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What exactly is the difference between "independent" and "dependent" [variables] ?

I could use all the examples I can get for this one.

thanx.

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- Thread starter Imparcticle
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- #1

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What exactly is the difference between "independent" and "dependent" [variables] ?

I could use all the examples I can get for this one.

thanx.

- #2

Zurtex

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[tex]P(A | B) = P(A)[/tex].

That is to say, that the probability of A given B, doesn’t really matter as B happening doesn’t affect A, so it is just equal to the probably of A.

You can derive from the above statement to say when the probability of A and B are independent then:

[tex]P(A \cap B) = P(A) P(B)[/tex]

- #3

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The probability of Event A happening is

How many times you take the chance divided by how many posible outcomes there are. Let's take the simplest example...

You flip a coin one time, there are two posible outcomes (heads or tails). [tex] \frac {1} {2} = .5[/tex]. Therefore that chance is .5, 50%, 1/2, whatever you wanna call it.

Now if you flip that coin two times, your chances are [tex] \frac {2} {2} = 1 [/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice).

This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?

How many times you take the chance divided by how many posible outcomes there are. Let's take the simplest example...

You flip a coin one time, there are two posible outcomes (heads or tails). [tex] \frac {1} {2} = .5[/tex]. Therefore that chance is .5, 50%, 1/2, whatever you wanna call it.

Now if you flip that coin two times, your chances are [tex] \frac {2} {2} = 1 [/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice).

This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?

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eNathan said:The probability of Event A happening is

How many times you take the chance divided by how many posible outcomes there are. Let's take the simplest example...

You flip a coin one time, there are two posible outcomes (heads or tails). [tex] \frac {1} {2} = .5[/tex]. Therefore that chance is .5, 50%, 1/2, whatever you wanna call it.

Now if you flip that coin two times, your chances are [tex] \frac {2} {2} = 1 [/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice).

This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?

But why does it matter if you flip the coin twice ?

it would still be 100 % tails or heads regardless of number of tosses no ?

and that statement obviously doesnt work in practice, because as you said, it wont definitely land on heads if you tried it.

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- #5

Zurtex

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Lets look at a coin problem.

Lets call event A the probability of getting a heads on the first coin and event B the probability of getting a heads on the on the 2nd coin. Now:

[tex]P(A) = \frac{1}{2} \; \text{and} \; P(B) = \frac{1}{2}[/tex]

If event A happens then P(B) = 1/2. Or that is to say that given event A then the probability of B is still the same, or:

[tex]P(B | A) = P(B)[/tex]

This tells us the two events A and B are independent, from this we can say that the probability of both the A and B occurring is equal to the probability of A occurring multiplied by the probability of B occurring. Or:

[tex]P(A \cap B) = P(A) P(B) = \frac{1}{4}[/tex]

Try and tackle every problem with rigorous use of axiomatic probability rules if you don’t have understanding of probability, otherwise more often than not you just tend to get confused.

- #6

HallsofIvy

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Where to begin! In addition to having absolutely nothing to do with the original question about the difference between "dependent" and "independent" events, eNathan's answer is simple nonsense!

"The probability of Event A happening is

How many times you take the chance divided by how many posible outcomes there are. "

No, it has nothing to do with "how many times you take the chance". In simple discrete probability it is the number of outcomes that correspond to event A divided by the total number of possible outcomes.

"Now if you flip that coin two times, your chances are [tex]\frac{2}{2}= 1[/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice)."

Number of "times you flip" divided by number of outcomes (head or tail: 2)! But as I said, that's completely wrong. The correct calculation here is: if you flip a coin twice, there are**four** possible outcomes: (H,H) (heads on both first and second flips), (H,T) (heads on first flip, tails on second), (T,H) (tails on first flip, heads on second), (T,T) (tails on both first and second flip). I'm not sure what he means by "getting either a head or a tail. If you flip a coin ONCE you are certain to get **either** heads of tails! eNathan may be saying that since the probability of getting heads on one flip is 1/2 and the probability of getting heads on one flip is 1/2, the probabililty of getting "heads **and** tails" (not "or") is 1/2+ 1/2= 1. No, the probability of getting heads on the first flip and heads on the second is (1/2)*(1/2)= 1/4- you multiply, not add.

Or, he may be arguing that since the probability of "a or b" is prob(a)+ prob(b) (for "mutally exclusive" events- the more general formula is prob(a)+prob(b)-prob(a**and** b)), the probability of getting heads **or** tails is 1/2+ 1/2. That's true- on **one** flip, not two. Since the coin can only come up heads or tails, the probability of getting heads **or** tails on one flip 1- it's certain to happen.

"This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?"

