André, Bianca, Carlos, and Dalva want to draw a book among themselves. For this, they put 3 white balls and 1 black ball in a box and agree that, in alphabetical order of their names, each will draw a ball, without returning it to the box. The one who draws the black ball will win the book.

a) What is the probability that André will win the book?

b) What is the probability that Dalva will win the book?

To draw another book among them, André suggested using 2 black balls and 6 white balls. As before, the first to draw a black ball will win the book; if the first four balls drawn are white, they will continue to draw balls, in the same order. In this new draw:

c) What is the probability that André will win the book?

d) What is the probability that Dalva will win the book?
Solution

a) For André to win the book, he must draw the black ball. Since the box contains four balls, of which only one is black, the probability of him drawing the black ball is $\frac{1}{4}$.

Another solution appears in the $2^{\text{nd}}$ solution of item $b$).

b) 1st solution: For Dalva to win the book, André, Bianca, and Carlos must draw white balls. Initially, the box contains 3 white balls, so the probability of André drawing a white ball is $\frac{3}{4}$. Assuming André draws a white ball, there will be 2 white balls and 1 black ball left in the box; thus, the probability of Bianca drawing a white ball is $\frac{2}{3}$. Similarly, if André and Bianca draw white balls, the probability of Carlos drawing a white ball will be $\frac{1}{2}$. Therefore, the probability of André, Carlos, and Bianca drawing white balls is $\frac{3}{4} \times \frac{2}{3} \times \frac{1}{2}=\frac{1}{4}$, which is the probability of Dalva winning the book. Similar reasoning shows that the probability of any of the friends winning the book is $\frac{1}{4}$, meaning the draw is fair and the order in which they draw the balls does not matter. To better understand this, see the following solution.

$2^{\text{nd}}$ solution: Keeping the rules of the draw, let's paint one white ball blue and another red; we then have four different balls in the box. The number of possible draws becomes $4 \times 3 \times 2 \times 1=24$; of these, Dalva wins the book when André, Bianca, and Carlos get the white, blue, and red balls, which can happen in $3 \times 2 \times 1=6$ different ways. Thus, the probability of Dalva winning the book is $\frac{6}{24}=\frac{1}{4}$. This reasoning applies to any of the friends, justifying the previous comment about the fairness of the draw.

c) $1^{\text{st}}$ solution: André can win the book in two ways, namely, when the first ball drawn is black or when the first four balls drawn are white and the fifth is black. The probability in the first case is $\frac{2}{8}=\frac{1}{4}$ and in the second is $\frac{6}{8} \times \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4}=\frac{3}{28}$. Thus, the sought probability is $\frac{1}{4}+\frac{3}{28}=\frac{5}{14}$.

$2^{\text{nd}}$ solution: The probability that André wins the book in the first round, as seen above, is $\frac{1}{4}$. To calculate the probability that he wins the book in the second round, we will calculate the possible cases and the favorable cases. The first five balls can be drawn in $8 \times 7 \times 6 \times 5 \times 4$ ways. For André to win the book on the fifth ball, the first four balls must be white and the fifth black, which can happen in $6 \times 5 \times 4 \times 3 \times 2$ ways. Thus, the probability that André wins the book on the fifth ball drawn is $\frac{6 \times 5 \times 4 \times 3 \times 2}{8 \times 7 \times 6 \times 5 \times 4}=\frac{3}{28}$. Therefore, the sought probability is $\frac{1}{4}+\frac{3}{28}=\frac{5}{14}$.

d) $1^{\text{st}}$ solution: Dalva will only win the book in the case where the first three balls drawn are white and the fourth is black; indeed, if the first four balls drawn are white, there will be two white and two black balls left in the box, and a black ball will be drawn before it is her turn. Thus, the probability that Dalva wins the book is $\frac{6}{8} \times \frac{5}{7} \times \frac{4}{6} \times \frac{2}{5}=\frac{1}{7}=\frac{2}{14}$.

$2^{\text{nd}}$ solution: Dalva can only win the book in the case where the first three balls drawn are white and the fourth is black. The first four balls can be drawn in $8 \times 7 \times 6 \times 5$ ways. For Dalva to win the book, the first three must be white and the fourth black, which can happen in $6 \times 5 \times 4 \times 2$ ways. Thus, the probability that Dalva wins the book is $\frac{6 \times 5 \times 4 \times 2}{8 \times 7 \times 6 \times 5}=\frac{1}{7}=\frac{2}{14}$.

It is left as an exercise for the reader to show that the probabilities of Bianca and Carlos winning the book are, respectively, $\frac{4}{14}$ and $\frac{3}{14}$. André was quite clever in proposing this new draw!

Note: We wrote all probabilities as fractions with the same denominator to compare them more quickly and also to facilitate the verification that their sum is equal to 1.