# Why does drawing without replacement result in dependent events when condition for independent events is met?

An example of this would be if we considered a bag which has x white balls, y red balls and z blue balls. Let the event of first ball being drawn be white be A, and second ball drawn be white be B.

The necessary and sufficient conditions for independence is $$P(A \cap B) = P(A)P(B)$$

Then, the probability of drawing 2 white balls which is $$P(A\cap B)$$ would be $$\frac{P^x_2}{P^{x+y+z}_2} = \frac{x(x-1)}{(x+y+z)(x+y+z-1)}$$

Now, $$P(A)P(B) = (\frac{x}{x+y+z})(\frac{x-1}{x+y+z-1})$$ which is the same thing as we found for $$P(A \cap B)$$

The necessary condition for independent events is satisfied, but the two events are said to be dependent which I don't understand.

Edit: I do understand the logic with which you can say they are dependent, which is when the first white ball is drawn, the probability of drawing the second white ball changes from $$\frac{x}{x+y+z}$$ to $$\frac{x-1}{x+y+z-1}$$ and so the probability is dependent on the drawing of the first ball. However, I am confused that the conditions of independent events are being met regardless.

They do not satisfy the condition for independence, because $$P(B)$$ is the probability of the second ball being white, which is $$\frac{x}{x+y+z}$$ because there are $$x+y+z$$ possible second balls, of which $$x$$ are white. Now $$P(A\cap B)\neq P(A)P(B)$$.
What you have worked out is that the probability of $$B$$, given that the first ball is white, is $$\frac{x-1}{x+y+z-1}$$. But this is not $$P(B)$$, it is $$P(B\mid A)$$. And so the relation you have is just $$P(A\cap B)=P(A)P(B\mid A)$$, which is true for any events $$A,B$$ (provided $$P(A)\neq 0$$).
• I don't understand one thing - since $P(B)$ is the "second ball being drawn", then would the sample space of balls not have decreased by one? Though I do realize the flaw wherein I changed $x$ to $x-1$. Oct 9 '20 at 19:20
• @Confused just think of putting the balls in order at random to start with. Every ball has the same probability of being white. Alternatively, you can work it out as $\frac{x}{x+y+z}\times\frac{x-1}{x+y+z-1}+\frac{y+z}{x+y+z}\times\frac{x}{x+y+z-1}$ and simplify to the same value, i.e. add up the probabilities for (white,white) and (not white,white) to get the total probability of (anything,white). Oct 10 '20 at 7:35