Two events are independent if the outcome of one doesn’t affect the outcome of the other. For example, when we were talking about the chances of rain on Tuesday and Wednesday above, we were assuming that the weather on Tuesday doesn’t affect the weather on Wednesday: they are independent of each other. The “and means times” rule only applies to independent events.

Ironically, the “or means plus” rule only applies to dependent events. In our example above, the chance of rain Friday absolutely does affect the chance of having it not rain on Friday.

So what do you do if the situation doesn’t let you use a rule? Believe it or not, the answer is that you change the way you are looking at the situation so that the rules apply again.

Let’s look at an example of using these various rules wrong, so that you’ll be more able to avoid these mistakes. Along the way, we’ll learn how to change our point of view to allow us to use the rules when we need to.

Example

If there’s a 1/3 chance that it will rain on Tuesday, and a 1/3 chance that it will rain on Wednesday, then what is the chance that it will rain on Tuesday or Wednesday (or both)?

Wrong solution: or means plus, so the chance of rain on Tuesday or Wednesday is 1/3 + 1/3 = 2/3. This is wrong; the events are independent.

Second wrong solution: or means plus, so the chance of rain on Tuesday or Wednesday or both is 1/3 + 1/3 + 1/9 = 7/9. (The 1/9 was the chance that it rains on both days.) Still wrong; although the chance of rain on both days is not independent of the rain on either day (good!) it’s still a problem that rain Tuesday is independent of rain Wednesday.

Better solution: Either it rains on Tuesday or not, and either it rains on Wednesday or not. So there are four different (independent) events that could happen:

Rain Tuesday and Wednesday
Rain Tuesday but not Wednesday
No rain Tuesday, but rain Wednesday
No rain either day

Within each row, the two events are independent, so we can use the “and means times” rule within each row. But each row affects the others, since if row #2 is true, the other rows cannot possibly be true. So we can use the “or means plus” between the rows.

We calculate the chance of each row:

1/3 * 1/3 = 1/9
1/3 * 2/3 = 2/9
2/3 * 1/3 = 2/9
2/3 * 2/3 = 4/9

Notice that only one of the numbered events can happen, and one of them must happen. (We say one and only one will happen.) So it is guaranteed to be either #1, #2, #3, or #4. And sure enough, the chances of the four events add up to 100%: 1/9 + 2/9 + 2/9 + 4/9 = 9/9 = 100%.

So this all works correctly.

The chance that it will rain on Tuesday or Wednesday (or both) is 2/9 + 2/9 + 1/9 = 5/9.

Let’s try another example. See if you can solve it yourself!

All telephone numbers are 7-digit whole numbers. Assume that every 7-digit whole number is a possible telephone number except those that begin with 0 or 1. What fraction of telephone numbers begin with 9 and end with 0?

A. 1/63
B. 1/80
C. 1/81
D. 1/90
E. 1/100

(spoiler)

An equivalent problem is finding the probability that a randomly selected telephone number begins with 9 and ends with 0.

There are $10 - 2 = 8$ possibilities for the first digit in total, and only $1$ that works, so the probability the number begins with $9$ is $1/8$ .

Out of the $10$ possibilities for the last digit, only $1$ works (i.e. $0$ ), so the probability the number ends with $0$ is $1/10$ .

Since these are independent events, the probably that both happens is:

$\frac{1}{8} * \frac{1}{10} = \frac{1}{10}$