# balls and two boxes problem in probability

We have two urns $$U_1$$ and $$U_2$$.

$$U_1$$ contains $$4$$ white and $$2$$ black balls.

$$U_2$$ contains $$3$$ white and $$4$$ black balls.

We draw $$2$$ balls from $$U_1$$ successively without returning and put them in $$U_2$$ then we draw $$1$$ ball from $$U_2$$.

We are asked to calculate the probabilities of the two following events :

$$E_1$$ : " Getting three balls of the same colour "

$$E_2$$ : " Getting exactly two white balls "

My approach for $$E_1$$ :

Let $$W_1$$, $$B_1$$, $$W_2$$ and $$B_2$$ be the events :

$$W_1$$ : " the two balls drawn from $$U_1$$ are white "

$$B_1$$ : " the two balls drawn from $$U_1$$ are black "

$$W_2$$ : " the ball drawn from $$U_2$$ is white "

$$B_2$$ : " the ball drawn from $$U_2$$ is black "

We have

\begin{align*}p(E_1) & = p_{W_1}(W_2)\times p(W_1) + p_{B_1}(B_2)\times p(B_1) \\ & = \dfrac 59\times \dfrac{P^2_4}{P_6^2} + \dfrac 39\times \dfrac{P^2_2}{P_6^2}\end{align*}

Is this correct and is there an easier approach ?

My approach for $$E_2$$

I introduce a new event :

$$W_1^1$$ : " Drawing exactly one white ball from urn $$U_1$$ "

We then have

\begin{align*}p(E_2) & = p_{W_1^1}(W_2)\times p(W_1^1) + p_{W_1}(B_2)\times p(W_1) \\ & = \dfrac 49\times\dfrac{2\times P^1_4\times P^1_2}{P_6^2} + \dfrac 49\times\dfrac{P^2_2}{P_6^2} \end{align*}

Is this correct and is there an easier approach ?

One last question :

in a finite probability space, is the following implication true :

$$P(A) = 0 \implies A = \emptyset$$

I'm aware of it being incorrect in general but is there a counterexample when $$X(\Omega)$$ is finite.

Thanks.

• I don't know what you mean by $A_6^2$ and so on. – saulspatz May 9 at 8:51
• I have not seen that notation before; I have previously seen the notation $P_n^k = \frac{n!}{(n-k}!}$ where the "P" is the initial letter of "permutations." You can use $A_n^k$ but it would be advisable to copy your comment into the question text itself where the notation is first used. – David K May 9 at 12:11
• To add to the confusion, the English practice also uses "permutation" in the way the French practice does. The only way to tell which kind of "permutation" is meant is to notice whether two numbers are specified as parameters or just one number is specified. – David K May 9 at 12:23
• By the way, the calculations look correct, though I think you swapped the order of the factors in some terms (no effect on the answer, of course, but more difficult to check). The last question seems to have obvious counterexamples but you could easily rewrite the definitions to rule them out, so I suppose you really need to refer to the definitions you were given. – David K May 9 at 12:27
• There are only eight outcomes to this experiment so drawing a tree makes it easy to compute the probabilities of all outcomes. Then the questions become as easy as adding the probabilities of the appropriate leaves. – John Douma May 9 at 19:07