Problem solving in probabilities

Question

I have a problem and a solution to it that I don't quite understand.

Question: A monkey is to demonstrate that she recognizes colors by tossing one red, one black, and one white ball into boxes of the same respective colors, one ball to a box. If the monkey has not learned the colors and merely tosses one ball into each box at random, find the probabilities of the following results:

a. There are no color matches.

b. There is exactly one color match.

Solution: This problem can be solved by listing sample points because only three balls are involved, but a more general method will be illustrated. Define the following events:

A1: A color match occurs in the red box.

A2: A color match occurs in the black box.

A3: A color match occurs in the white box.

There are 3! = 6 equally likely ways of randomly tossing the balls into the boxes with one ball in each box. Also, there are only 2! = 2 ways of tossing the balls into the boxes if one particular box is required to have a color match. Hence,

$P(A1) = P(A2) = P(A3)$ = 2/6 = 1/3

Similarly it follows that,

$P(A1 \cap\ A2) = P(A2 \cap\ A3) = P(A3 \cap\ A3) = P(A1 \cap\ A2 \cap\ A3)$ = 1/6 (How?)

$\vdots$

$P(exactly\,one\,match) = P(A1) + P(A2) + P(A3)−2[P(A1 \cap\ A2) + P(A1 \cap\ A3) + P(A2 \cap\ A3)] +3[P(A1 \cap\ A2 \cap\ A3)]$ (How?)

Any explanation would be greatly appreciated!

Answer 1

One the red box is matched, there is a probability of $\frac{1}{2}$ that the black box is also matched. So $$ P(A_1 \cap A_2) =P (A_2 \mid A_1) P(A_1) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} $$

The second line is an application of the inclusion-exclusion principle to three sets. The event that exactly one of $A$ and $B$ occurs is $A \mathrel{\triangle} B = (A \cup B) \setminus(A\cap B)$. First, prove that $$ P(A \mathrel{\triangle} B) = P(A)+P(B)-2P(A\cap B) $$ This is like $$ P(A \cup B) = P(A)+P(B)-P(A\cap B) $$ where we subtract $P(A\cap B)$ to correct that fact that we over-counted the intersection by 1. However, with $P(A \mathrel{\triangle} B)$ we don't want to count the intersection at all, so we subtract double the probability of the intersection.

Apply this to three sets: \begin{align*} P(A \mathrel{\triangle} (B\mathrel{\triangle} C)) &= P(A) + P(B \mathrel{\triangle} C) - 2 P(A \cap (B\mathrel{\triangle}C)) \\ &= P(A) + P(B) + P(C) - 2 P(B \cap C) - 2 P((A \cap B) \mathrel{\triangle} (A \cap C)) \\ &= P(A) + P(B) + P(C) - 2 P(B \cap C) - 2 P(A \cap B) - 2 P(A \cap C) + 4 P(A \cap B \cap C) \end{align*} OK, except for three events, $A \mathrel{\triangle} (B \mathrel{\triangle} C)$ overcounts the three-fold intersection. So we subtract it off again, giving the correct coefficient of $+3$ for $P(A \cap B \cap C)$.

Answer 2

Since the monkey cannot throw two objects into the same box, the events $A1$ , $A2$, and $A3$ are not independent events.

In particular, once one object is thrown into the right box, then the probability of a second object being thrown into the right box becomes $\frac{1}{2}$

Hence:

$$P(A1 \cap A2) = P(A1|A2)\cdot P(A2) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$$

As for getting exactly one object into the right box, consider the following Venn Diagram:

Regions $I$, $V$, and $VII$ represent the $3$ ways in which you would get exactly one object into the right box. Now, if we add the regions for $A1$, $A2$, and $A3$, then we add those $3$ regions $I$, $V$, and $VII$, but we would also have added the regions $II$, $III$, and $VI$ twice, and three times the region $IV$. Now, by twice subtracting $A1 \cap A2$, $A1 \cap A3$, and $A2 \cap A3$, we subtract the regions $II$, $III$, and $VI$ twice, but we also end up subtracting the region $IV$ six times. ... which means the grand total is now $I$, $V$, and $VII$ minus $3$ times $IV$. But $IV$ represents $A1 \cap A2 \cap A3$, so just add that one back $3$ times. In sum:

$$P(exactly\,one\,match) =$$

$$P(A1) + P(A2) + P(A3)$$

$$−2[P(A1 \cap\ A2) + P(A1 \cap\ A3) + P(A2 \cap\ A3)]$$

$$+3[P(A1 \cap\ A2 \cap\ A3)]$$

Answer 3

1. Note that $P(A_1 \cap A_2)$ means that both the black and the red balls matched the boxes, and then only one left, thus the white ball got into the right box as well. The same is true for any $A_j$ and $A_i$. For the calculated probability, note that the probability of the event of red ball matches the red box is $1/3$, then you will be left with two boxes and two balls, hence the probability that the black ball will match the black box is $1/2$. And for the last one you have only one option, i.e., $$ P(A_1 \cap A_2) = P(A_1 \cap A_2 \cap A_3 ) = \frac{1}{3}\times\frac{1}{2}\times 1 = \frac{1}{6}. $$

2. To get exactly one right, let us say, the red one, you should subtract from $P(A_1)$ the probability of two and three right matches, thus $P(A_1) - P(A_1 \cap A_2)-P(A_1 \cap A_3)$, but now you have subtracted two times the triple $P(A_1 \cap A_2 \cap A_3)$, hence you have to add it back ones. It is called the inclusion-exclusion principle.