Counter-intuitive independent and dependent events
up vote
1
down vote
favorite
In elementary school, I was taught about "independent" and "dependent" events with examples of taking two balls from a box. Assume there is a box containing 2 blue balls and 3 red balls. Two balls are drawn from the box, one is drawn after the other draw. Let $A$ and $B$ represent the probability of taking a blue ball in the first draw and that of taking a blue ball in the second draw, respectively.
Independent
If the ball taken in the first draw is returned before taking the second ball, find the probability that both balls are blue.
It is clear enough that both events are independent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{2}{5}\
&= frac{4}{25}
end{align}
Dependent
If the ball taken in the first draw is not returned before taking the second ball, find the probability that both balls are blue.
It is also clear enough that both events are dependent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{1}{4}\
&= frac{1}{10}
end{align}
Here the independence and dependency are extremely obvious.
Today I found a counter-intuitive example as follows.
One card is drawn from a deck of 52 cards. Let $X$ be the event that a club is selected, let $Y$ be the event that a face card (Jack, Queen, or King) is drawn, and let $Z$ be the event that a Jack is drawn.
The solution uses
$$P(X|Y)=P(X)$$
to judge that $X$ and $Y$ are independent. This equation is understandable and I know how it is derived.
The solution also shows that $$P(Z|Y)not =P(Z)$$ so $Y$ and $Z$ are dependent.
Question
Let's forget about the equations and just use our intuition.
How can we explain that
$X$ (taking a club) and $Y$ (taking a face) are independent?
$Y$ (taking a face) and $Z$ (taking a Jack) are dependent?
My intuition does work only for the first case about taking two balls because it is so obvious as follows.
$A$ (taking a blue) and $B$ (taking a blue) with replacement are independent.
$A$ (taking a blue) and $B$ (taking a blue) without replacement are dependent.
probability
add a comment |
up vote
1
down vote
favorite
In elementary school, I was taught about "independent" and "dependent" events with examples of taking two balls from a box. Assume there is a box containing 2 blue balls and 3 red balls. Two balls are drawn from the box, one is drawn after the other draw. Let $A$ and $B$ represent the probability of taking a blue ball in the first draw and that of taking a blue ball in the second draw, respectively.
Independent
If the ball taken in the first draw is returned before taking the second ball, find the probability that both balls are blue.
It is clear enough that both events are independent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{2}{5}\
&= frac{4}{25}
end{align}
Dependent
If the ball taken in the first draw is not returned before taking the second ball, find the probability that both balls are blue.
It is also clear enough that both events are dependent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{1}{4}\
&= frac{1}{10}
end{align}
Here the independence and dependency are extremely obvious.
Today I found a counter-intuitive example as follows.
One card is drawn from a deck of 52 cards. Let $X$ be the event that a club is selected, let $Y$ be the event that a face card (Jack, Queen, or King) is drawn, and let $Z$ be the event that a Jack is drawn.
The solution uses
$$P(X|Y)=P(X)$$
to judge that $X$ and $Y$ are independent. This equation is understandable and I know how it is derived.
The solution also shows that $$P(Z|Y)not =P(Z)$$ so $Y$ and $Z$ are dependent.
Question
Let's forget about the equations and just use our intuition.
How can we explain that
$X$ (taking a club) and $Y$ (taking a face) are independent?
$Y$ (taking a face) and $Z$ (taking a Jack) are dependent?
My intuition does work only for the first case about taking two balls because it is so obvious as follows.
$A$ (taking a blue) and $B$ (taking a blue) with replacement are independent.
$A$ (taking a blue) and $B$ (taking a blue) without replacement are dependent.
probability
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In elementary school, I was taught about "independent" and "dependent" events with examples of taking two balls from a box. Assume there is a box containing 2 blue balls and 3 red balls. Two balls are drawn from the box, one is drawn after the other draw. Let $A$ and $B$ represent the probability of taking a blue ball in the first draw and that of taking a blue ball in the second draw, respectively.
