You're wondering whether you should go see a new action movie, Muscle Man: How One Man's Muscles Save the World (again). Now, you're not the biggest fan of action movies, but you do enjoy one from time to time. So you figure it'll be worth it if you can get a good seat or if the action sequences are amazing. If neither of those things happens, then it's not worth going for you. Now, you're wondering, should I go see the movie?
Well, something that matters to you is the likelihood of: (i) I get a good seat, or (ii) the action sequences are amazing. Let A = "I get a good seat" and B = "The action sequences are amazing." You really want to know the value of P(A or B). If it's really high, then the movie is probably worth it. If it's really low, then the movie probably isn't (and buying the ticket and so on just isn't worth it).
In this section, we'll talk about calculating P(A or B), which we read as "The probability that A or B occurs." But before we do so, we need to issue an important clarification: A or B means that A occurs or that B occurs or that both occur. In other words, if A and B both happen, then 'A or B' happens as well. This is somewhat at odds with how we often use the word "or," as in sentences like, "You can study hard or you can fail the test," with the implication being that you cannot do both. Excise that meaning from your mind; in probability, we say that "A or B" occurs if A occurs, or if B occurs, or if A and B occur.
Now, in order to calculate P(A or B), it will help to introduce the idea of mutually exclusive events:
Definition: Two events, A and B, are mutually exclusive if it is impossible for them to both occur.
Another way to put this (symbolically): P(A and B) = 0.
Here are some examples of mutually exclusive events:
- When flipping a coin, getting heads and getting tails are mutually exclusive events. It is impossible to get heads and tails from the same flip of a coin.
- Suppose you and your friend have both entered into a raffle that only picks one winner. Then, the event of you winning is mutually exclusive with your friend winning.
- Suppose you are taking a class in college and you need an A or B to graduate with honors. The event where you get an A is mutually exclusive with the event where you get a B.
The point of talking about mutually exclusive events is to make it easier to calculate probabilities of one event OR another event occurring. We can do such calculations via the following rule:
Mutually Exclusive Rule for P(A or B)
Let A, B be mutually exclusive events. Then, P(A or B) = P(A) + P(B).
In other words, the probability that A or B occurs is equal to the probability that A occurs plus the probability that B occurs.
And if you like to get a sense for why such rules work (rather than simply memorize the formula), see here for an illustration that helps make the rule more intuitive. Now, let's see this rule in action:
In rolling a fair, six-sided die, what is the probability that you will get a 1 or a 4?
We know that rolling a 1 and rolling a 4 are mutually exclusive events, since it is impossible for them both to occur. We know that P(Rolling a 1) = ⅙ and that P(Rolling a 4) = ⅙. Thus, by our above rule, P(Rolling a 1 or Rolling a 4) = P(Rolling a 1) + P(Rolling a 4) = + ⅙ + ⅙ = ