds true for all cases).
The first step to find a probability of any event is to find the set that represents the Whole (i.e. the scope of all the possible outcomes, when you roll a die of 6 sides the Whole set can be represented as the numbers from 1 through 6, since they form all the possible outcomes), the second step is to find the Part (i.e. the set containing the subset of outcomes that you are trying to find the likelihood of occurrence).
Every probability calculation can be seen with the lens of a conditional probability (What!? Keep reading :) )
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory A and B = A when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set.
As an example, say we are interested in knowing the probability of getting an even number when rolling a die of 6 sides, you can always formulate the problem thinking about conditional probability, when we defined that we are going to be rolling a die of six sides we just implicitly defined the Whole set (i.e the possible outcomes are numbers from 1 to 6), the second piece of information is the event itself that we would like to obtain the probability (even number), that is the Part. An important subtlety here, the Part you are interested here in the first place, is not ALL the even numbers but the ones that make part of your Whole set (i.e. subset of the Whole). How do you find that subset? The intersection of the two sets? precisely.
So the intersection gives you the Part that you are interested in, and once you have the Whole and Part you divide the two and find the probability (frequency of occurrence, or the comparative relation between the Part and the Whole).
If the outcomes or event that you are trying to find the likelihood does not make any part of the Whole, in other words, if the intersection of both sets is empty then we know that the probability of such event is zero, the event is simply impossible to occur, there is no way you will get a 7 if you roll a die of 6 sides.
Always remember when you see P(A|B) we are already explicitly saying that B is the Whole or universal set, all you have to do now is to find the comparative relation between the Part and Whole..
To end, let me clarify a few common misconceptions:
On the context of conditional probability, aside from the relationship between the
WholeandPart, the probability of what you are judging to be theWholeset does not matter by itself! As an example, the probability of someone having a tumor is irrelevant, if what we are trying to find is the probability of someone having a benign tumor given the person is already diagnosed with a tumor.P(A|B)is not the same asP(B|A): The probability of someone having a benign tumor given the person is already diagnosed with a tumor, is not the same as the probability of someone being diagnosed with a tumor given the fact the person has a benign tumor.Shouldn't
P(A|B)be equal toP(A and B)?
This is a common misinterpretation of what conditional probability means, we are not interested in the probability of A and B happening (i.e. $ frac{P(A cap B)}{P(U)} $), that's called joint probability and is a separate (and related) subject. Conditional probability means, now thatBALREADY HAPPENED, what is the probability ofAalso happening?
answered Mar 10 at 2:30
bsdbsd
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3 Answers
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3 Answers
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$begingroup$
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $frac{2}{3}$ and the probability of A and B occurring is $frac{1}{3}$ therefore the probability of A given B is: $frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y neq x) = frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have ${HH, TT, HT, TH}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $frac{2}{4} = frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $frac{1}{4}$, divided by the probability of getting heads the first time, $frac{2}{4}$: $frac{frac{1}{4}}{frac{2}{4}} = frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
$endgroup$
add a comment |
$begingroup$
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $frac{2}{3}$ and the probability of A and B occurring is $frac{1}{3}$ therefore the probability of A given B is: $frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y neq x) = frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have ${HH, TT, HT, TH}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $frac{2}{4} = frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $frac{1}{4}$, divided by the probability of getting heads the first time, $frac{2}{4}$: $frac{frac{1}{4}}{frac{2}{4}} = frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
$endgroup$
add a comment |
$begingroup$
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $frac{2}{3}$ and the probability of A and B occurring is $frac{1}{3}$ therefore the probability of A given B is: $frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y neq x) = frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have ${HH, TT, HT, TH}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $frac{2}{4} = frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $frac{1}{4}$, divided by the probability of getting heads the first time, $frac{2}{4}$: $frac{frac{1}{4}}{frac{2}{4}} = frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
$endgroup$
"If B happens given A, didn't A and B happen at the same time?" Yes that is true (or it could be true--it could also be considering sequential events--not necessarily simultaneous ones). However, A and B happening together might be "unlikely" (really it just means they might not always happen together).
So consider that you have three outcomes and exactly two always occur together. That is there are three possible outcomes: A and B, A and C, or B and C. Now what is the probability that A occurs given that B occurs? Well the probability that they both occur is $frac{1}{3}$ (since A and B is only one possible outcome out of three). But what is the probability that given that B occurs that A occurs? Well there are only two events such that B occurs: either A and B or B and C. Therefore given that B occurs the probability that A also occurs is $frac{1}{2}$ (because you have excluded the possibility that A and C occur since you assumed B occurred).
This is indeed inline with Bayes' Theorem. The probability of B occurring is $frac{2}{3}$ and the probability of A and B occurring is $frac{1}{3}$ therefore the probability of A given B is: $frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}$. (this is probably a poor example of Bayes' Theorem because, in this case, if you chose any two events, you'd always get $p(x|y neq x) = frac{1}{2}$)
Second Example: Independent Events
Consider that you flip a (fair) coin twice, sequentially. What are the possible outcomes? You can have ${HH, TT, HT, TH}$ (since it's a fair coin, each of these possibilities are equally likely--that's important here although we could extend to situations where they are not equally likely--but then we start getting more into the Bernoulli random variable rather than conditional probabilities). What is the probability that you get a heads on the first flip? Clearly from the above we get $frac{2}{4} = frac{1}{2}$.
Now, what is the probability that you flip a heads given that the first flip is heads? There is only a single outcome that satisfies this (i.e. that you get two heads in a row). However there is only one other event, $HT$, such that heads was rolled first, therefore the probability that you get a heads given that the first flip was heads is $frac{1}{2}$. Again, this is the same as the probability of two heads (heads both times), $frac{1}{4}$, divided by the probability of getting heads the first time, $frac{2}{4}$: $frac{frac{1}{4}}{frac{2}{4}} = frac{1}{2}$.
Indeed we see by the above that the probability of getting heads given an initial flip of heads is the same as the probability of getting heads on the second roll--both are $frac{1}{2}$. This tells us that each flip is independent of the other since the probability of getting heads isn't affected by the previous flip.
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
edited Jul 4 '15 at 7:32
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
answered Jul 4 '15 at 6:47
JaredJared
5,20311116
5,20311116
add a comment |
add a comment |
$begingroup$
The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
$endgroup$
add a comment |
$begingroup$
The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
$endgroup$
add a comment |
$begingroup$
The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
$endgroup$
The difference is that in one case, you already know B has happened. This is not subject to chance anymore, it's a fact. So you compute the probability of A happening, knowing for a fact that B happened
For the probability in the intersection, on the other hand, you don't know that B has happened and have to calculate the probability that both A and B occur
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
answered Jul 4 '15 at 6:43
AntAnt
17.5k22974
17.5k22974
add a comment |
add a comment |
$begingroup$
In summary, the essence of conditional probability is in every probability calculation, find the Whole (i.e. universal set), find the Part (intersection between the event you are trying to calculate the probability and the universal set) and divide the Part by the Whole to find the probability of given event (i.e the comparative relation between Part and Whole).
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory $ A cap B = A $ when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set (U).
Why do we need to divide?
Proportions: Finding the likelihood or probability of some event can be interpreted as a proportion problem, which implies the existence of a Part and a Whole. The way that you find how much a Part represent from the Whole is by dividing the Part by the Whole: $ frac{Part}{Whole} $ (if you are still uncomfortable with the concept and for some reason can't visualize why division would provide you with such information, just take it as a rule for now, at some point the concept will click). So whenever you are thinking about the probability or likelihood of an event always think about the Part and the Whole, which leads to the next question: How do we find the Part and the Whole? We will need to understand how we represent events to answer that question.
Sets, Subsets and Cardinalities: An event is a set of outcomes, the cardinality of a set is the measure of the size of that same set. Now, for the sake of our probabilistic theory discussion, you will see that a Part is always a subset of the Whole (you may be thinking of some edge cases that do not fit this rule, keep reading and you will see that this idea indeed holds true for all cases).
The first step to find a probability of any event is to find the set that represents the Whole (i.e. the scope of all the possible outcomes, when you roll a die of 6 sides the Whole set can be represented as the numbers from 1 through 6, since they form all the possible outcomes), the second step is to find the Part (i.e. the set containing the subset of outcomes that you are trying to find the likelihood of occurrence).
Every probability calculation can be seen with the lens of a conditional probability (What!? Keep reading :) )
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory A and B = A when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set.
As an example, say we are interested in knowing the probability of getting an even number when rolling a die of 6 sides, you can always formulate the problem thinking about conditional probability, when we defined that we are going to be rolling a die of six sides we just implicitly defined the Whole set (i.e the possible outcomes are numbers from 1 to 6), the second piece of information is the event itself that we would like to obtain the probability (even number), that is the Part. An important subtlety here, the Part you are interested here in the first place, is not ALL the even numbers but the ones that make part of your Whole set (i.e. subset of the Whole). How do you find that subset? The intersection of the two sets? precisely.
So the intersection gives you the Part that you are interested in, and once you have the Whole and Part you divide the two and find the probability (frequency of occurrence, or the comparative relation between the Part and the Whole).
If the outcomes or event that you are trying to find the likelihood does not make any part of the Whole, in other words, if the intersection of both sets is empty then we know that the probability of such event is zero, the event is simply impossible to occur, there is no way you will get a 7 if you roll a die of 6 sides.
Always remember when you see P(A|B) we are already explicitly saying that B is the Whole or universal set, all you have to do now is to find the comparative relation between the Part and Whole..
To end, let me clarify a few common misconceptions:
On the context of conditional probability, aside from the relationship between the
WholeandPart, the probability of what you are judging to be theWholeset does not matter by itself! As an example, the probability of someone having a tumor is irrelevant, if what we are trying to find is the probability of someone having a benign tumor given the person is already diagnosed with a tumor.P(A|B)is not the same asP(B|A): The probability of someone having a benign tumor given the person is already diagnosed with a tumor, is not the same as the probability of someone being diagnosed with a tumor given the fact the person has a benign tumor.Shouldn't
P(A|B)be equal toP(A and B)?
This is a common misinterpretation of what conditional probability means, we are not interested in the probability of A and B happening (i.e. $ frac{P(A cap B)}{P(U)} $), that's called joint probability and is a separate (and related) subject. Conditional probability means, now thatBALREADY HAPPENED, what is the probability ofAalso happening?
answered Mar 10 at 2:30
bsdbsd
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In summary, the essence of conditional probability is in every probability calculation, find the Whole (i.e. universal set), find the Part (intersection between the event you are trying to calculate the probability and the universal set) and divide the Part by the Whole to find the probability of given event (i.e the comparative relation between Part and Whole).
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory $ A cap B = A $ when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set (U).
Why do we need to divide?
Proportions: Finding the likelihood or probability of some event can be interpreted as a proportion problem, which implies the existence of a Part and a Whole. The way that you find how much a Part represent from the Whole is by dividing the Part by the Whole: $ frac{Part}{Whole} $ (if you are still uncomfortable with the concept and for some reason can't visualize why division would provide you with such information, just take it as a rule for now, at some point the concept will click). So whenever you are thinking about the probability or likelihood of an event always think about the Part and the Whole, which leads to the next question: How do we find the Part and the Whole? We will need to understand how we represent events to answer that question.
Sets, Subsets and Cardinalities: An event is a set of outcomes, the cardinality of a set is the measure of the size of that same set. Now, for the sake of our probabilistic theory discussion, you will see that a Part is always a subset of the Whole (you may be thinking of some edge cases that do not fit this rule, keep reading and you will see that this idea indeed holds true for all cases).
The first step to find a probability of any event is to find the set that represents the Whole (i.e. the scope of all the possible outcomes, when you roll a die of 6 sides the Whole set can be represented as the numbers from 1 through 6, since they form all the possible outcomes), the second step is to find the Part (i.e. the set containing the subset of outcomes that you are trying to find the likelihood of occurrence).
Every probability calculation can be seen with the lens of a conditional probability (What!? Keep reading :) )
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory A and B = A when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set.
As an example, say we are interested in knowing the probability of getting an even number when rolling a die of 6 sides, you can always formulate the problem thinking about conditional probability, when we defined that we are going to be rolling a die of six sides we just implicitly defined the Whole set (i.e the possible outcomes are numbers from 1 to 6), the second piece of information is the event itself that we would like to obtain the probability (even number), that is the Part. An important subtlety here, the Part you are interested here in the first place, is not ALL the even numbers but the ones that make part of your Whole set (i.e. subset of the Whole). How do you find that subset? The intersection of the two sets? precisely.
So the intersection gives you the Part that you are interested in, and once you have the Whole and Part you divide the two and find the probability (frequency of occurrence, or the comparative relation between the Part and the Whole).
If the outcomes or event that you are trying to find the likelihood does not make any part of the Whole, in other words, if the intersection of both sets is empty then we know that the probability of such event is zero, the event is simply impossible to occur, there is no way you will get a 7 if you roll a die of 6 sides.
Always remember when you see P(A|B) we are already explicitly saying that B is the Whole or universal set, all you have to do now is to find the comparative relation between the Part and Whole..
To end, let me clarify a few common misconceptions:
On the context of conditional probability, aside from the relationship between the
WholeandPart, the probability of what you are judging to be theWholeset does not matter by itself! As an example, the probability of someone having a tumor is irrelevant, if what we are trying to find is the probability of someone having a benign tumor given the person is already diagnosed with a tumor.P(A|B)is not the same asP(B|A): The probability of someone having a benign tumor given the person is already diagnosed with a tumor, is not the same as the probability of someone being diagnosed with a tumor given the fact the person has a benign tumor.Shouldn't
P(A|B)be equal toP(A and B)?
This is a common misinterpretation of what conditional probability means, we are not interested in the probability of A and B happening (i.e. $ frac{P(A cap B)}{P(U)} $), that's called joint probability and is a separate (and related) subject. Conditional probability means, now thatBALREADY HAPPENED, what is the probability ofAalso happening?
answered Mar 10 at 2:30
bsdbsd
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In summary, the essence of conditional probability is in every probability calculation, find the Whole (i.e. universal set), find the Part (intersection between the event you are trying to calculate the probability and the universal set) and divide the Part by the Whole to find the probability of given event (i.e the comparative relation between Part and Whole).
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory $ A cap B = A $ when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set (U).
Why do we need to divide?
Proportions: Finding the likelihood or probability of some event can be interpreted as a proportion problem, which implies the existence of a Part and a Whole. The way that you find how much a Part represent from the Whole is by dividing the Part by the Whole: $ frac{Part}{Whole} $ (if you are still uncomfortable with the concept and for some reason can't visualize why division would provide you with such information, just take it as a rule for now, at some point the concept will click). So whenever you are thinking about the probability or likelihood of an event always think about the Part and the Whole, which leads to the next question: How do we find the Part and the Whole? We will need to understand how we represent events to answer that question.
Sets, Subsets and Cardinalities: An event is a set of outcomes, the cardinality of a set is the measure of the size of that same set. Now, for the sake of our probabilistic theory discussion, you will see that a Part is always a subset of the Whole (you may be thinking of some edge cases that do not fit this rule, keep reading and you will see that this idea indeed holds true for all cases).
The first step to find a probability of any event is to find the set that represents the Whole (i.e. the scope of all the possible outcomes, when you roll a die of 6 sides the Whole set can be represented as the numbers from 1 through 6, since they form all the possible outcomes), the second step is to find the Part (i.e. the set containing the subset of outcomes that you are trying to find the likelihood of occurrence).
Every probability calculation can be seen with the lens of a conditional probability (What!? Keep reading :) )
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory A and B = A when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set.
As an example, say we are interested in knowing the probability of getting an even number when rolling a die of 6 sides, you can always formulate the problem thinking about conditional probability, when we defined that we are going to be rolling a die of six sides we just implicitly defined the Whole set (i.e the possible outcomes are numbers from 1 to 6), the second piece of information is the event itself that we would like to obtain the probability (even number), that is the Part. An important subtlety here, the Part you are interested here in the first place, is not ALL the even numbers but the ones that make part of your Whole set (i.e. subset of the Whole). How do you find that subset? The intersection of the two sets? precisely.
So the intersection gives you the Part that you are interested in, and once you have the Whole and Part you divide the two and find the probability (frequency of occurrence, or the comparative relation between the Part and the Whole).
If the outcomes or event that you are trying to find the likelihood does not make any part of the Whole, in other words, if the intersection of both sets is empty then we know that the probability of such event is zero, the event is simply impossible to occur, there is no way you will get a 7 if you roll a die of 6 sides.
Always remember when you see P(A|B) we are already explicitly saying that B is the Whole or universal set, all you have to do now is to find the comparative relation between the Part and Whole..
To end, let me clarify a few common misconceptions:
On the context of conditional probability, aside from the relationship between the
WholeandPart, the probability of what you are judging to be theWholeset does not matter by itself! As an example, the probability of someone having a tumor is irrelevant, if what we are trying to find is the probability of someone having a benign tumor given the person is already diagnosed with a tumor.P(A|B)is not the same asP(B|A): The probability of someone having a benign tumor given the person is already diagnosed with a tumor, is not the same as the probability of someone being diagnosed with a tumor given the fact the person has a benign tumor.Shouldn't
P(A|B)be equal toP(A and B)?
This is a common misinterpretation of what conditional probability means, we are not interested in the probability of A and B happening (i.e. $ frac{P(A cap B)}{P(U)} $), that's called joint probability and is a separate (and related) subject. Conditional probability means, now thatBALREADY HAPPENED, what is the probability ofAalso happening?
answered Mar 10 at 2:30
bsdbsd
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In summary, the essence of conditional probability is in every probability calculation, find the Whole (i.e. universal set), find the Part (intersection between the event you are trying to calculate the probability and the universal set) and divide the Part by the Whole to find the probability of given event (i.e the comparative relation between Part and Whole).
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory $ A cap B = A $ when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set (U).
Why do we need to divide?
Proportions: Finding the likelihood or probability of some event can be interpreted as a proportion problem, which implies the existence of a Part and a Whole. The way that you find how much a Part represent from the Whole is by dividing the Part by the Whole: $ frac{Part}{Whole} $ (if you are still uncomfortable with the concept and for some reason can't visualize why division would provide you with such information, just take it as a rule for now, at some point the concept will click). So whenever you are thinking about the probability or likelihood of an event always think about the Part and the Whole, which leads to the next question: How do we find the Part and the Whole? We will need to understand how we represent events to answer that question.
Sets, Subsets and Cardinalities: An event is a set of outcomes, the cardinality of a set is the measure of the size of that same set. Now, for the sake of our probabilistic theory discussion, you will see that a Part is always a subset of the Whole (you may be thinking of some edge cases that do not fit this rule, keep reading and you will see that this idea indeed holds true for all cases).
The first step to find a probability of any event is to find the set that represents the Whole (i.e. the scope of all the possible outcomes, when you roll a die of 6 sides the Whole set can be represented as the numbers from 1 through 6, since they form all the possible outcomes), the second step is to find the Part (i.e. the set containing the subset of outcomes that you are trying to find the likelihood of occurrence).
Every probability calculation can be seen with the lens of a conditional probability (What!? Keep reading :) )
Let A be any given event and U be the set of all possible events or outcomes:
$ P(A)=frac{P(A)}{P(U)} = frac{P(A cap U)}{P(U)} = P(A|U) $
If you are struggling to see the full equivalence above, remember that in set theory A and B = A when A is contained in B, which is always the case for any set when are trying to find the intersection with the Universal set.
As an example, say we are interested in knowing the probability of getting an even number when rolling a die of 6 sides, you can always formulate the problem thinking about conditional probability, when we defined that we are going to be rolling a die of six sides we just implicitly defined the Whole set (i.e the possible outcomes are numbers from 1 to 6), the second piece of information is the event itself that we would like to obtain the probability (even number), that is the Part. An important subtlety here, the Part you are interested here in the first place, is not ALL the even numbers but the ones that make part of your Whole set (i.e. subset of the Whole). How do you find that subset? The intersection of the two sets? precisely.
So the intersection gives you the Part that you are interested in, and once you have the Whole and Part you divide the two and find the probability (frequency of occurrence, or the comparative relation between the Part and the Whole).
If the outcomes or event that you are trying to find the likelihood does not make any part of the Whole, in other words, if the intersection of both sets is empty then we know that the probability of such event is zero, the event is simply impossible to occur, there is no way you will get a 7 if you roll a die of 6 sides.
Always remember when you see P(A|B) we are already explicitly saying that B is the Whole or universal set, all you have to do now is to find the comparative relation between the Part and Whole..
To end, let me clarify a few common misconceptions:
On the context of conditional probability, aside from the relationship between the
WholeandPart, the probability of what you are judging to be theWholeset does not matter by itself! As an example, the probability of someone having a tumor is irrelevant, if what we are trying to find is the probability of someone having a benign tumor given the person is already diagnosed with a tumor.P(A|B)is not the same asP(B|A): The probability of someone having a benign tumor given the person is already diagnosed with a tumor, is not the same as the probability of someone being diagnosed with a tumor given the fact the person has a benign tumor.Shouldn't
P(A|B)be equal toP(A and B)?
This is a common misinterpretation of what conditional probability means, we are not interested in the probability of A and B happening (i.e. $ frac{P(A cap B)}{P(U)} $), that's called joint probability and is a separate (and related) subject. Conditional probability means, now thatBALREADY HAPPENED, what is the probability ofAalso happening?
answered Mar 10 at 2:30
bsdbsd
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 16:10
answered Mar 10 at 2:30
bsdbsd
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 2:30
bsdbsd
1011
answered Mar 10 at 2:30
bsdbsd
1011
1011
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
bsd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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$begingroup$
The formula is right. But the statement is not right. It has to be: "Probability of $A$ happening given $B$ is equal to the probability of A and B both happening divided by B happening."
$endgroup$
– callculus
Jul 4 '15 at 6:42
1
$begingroup$
$P(A|B)$ isn´t smaller than $P(Acap B)$. $P(B)cdot P(A|B)=P(A cap B)$ and $P(B) leq 1$. Thus $P(A|B)geq P(A cap B)$
$endgroup$
– callculus
Jul 4 '15 at 7:02
$begingroup$
@calculus I am aware that the probability of B given some observations is going to reduce chances of A happening given B as a condition condition, but the problem is the play of words when stating the formula such as "both happening" and "this happening given that" is what I find confusing.
$endgroup$
– pwned
Jul 4 '15 at 7:23
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I understood your questions. I only make comments on you false statements. I hope it is clear now.
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– callculus
Jul 4 '15 at 7:27
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@pwned calculus's point is a good one. Your confusion seemed to be about the difference between "A and B" vs. "A given B". You noted that "A given B" sort of entails "A and B". calculus explained that your intuition is partly correct which is that "A given B" must be at least as probable as "A and B". The point is that, in a sense, "A given B" is a "weaker" condition than "A and B".
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– Jared
Jul 4 '15 at 7:37