Independent
If the ball taken in the first draw is returned before taking the second ball, find the probability that both balls are blue.
It is clear enough that both events are independent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{2}{5}\
&= frac{4}{25}
end{align}
Dependent
If the ball taken in the first draw is not returned before taking the second ball, find the probability that both balls are blue.
It is also clear enough that both events are dependent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{1}{4}\
&= frac{1}{10}
end{align}
Here the independence and dependency are extremely obvious.
Today I found a counter-intuitive example as follows.
One card is drawn from a deck of 52 cards. Let $X$ be the event that a club is selected, let $Y$ be the event that a face card (Jack, Queen, or King) is drawn, and let $Z$ be the event that a Jack is drawn.
The solution uses
$$P(X|Y)=P(X)$$
to judge that $X$ and $Y$ are independent. This equation is understandable and I know how it is derived.
The solution also shows that $$P(Z|Y)not =P(Z)$$ so $Y$ and $Z$ are dependent.
Question
Let's forget about the equations and just use our intuition.
How can we explain that
$X$ (taking a club) and $Y$ (taking a face) are independent?
$Y$ (taking a face) and $Z$ (taking a Jack) are dependent?
My intuition does work only for the first case about taking two balls because it is so obvious as follows.
$A$ (taking a blue) and $B$ (taking a blue) with replacement are independent.
$A$ (taking a blue) and $B$ (taking a blue) without replacement are dependent.
probability
In elementary school, I was taught about "independent" and "dependent" events with examples of taking two balls from a box. Assume there is a box containing 2 blue balls and 3 red balls. Two balls are drawn from the box, one is drawn after the other draw. Let $A$ and $B$ represent the probability of taking a blue ball in the first draw and that of taking a blue ball in the second draw, respectively.
Independent
If the ball taken in the first draw is returned before taking the second ball, find the probability that both balls are blue.
It is clear enough that both events are independent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{2}{5}\
&= frac{4}{25}
end{align}
Dependent
If the ball taken in the first draw is not returned before taking the second ball, find the probability that both balls are blue.
It is also clear enough that both events are dependent. So the required probability is
begin{align}
P(Acap B) &=P(A) times P(B)\
&= frac{2}{5} times frac{1}{4}\
&= frac{1}{10}
end{align}
Here the independence and dependency are extremely obvious.
Today I found a counter-intuitive example as follows.
One card is drawn from a deck of 52 cards. Let $X$ be the event that a club is selected, let $Y$ be the event that a face card (Jack, Queen, or King) is drawn, and let $Z$ be the event that a Jack is drawn.
The solution uses
$$P(X|Y)=P(X)$$
to judge that $X$ and $Y$ are independent. This equation is understandable and I know how it is derived.
The solution also shows that $$P(Z|Y)not =P(Z)$$ so $Y$ and $Z$ are dependent.
Question
Let's forget about the equations and just use our intuition.
How can we explain that
$X$ (taking a club) and $Y$ (taking a face) are independent?
$Y$ (taking a face) and $Z$ (taking a Jack) are dependent?
My intuition does work only for the first case about taking two balls because it is so obvious as follows.
$A$ (taking a blue) and $B$ (taking a blue) with replacement are independent.
$A$ (taking a blue) and $B$ (taking a blue) without replacement are dependent.
probability
probability
edited Nov 17 at 7:47
asked Nov 17 at 7:34
Artificial Stupidity
336110
336110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.
Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.
Indeed, both intuitions are true.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.
Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.
Indeed, both intuitions are true.
add a comment |
up vote
1
down vote
accepted
A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.
Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.
Indeed, both intuitions are true.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.
Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.
Indeed, both intuitions are true.
A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.
Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.
Indeed, both intuitions are true.
answered Nov 17 at 7:47
Parcly Taxel
41k137198
41k137198
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002068%2fcounter-intuitive-independent-and-dependent-events%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown