# Independent event but yet not independent

• I
• rajeshmarndi
In summary: Here is an example of a sample space that is not uniform: Suppose you are playing a game where you draw two cards from a deck and win if the two cards are the same color. Then the sample space is all of the possible two-card combinations (52*52). Suppose you are playing the game with a deck of 52 cards and you know that one of the cards is the Ace of Spades. Then the sample space is all combinations of cards except the Ace of Spades (51*52). The "event" of drawing a red card has a probability of 1/2 in the first game, but it has
rajeshmarndi
These are two example where it seems they are both independent but the second one is not. Why?

First problem.

There are 100 balls numbered 1 through 100 in a bag and one ball is randomly picked and returned to the bag for the next pick.

There is event A that the ball picked is divisible by 2 and event B that the ball picked is divisible by 5. It is given that both are independent event which seems very much obvious because they cannot affect the next pick, since the ball is placed back.

Also P(A)=50/100 = 1/2, P(B)=20/100 = 1/5 and P(A ∩ B) = 10/100=1/10

i.e P(A ∩ B)=P(A).P(B)

I too, do not understand how P(A ∩ B) decide independent event?

Second problem.

Only difference, is that, the bag now contain 65 ball.

And now these event A and B become non-independent event. How?

I know now,

P(A ∩ B) ≠ P(A).P(B).

Again as I stated above, how the above equation, is able to determine independent event.

I know P(A ∩ B) = P(A).P(B) . But how does this determine independent event?

Here too in the second problem, both A and B seems to be, not affected by each other, just the same way, they seem obviously independent event as in the first problem. But yet, they are now non-independent event.

Thanks.

[edit:]
The exact phrase from the question.

First problem.
There are 100 tickets in a bag numbered 1 through 100 and a ticket is picked at random. Let A be the event that the number on the ticket is divisible by 2 and let B be the event that number on the ticket is divisible by 5. Show that A and B are independent.

Second problem.
Let A and B be defined as in First problem above where a ticket is picked at random from a bag containing 65 tickets numbered 1 through 65. In this case A and B are not independent.

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rajeshmarndi said:
Second problem.

Only difference, is that, the bag now contain 65 ball.

And now these event A and B become non-independent event. How?

I know now,

P(A ∩ B) ≠ P(A).P(B).
Why do you say this? I don't know any way to compute P(A ∩ B) other than by computing P(A).P(B).

FactChecker said:
Why do you say this? I don't know any way to compute P(A ∩ B) other than by computing P(A).P(B).
In the second problem, P(A ∩ B) ≠ P(A).P(B) .
P(A)=32/65, P(B)=13/65 and P(A ∩ B)=6/65. Hence P(A ∩ B) ≠ P(A).P(B) .

Both event A and B seems to be independent, just like in the first problem. Because , I could not see here, how does event A and B affect each other. Also I am unable to comphrehend, what is the relation between P(A ∩ B) and independent event. That is, why if P(A ∩ B) = P(A).P(B) , make both event A and B independent.

rajeshmarndi said:
In the second problem, P(A ∩ B) ≠ P(A).P(B) .
P(A)=32/65, P(B)=13/65 and P(A ∩ B)=6/65. Hence P(A ∩ B) ≠ P(A).P(B) .
In the second problem by using 6/65, you are interpreting ##A \cap B## to mean the same ball is both divisible by 2 and also divisible by 5. In the first problem, you interpreted ##A \cap B## to mean ball A is divisible by 2 and a possibly different ball B is divisible by 5.

FactChecker
Here is a simpler example that may clarify things:
We all agree that two coin tosses are independent, so the events of getting a head on the first toss and getting a tails on the second toss are independent. But the probability of getting both a heads and a tails on one toss is 0. This is similar to what you are doing in your second example.

rajeshmarndi said:
In the second problem, P(A ∩ B) ≠ P(A).P(B) .
P(A)=32/65, P(B)=13/65 and P(A ∩ B)=6/65. Hence P(A ∩ B) ≠ P(A).P(B).
You are calculating P(A ∩ B) wrong, but I am not sure where you are getting those numbers, so I can't tell where your mistake is.

There are 65 ways to draw the first ball and 65 ways to draw the second ball so there are 4225 total possibilities. So the denominator should be 4225, not 65. Of those there are 416 combinations that satisfy the condition A ∩ B. So P(A ∩ B) = 416/4225 = P(A).P(B)

You made a similar mistake above. In the first experiment you should have had P(A ∩ B) = 1000/10000 since there were 10000 possibilities and 1000 of them were A ∩ B.

Dale said:
You are calculating P(A ∩ B) wrong.There are 65 ways to draw the first ball and 65 ways to draw the second ball
I apolize I mentioned for the next pick, it sound like there is two draw. Actually, there is one draw. And the probability in first problem , of event A and B is therefore P(A), P(B) and P(A ∩ B). Similarly in the second problem.

rajeshmarndi said:
I apolize I mentioned for the next pick, it sound like there is two draw. Actually, there is one draw. And the probability in first problem , of event A and B is therefore P(A), P(B) and P(A ∩ B). Similarly in the second problem.
Then, yes. You are correct. In the first example the events are independent and in the second example, they are not.
EDIT: Independence is very dependent (Hahaha) on the sample space.

I have edited my question in the beginning, to the exact phrase as in the book.

FactChecker said:
Then, yes. You are correct. In the first example the events are independent and in the second example, they are not.
This is what I couldn't understand. Why in the second example they are not.

rajeshmarndi said:
This is what I couldn't understand. Why in the second example they are not.
Independence is very dependent (Hahaha) on the sample space.

This is an interesting question. The "independence" of events in a physical situation is often explained by the two events being remote from each other or not casually related. I see no such physical explanation that applies in this case. In fact, this case shows that events can be independent and still have some physical effect on each other.

This case is better explained by the equivalent definition of independence that events ##A## and ##B## are independent iff the probability of "##A## given ##B##" is equal to the probability of ##A##. We ask whether information about the outcome of ##B## gives us an information that influences the probability of ##A## - not necessarily whether the information changes what we know about the specific ways in which the event ##A## can occur.

It just happens that in the first problem, given the information "The ball is divisible by 5" we don't gain any information about the probability that the ball is also divisible by 2 even though given the information that the ball is divisible by 5 does change the possible set of balls that were drawn. For example, without the information "The ball is divisible by 5" the event "the ball is divisible by 2" may occur if we draw ball number 4. ##\ ## Given the information "the ball's number is divisible by 5" the event "the ball is divisible by 2" cannot occur by drawing ball number 4.

So we can say the information "the ball is divisible by 5" does change the physical possibilities for which ball is drawn. However, the definition of "independent" is not a definition that concerns which physical outcomes are specified. The definition of "independent" only concerns how information affects a probability - typically the probability is given as a fraction of outcomes that have a certain property. So there will be cases (such as 100 total balls) where information changes which outcomes can happen , but does not change the fraction of those outcomes that have some property.

rajeshmarndi said:
Actually, there is one draw
That is a very different problem. Yes, they are not independent. I am not sure why you think they should be.

Here is an example that makes it more obvious: pick a number between 1 and 100. Let A be the probability that the number is odd and let B be the probability that the same number is prime. If you know that a number is prime, then you have a very good chance that the number is odd. So knowing B gives you a lot of information about A.

Dale said:
Yes, they are not independent. I am not sure why you think they should be.
What is the difference when the number of ball is changed to 65. Also why do we need to use P(A ∩ B) = P(A).P(B) to determine they are independent. I mean it can be seen it is very obvious, suppose we get now 15 "divisible by 5". The future pick cannot be dependent on this pick or any earlier pick, anyway.