No, if you flip a coin twice, the probability of getting**a** head (at least one) is 75%: you could get two heads:probability (1/2)(1/2)= 1/4. You could get a head on the first flip and tail on the second: probability (1/2)(1/2)= 1/4. You could get a tail on the first flip and a head on the second: probability (1/2)(1/2)= 1/4. Probability of getting at least one head is 1/4+ 1/4+ 1/4= 3/4. (Probability of getting **exactly** one head, if that was what was meant, if 1/4+1/4= 1/2.)

eNathan- if the result of your reasoning is non-sense, you should at least consider the possibility (if not "probability") that it is your reasoning that is at fault rather than mathematics!

"The probability of Event A happening is

How many times you take the chance divided by how many posible outcomes there are. "

No, it has nothing to do with "how many times you take the chance". In simple discrete probability it is the number of outcomes that correspond to event A divided by the total number of possible outcomes.

"Now if you flip that coin two times, your chances are [tex]\frac{2}{2}= 1[/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice)."

Number of "times you flip" divided by number of outcomes (head or tail: 2)! But as I said, that's completely wrong. The correct calculation here is: if you flip a coin twice, there are

Or, he may be arguing that since the probability of "a or b" is prob(a)+ prob(b) (for "mutally exclusive" events- the more general formula is prob(a)+prob(b)-prob(a

"This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?"

No, if you flip a coin twice, the probability of getting

eNathan- if the result of your reasoning is non-sense, you should at least consider the possibility (if not "probability") that it is your reasoning that is at fault rather than mathematics!

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A dependant variable is a variable whose value is determined by the value of an independent variable, simply the dependant variable is a function (x) of the independant variable (y). f(x) = y

An independant variable is the observed variable, and is not dependant on any other variables in the environment.

NewScientist

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Thats right, but, i believe when the OP meant variables, he/she meant "random variables" (given that the subject was probability), which arent exactly defined the way you have done here.NewScientist said:

A dependant variable is a variable whose value is determined by the value of an independent variable, simply the dependant variable is a function (x) of the independant variable (y). f(x) = y

An independant variable is the observed variable, and is not dependant on any other variables in the environment.

NewScientist

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The dependant variable is not a random variable, it is a function of other variables - it DEPENDS on their inputs.

In probability, especially forcasting or doing simulations, some data is input as independat variables, and the dependant variables follow.

- #10

HallsofIvy

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Here's an example that I thought of when I was reading eNathan's ravings: Suppose a coin has probility 0.5 of either heads or tails (a fair coin):P(H)= 1/2. I flip the coin twice. Since what happens on the first flip does not affect the second flip, they are "independent". The probability that the first flip is heads and the second tails is (1/2)(1/2)= 1/4. The probability of "first tails, then heads" is also 1/4 and the probability of "one heads, the other tails, in any order" is 1/4+ 1/4= 1/2.

Now suppose I have an "unfair" coin that has probability 1/3 of coming up heads, 2/3 of coming up tails:P(H)= 1/3, P(T)= 2/3. The probabilty that, when I flip it twice, the first flip is heads and the second tails is (1/3)(2/3)= 2/9. The two flips are still independent. The probablity of "first tails, second heads" is still 2/9 and the probability of "one heads, the other tails, in any order" is 2/9+ 2/9= 4/9.

NOW, I flip my fair coin. If it comes up heads I will flip it again, if it comes up tails, I will flip the unfair coin. The probability it comes up heads on the first flip is 1/2 and then I flip it again. Now the probability of getting tails on the second flip is also 1/2 and the probability of getting "first heads, then tails" is (1/2)(1/2)= 1/4. But there is a 1/2 probability that the fair coin will come up tails. If that happens, I flip the unfair coin and now the probability that will come up heads is only 1/3. The probability of getting "first tails, then heads" is (1/2)(1/3)= 1/6. The probabilities on the second flip depend on what happens on the first flip. The probability of getting "one heads, one tails, in any order" is now 1/4+ 1/6= 5/12.

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My math book defines the basic counting principle the following way:

Suppose an event can occur in p different ways. Another event can occur in q different ways. There are (p)(q) ways both events can occur.

The book then pointed out that this principle can be extended to accomadate those events which are defined as being either dependent or independent. I thought I understood how this could be done after looking at the examples but then became lost when I had to do the practice problems.

For example, here is one problem that I do not understand:

1.) How many 4 digit patterns are there in which all the digits are different?

answer: 480

When I look at this problem, all I can come up with is this:

The question can be restated as "How many 4 digit patterns can be formed from the numbers 0-9 using each number exactly once?"

I would thus do the problem by finding the quotient of [9!/(4-1)!] , in which case I'd get 60480.

I merely chose this method because it worked on the previous problem:

How many different 4 letter patterns can be formed frm the letters a, e,i,o, r,s and t if no letter occurs more than once?

answer:840

how to get the answer: [7!/(4-1)!]. where 7 represents the number of letters and 4 is the number of letter patterns. Why the minus 1 ? It just worked so I accepted it.

If there is anything I abhor, it is doing math using algorythms (<--gee, great speller I am!) that I memorize, instead of understand. So please don't give me formulae, but just some conceptual examples that'll make this as close to intuitive as it can be.

thanx again.

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HallsofIvy said:Hmmm, looking back I see that the original post did say "(variables)" but he also was talking about probability. I suspect he really did mean dependent and independent "events" rather than "variables".

Here's an example that I thought of when I was reading eNathan's ravings: Suppose a coin has probility 0.5 of either heads or tails (a fair coin):P(H)= 1/2. I flip the coin twice. Since what happens on the first flip does not affect the second flip, they are "independent". The probability that the first flip is heads and the second tails is (1/2)(1/2)= 1/4. The probability of "first tails, then heads" is also 1/4 and the probability of "one heads, the other tails, in any order" is 1/4+ 1/4= 1/2.

Now suppose I have an "unfair" coin that has probability 1/3 of coming up heads, 2/3 of coming up tails:P(H)= 1/3, P(T)= 2/3. The probabilty that, when I flip it twice, the first flip is heads and the second tails is (1/3)(2/3)= 2/9. The two flips are still independent. The probablity of "first tails, second heads" is still 2/9 and the probability of "one heads, the other tails, in any order" is 2/9+ 2/9= 4/9.

so if you flip it three times, you would multiply (1/2)(1/2)(1/2) (if it is a fair coin)?

Why does this (the one u explained) work?

- #13

HallsofIvy

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One way to see that is to write B as "A does not happen" so this becomes a binomial distribution. The probability of A is 1/2, so the probability of B is also 1/2. Doing the experiment n times means that there are 2

(from AAA... A to BBB... B: there are two things (A or B) that can happen each time so there are 2

"AAA...A" so the probabilty is 1/2

If you flip a fair coin n times, the probability of getting i heads and j= n-i tails (in any order) is

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HallsofIvy said:Where to begin! In addition to having absolutely nothing to do with the original question about the difference between "dependent" and "independent" events, eNathan's answer is simple nonsense!

"The probability of Event A happening is

How many times you take the chance divided by how many posible outcomes there are. "

No, it has nothing to do with "how many times you take the chance". In simple discrete probability it is the number of outcomes that correspond to event A divided by the total number of possible outcomes.

"Now if you flip that coin two times, your chances are [tex]\frac{2}{2}= 1[/tex] So you have a 100% chance of getting either a heads or a tails (if you flip the coin twice)."

Number of "times you flip" divided by number of outcomes (head or tail: 2)! But as I said, that's completely wrong. The correct calculation here is: if you flip a coin twice, there arefourpossible outcomes: (H,H) (heads on both first and second flips), (H,T) (heads on first flip, tails on second), (T,H) (tails on first flip, heads on second), (T,T) (tails on both first and second flip). I'm not sure what he means by "getting either a head or a tail. If you flip a coin ONCE you are certain to geteitherheads of tails! eNathan may be saying that since the probability of getting heads on one flip is 1/2 and the probability of getting heads on one flip is 1/2, the probabililty of getting "headsandtails" (not "or") is 1/2+ 1/2= 1. No, the probability of getting heads on the first flip and heads on the second is (1/2)*(1/2)= 1/4- you multiply, not add.

Or, he may be arguing that since the probability of "a or b" is prob(a)+ prob(b) (for "mutally exclusive" events- the more general formula is prob(a)+prob(b)-prob(aandb)), the probability of getting headsortails is 1/2+ 1/2. That's true- ononeflip, not two. Since the coin can only come up heads or tails, the probability of getting headsortails on one flip 1- it's certain to happen.

"This reminds me...I have been wanting to post a thread on this question, but I suppose it wont harm to throw it in right now. If you flip a coin twice, your chances of getting a heads (for instsance) is 100%, but there still IS a chance that you wont get heads. Is this just how the mathematics of chance operate?"

No, if you flip a coin twice, the probability of gettingahead (at least one) is 75%: you could get two heads:probability (1/2)(1/2)= 1/4. You could get a head on the first flip and tail on the second: probability (1/2)(1/2)= 1/4. You could get a tail on the first flip and a head on the second: probability (1/2)(1/2)= 1/4. Probability of getting at least one head is 1/4+ 1/4+ 1/4= 3/4. (Probability of gettingexactlyone head, if that was what was meant, if 1/4+1/4= 1/2.)

eNathan- if the result of your reasoning is non-sense, you should at least consider the possibility (if not "probability") that it is your reasoning that is at fault rather than mathematics!

Yes, you are right. I forgot it :lol: I dont know how I forgot that, I just learned it a year ago.

