# Better ways to explain mutually exclusiveness and dependency of events

I am teaching probability on mutually exclusiveness and dependency of events. Let me take a simple example as follows.

A box contains 2 red balls and 3 purple balls. They are identical except for their color. A ball is randomly taken, the color is noted, and it is replaced back to the box. We repeat this action twice.

• Let $$A$$ be the event that a red ball is taken in the first draw.
• Let $$B$$ be the event that a purple ball is taken in the second draw.
• Let $$C$$ be the event that there are 2 red balls in the 2 draws.
• Let $$D$$ be the event that there are 2 purple balls in the 2 draws.

# First Fallacy

A student answers that $$A$$ and $$B$$ are mutually exclusive because there is no intersection between $$A=\{R_1,R_2\}$$ and $$B=\{P_1,P_2,P_3\}$$. He then calculates the probability that the first ball is red or the second ball is purple is $$p(A\cup B) = 2/5 +3/5 =1$$

# Second Fallacy

A student answers that $$C$$ and $$D$$ are independent because $$C$$ and $$D$$ don't interfere each other. He then calculates $$p(C\cap D) = 4/25 \times 9/25 = 36/625$$

# Question

How do we correct him with easy-to-digest explanation?

• It's always a good idea to mention the age group that you're teaching.. – yathish Dec 2 '18 at 4:25

## 2 Answers

Emphasize to the student that

1. Every probability has an associated experiment.
2. Every experiment has an associated sample space.
3. Every event is a subset of the sample space.

In this case, the associated experiment is "randomly take one ball from a box, note the color, then return the ball to the box. Repeat." The associated sample space is

$$\{P_1P_1,P_1P_2,P_1P_3,P_1R_1,P_1R_2,\\ P_2P_1,P_2P_2,P_2P_3,P_2R_1,P_2R_2,\\ P_3P_1,P_3P_2,P_3P_3,P_3R_1,P_3R_2,\\ R_1P_1,R_1P_2,R_1P_3,R_1R_1,R_1R_2,\\ R_2P_1,R_2P_2,R_2P_3,R_2R_1,R_2R_2\}$$.

The event that a red ball is taken in the first draw is not $$\{R_1,R_2\}$$ because it is not a subset of the sample space. Similarly, the event that a purple ball is taken in the second draw is not $$\{P_1,P_2,P_3\}$$. It should be easy to see that

$$A=\{R_1P_1,R_1P_2,R_1P_3,R_1R_1,R_1R_2,R_2P_1,R_2P_2,R_2P_3,R_2R_1,R_2R_2\}$$

and

$$B=\{P_1P_1,P_1P_2,P_1P_3,P_2P_1,P_2P_2,P_2P_3,P_3P_1,P_3P_2,P_3P_3,R_1P_1,R_1P_2,R_1P_3,R_2P_1,R_2P_2,R_2P_3\}$$

are not mutually exclusive.

Your other concerns are addressed in a similar manner.

Its easy for even "seasoned" mathematicians to make mistakes in such straightforward probability questions. With that in mind I suggest the following -

1. Insist on solving the problems systematically (especially for beginners). This means - writing out the sample space (if not in its entirety, at least the pattern), rereading what the event is, choosing the outcomes according to the event only from the sample space and so on.
2. Make it clear from the beginning that the two formulas $$p(A\cup B) = p(A)+p(B)$$ and $$p(A\cap B)=p(A) \times p(B)$$ are both conditional, not universal. The former is conditional on the events being mutually exclusive and the latter on independency of the two events. So before applying each of the formulas, the conditions must be checked.
3. In order to check the mutual exclusiveness of two events, students may be encouraged to think inversely i.e., "Do any of the outcomes that are in favour of event $$A$$ are also favourable to event $$B$$?"
4. Take the aid of Venn diagrams to visualise the events. This will help see both mutual exclusion and independence. This approach is particularly useful to check whether events are truly independent.

You can use an easy-to-digest explanation for a few problems initially, but if students do not practice such problems systematically until they get an intuition for it, they are bound to struggle with the concept.

Lastly, do encourage the student(s) for their efforts and specifically point out that it can be a hard concept to learn in the beginning but becomes easier with practice.

• Please choose another word instead of conditional in the context that I used it here. Otherwise it may lead to unnecessary confusion when they'll be introduced to conditional probability. – yathish Dec 2 '18 at 4:29