I don't think W/O replacement events are dependent.

Question

Let say there're cards numbered 1 to 10. P(A) is the probability you pick a card whose number is the multiple of 2. P(B) is the probability you pick a card whose number is the multiple of 5. So now you draw two cards without replacement. You don't place the first card you just drew back in the original deck. Then, you draw again from that 9 remaining cards.

In the picture, P(A) is shown as the ordered pairs colored red, and P(B) is shown as the ordered pairs shaded with light blue.

Then I counted them all.

1. P(A) = (9+9+9+9+9) / (100-10) = 45/90 = 1/2
2. P(B) = (9+9) / (100-10) = 18/90 = 1/5
3. P(A∩B) = 9 / 90 = 1/10

So here, I found out that P(A∩B) is actually the multiple of P(A) and P(B), which shouldn't be if they are dependent, meaning they are independent. I've also found out, in a different model, that the probabilities could be dependent each other in the case of W/O replacement such as this one.

I know events with replacements are independent. But it feels like events without replacements doesn't mean dependency. It's more like the case of contraposition of the first one. Events with replacements mean independent, which means, by the contraposition rule, dependent events are the ones done without replacements.

But all the other internet pages say like "if they are done without replacement, then they are dependent". This claim is so prevalent. But one of us, they or I, must be wrong. What am I missing here?

Answer 1

first, "with\without replacement" always appears with "sampling" 。I think the event A and event B you described is not sampling itself.

Normally, if P(A=i) is the probability you pick a card whose number is i,P(B=j) is the probability you pick a card whose number is j. apperantly, P(B|A) != P(B) without replacement.

Answer 2

In the first version of your question, $P(A)$ only applied to the first card and $P(B)$ only applied to the second card. In that case, your workings are correct: $P(A)=\frac12,P(B)=\frac15,P(A\cap B)=\frac1{10}=P(A)P(B)$.

However, you then made this comment:

Let say you do the case A first, and then case B. If you draw 5 or 10 in your first drawing, P(B) is 2/10 times 1/9. And if you didn't draw 5 or 10 in your first drawing, P(B) is 8/10 times 2/9.

This suggests an alternate interpretation: $P(A)$ is the probability of picking at least one even card from the two draws, $P(B)$ is the probability of picking at least one multiple-of-$5$ card from the two draws.

Then $P(A)$ is the complement of the probability that you don't pick an even card in either draw: $$1-\frac5{10}\cdot\frac49=\frac79$$ $P(B)$ is similar, the complement of the probability of not picking a multiple-of-$5$ card in either draw: $$1-\frac8{10}\cdot\frac79=\frac{17}{45}$$ $P(A\cap B)$ is the probability of picking one multiple-of-$5$ card and one even card: $$\frac{2(2\cdot5-1)}{90}=\frac15$$ (There are $2$ ways to pick the multiple-of-$5$ card and $5$ ways to pick the even card, but then subtract $1$ for the impossible $(10,10)$ case. Then there are $2$ ways in which those cards can be drawn out, out of $90$ possible draws.)

Clearly, with this interpretation, $P(A\cap B)\ne P(A)P(B)$. Thus you cannot say straight away that "events that arise from sampling without replacement are dependent" – usually this is true, but your original formulation is a counterexample.

Answer 3

A=the first card is even;

B=the second card is 5 or 10.

$P(A)=\frac{1}{2}$ as you said, but $P(B)=P(B\cap A)+P(B\cap A^C)=P(A)P(B|A)+P(b\cap A^C)=\frac{1}{2}\frac{1}{5}+\frac{4}{5} \frac{2}{9}+\frac{1}{5} \frac{1}{9} =\frac{1}{10}+\frac{1}{5}=\frac{3}{10}$

I hope no errors were made.