# 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

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.

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 $$B$$?"