Doubt about a probability excercise

Question

I'm a statistics teacher at a college. One day a student came with a doubt about an exercise about probability. The text goes like this:

A person has two boxes $A$ and $B$. In the first one has $4$ white balls and $5$ black balls and in the second has $5$ white balls and $4$ black balls. This person takes randomly one ball from the first box and put it into the second box. After that he takes a ball from the second box. Find the probability of taking balls of the same color in this process (i.e, the one that is taken from box $A$ to $B$ and the one taken from box $B$).

The student made the following: Let $C$ be the event of taking balls of the same color in the process above described. Let $W$ be the event of taking White balls from both boxes and $Bl$ the event of taking black balls from both boxes. Let $Wb_1$ be the event of taking a white ball from the first box and $Wb_2$ the event of taking a white ball from the second box. The same with $Blb_1$ and $Blb_2$. Then,

\begin{align} \mathbb P(C)&=\mathbb P(W)+\mathbb P(Bl)\\ &= \mathbb P(Wb_1)\mathbb P(Wb_2)+\mathbb P(Blb_1)\mathbb P(Blb_2)\\&= \frac49 \cdot\frac6{10} + \frac59\cdot\frac5{10} \end{align}

I told the student the reasoning was wrong because he has to use conditional probability because events $Wb_1$ and $Wb_2$ as well as $Blb_1$ and $Blb_2$ are not independent. A probability teacher (his actual teacher) told the student he was right and that's why I make this post. ¿Who is right? Thanks!

Answer 1

The calculation on the right-hand side is correct; just the notation is bad, because as you say the right-hand factors in both terms are conditional probabilities. A better way to write this would be

$$ P(\text{C})=P(\text{W})+P(\text{Bl})= P(\text{Wb1})P(\text{Wb2}\mid\text{Wb1})+P(\text{Blb1})P(\text{Blb2}\mid\text{Blb1})=\frac49\cdot\frac6{10}+\frac59\cdot\frac5{10}\;. $$

Answer 2

Split it into disjoint events, and add up their probabilities:

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The probability of choosing white from the first box and then from the second box is:

$$\frac{4}{9}\cdot\frac{6}{10}=\frac{24}{90}$$

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The probability of choosing black from the first box and then from the second box is:

$$\frac{5}{9}\cdot\frac{5}{10}=\frac{25}{90}$$

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So the probability of choosing the same color is:

$$\frac{24}{90}+\frac{25}{90}=\frac{49}{90}$$

Answer 3

As has already been pointed out, the precise and careful choice of notation is paramount.

The event $C$ of choosing the same color is the disjoint union of two separate events, so it is better to choose the following notation: Let $(b_1, b_2)$ be the random outcome of the two ball draws in order, where $b_i \in \{W, B\}$ for $i = 1, 2$. Thus, we want: $$\begin{align*} \Pr[(W,W) \cup (B,B)] &= \Pr[(W,W)] + \Pr[(B,B)]. \end{align*}$$ Now consider the first term on the RHS: $$\Pr[(W,W)] = \Pr[b_1 = W \cap b_2 = W] = \Pr[b_2 = W \mid b_1 = W]\Pr[b_1 = W],$$ by the definition of conditional probability. Similarly, $$\Pr[(B,B)] = \Pr[b_2 = B \mid b_1 = B]\Pr[b_1 = B].$$ Now we easily see $$\Pr[b_1 = W] = \frac{4}{4+5} = \frac{4}{9},$$ and $$\Pr[b_1 = B] = \frac{5}{4+5} = \frac{5}{9},$$ and the conditional probabilities are $$\Pr[b_2 = W \mid b_1 = W] = \frac{5+1}{4+5+1} = \frac{6}{10},$$ and $$\Pr[b_2 = B \mid b_1 = B] = \frac{4+1}{4+5+1} = \frac{5}{10}.$$ Then we get the desired probability $$\Pr[(W,W) \cup (B,B)] = \frac{4(6) + 5(5)}{9(10)} = \frac{49}{90}.$$ Note how we have clearly defined random variables $b_1, b_2$. Precise notation is crucial not just for solving problems but also in making the solution intelligible and rigorous.