Actaully, I was just thinking about answer after I posted the question.No, if you flip a coin twice, the probability of gettingahead (at least one) is 75%:

Anyway, thanks for correcting me, ill make sure to double check what I write from now often. :grumpy:

Would this be correct. The chance of event A occuring is

[tex]P = 1 - \frac { (\frac {1} {O}) } {T} [/tex]

Where P is the probability, O is the posible number of outcomes, and T is how many times you take this chance?

- #15

EnumaElish

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Let us be careful with our vocabulary here. If you were doing regression analysis, just the opposite is true. In the regression equation y(t) = bNewScientist said:The dependant variable is not a random variable, it is a function of other variables - it DEPENDS on their inputs.

In probability, especially forcasting or doing simulations, some data is input as independat variables, and the dependant variables follow.

In general, if y = f(x,u) where x is either random or nonrandom and u is random, then y is random. In this general form, too, y is the dependent variable and x is the independent variable.

{P.S. If you know function f, then you can uniquely estimate the distribution of y from the distribution of u (assuming x is nonrandom, as usually assumed in regression analysis). Conversely, if you knew the distributions of both y and u (and provided one or two standard assumptions hold), then you can derive the function f uniquely (as a combination of b's and x's). This is the statistical theory of regression analysis explained in a few words while straining to sit upright in front of the computer screen.}

{P.P.S. Regression analysis uses "independent" and "dependent" differently from the way they are used in general probability theory. For this general usage, independence between two variables is defined in terms of conditional probabilities. Someone has already posted that definition above in this thread. An alternative definition of independence in that general sense is "two variables are independent if and only if their joint distribution is identical to the product of their individual (i.e. marginal) distributions." Moreover, linear independence is operationalized as the covariance between X and Y being zero. In other words, "two variables are linearly independent if and only if their covariance is zero." General independence implies linear independence; but the reverse is not true: two variables that have zero covariance may be dependent in a nonlinear way.}

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- #16

uart

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For example, here is one problem that I do not understand:

1.) How many 4 digit patterns are there in which all the digits are different?

answer: 480

When I look at this problem, all I can come up with is this:

The question can be restated as "How many 4 digit patterns can be formed from the numbers 0-9 using each number exactly once?"

I would thus do the problem by finding the quotient of [9!/(4-1)!] , in which case I'd get 60480.

No, neither of those answers is correct. I believe the answer in the book is incorrect (should be 5040).

You need to learn about combinations and permuations. There are general formuli for these but they are easily derived and understood. I'll take this "4 digits out of 10" example to help explain the formula.

Firstly assuming that each different odering of the same four digits is considered as being a different pattern (eg that 1,2,3,4 is considered a different pattern than 2,1,3,4) then the number of possible patterns is simply 10*9*8*7. That is, there are 10 ways that you can choose the first digit in each pattern (any one of 0..9) but only 9 ways you can choose the second digit (because the problem stipulates that each digit only occurs once), and only 8 way you can choose the third digit and so forth.

This type of counting problem occurs so often that with give it a name (permutation) and generalize the formula to a compact form. I think you can see that another way to write 10*9*8*7 is 10!/(10-4)!. This formula is then generalized for the number of permutation of "r" items drawn from a set of "n" items as :

A closely related problem is the case where each group of four digits is only counted once, regardless of how many ways that this group can be internally rearranged. That is, this time we dont count 1,2,3,4 as being a differnent pattern to 2,1,3,4. It is fairly easy to see that there are 4*3*2*1 = 4! ways of rearranging any particular 4 digits (4 way to choose the first digit of the arrangement, 3 ways to choose the second and so on) and hence there will be 4! less combinations of 4 the digits than there are permutations. In general there are r! less (I mean divide by r!) combinations than there are permutations, so the general formula for this is:

Hope that helps.

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- #17

uart

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Imparcticle said:How many different 4 letter patterns can be formed frm the letters a, e,i,o, r,s and t if no letter occurs more than once?

answer:840

how to get the answer: [7!/(4-1)!]. where 7 represents the number of letters and 4 is the number of letter patterns. Why the minus 1 ? It just worked so I accepted it.

Ok this is one example where your book does have the correct *answer*. Only it's not

- #18

HallsofIvy

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that is the same as (10*9*8*7*6*5*4*3*2*1)/(6*5*4*3*2*1)= 10!/6!.

In either case, that is 5040 as uart said.

For the other problem: "How many different 4 letter patterns can be formed frm the letters a, e,i,o, r,s and t if no letter occurs more than once?"

I would argue that there are 7 choices for the first letter, 6 for the second, 5 for the third, and 4 for the last. That means there are 7*6*5*4= (7*6*5*4*3*2*1)/(3*2*1)= 7!/3!= 7!/(7-4)!= 840.

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