I understand when a draw reduced the availability, the probability for the next draw is affected. Else any event should be independent.

rajeshmarndi said:
What is the difference when the number of ball is changed to 65.
The difference is that you have removed a different proportion of A than you have removed of B. So now knowing A gives you some information about B.

rajeshmarndi said:
The future pick cannot be dependent on this pick or any earlier pick, anyway.
It sounds like you are going back to the original problem where there were two successive draws with rsplacement. Please stick to one comparison. This is difficult to follow. Do you want to discuss two successive draws (independent, but your calculation is wrong) or one draw (can be dependent as you have correctly calculated)

rajeshmarndi said:
Also why do we need to use P(A ∩ B) = P(A).P(B) to determine they are independent.
That is the definition of independent. I don't understand the question. Are you really asking why you have to use the definition of "independent" to determine if something is independent? I mean, that is what a definition is for. You say "I am going to use the word [X] to mean [definition of X]" then if you want to determine if something is X you just check to see if the definition applies.

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Dale said:
The difference is that you have removed a different proportion of A then you have removed of B. So now knowing A gives you some information about B.
Suppose for example toss of a coin. Event A "getting a Head" and event B "getting a Tail". Then
P(A)=P(B)=.5 and P(A ∩ B)=0. Hence P(A ∩ B) ≠ P(A).P(B).

Isn't still event A and B are independent. That is, they do not affect the probability of the other.

rajeshmarndi said:
Isn't still event A and B are independent. That is, they do not affect the probability of the other
No, they are most definitely not independent. If you get heads then you cannot also have tails. Knowing one gives you complete information about the other.

Suppose that we were betting on a fair coin toss, you are betting on tails. If you don't know anything else then you might accept any odds better than 1:1. Do you understand the reasoning behind that?

Now, suppose that you additionally know that it is heads. Now, what odds would you accept? Would you still accept a 1:1 odds wager?

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Ok, thanks. I am getting some idea about independent events. If any doubts I will further post.

(Correcting an earlier post that I have deleted):
The attachment shows Venn diagrams where the area of an event is the probability of the event. The top figure shows two independent events A and B. P(A∩B)=P(A)×P(B). In the bottom figure, P(B) has been reduced in a way such that the reduction factor of P(B) is not the same as the reduction factor of P(A∩B). So A and B are not independent events in the bottom figure. This is similar to what you did by reducing 70 to 65. You reduced the even numbers by a different proportion from the reduction of (even and divisible by 5).

What is the reason, it become necessary to satisfy P(A∩B)=P(A)×P(B), for events to be independent?

What would happen to the independency, without them. I mean, isn't it enough, that neither affect the probability of the other to become independent events.

rajeshmarndi said:
I mean, isn't it enough, that neither affect the probability of the other to become independent events
Yes. You can use Bayes theorem to prove that P(A∩B)=P(A) P(B) implies that both P(A|B)=P(A) and P(B|A)=P(B). It is the same thing.

## What is an independent event?

An independent event is an event where the outcome of one event does not affect the outcome of another event. This means that the probability of one event occurring remains the same regardless of whether or not the other event has occurred.

## What is a dependent event?

A dependent event is an event where the outcome of one event does affect the outcome of another event. This means that the probability of one event occurring is dependent on whether or not the other event has occurred.

## Can an event be independent but also not independent?

No, an event cannot be both independent and not independent. An event is either independent or dependent, there is no middle ground. However, an event can appear independent at first but upon further analysis, it may be revealed to be dependent.

## What is an example of an independent event but yet not independent?

An example of an independent event but yet not independent is flipping a coin twice. Each individual coin flip is independent of the other, but if the first flip lands on heads, the probability of the second flip also landing on heads is affected. Therefore, the event of getting heads on the second flip is dependent on the outcome of the first flip.

## How can you determine if an event is independent or dependent?

To determine if an event is independent or dependent, you can use the formula P(A and B) = P(A) x P(B). If the probability of event A and event B occurring together is equal to the product of the individual probabilities of each event occurring, then the events are independent. If the probability is not equal, then the events are dependent.

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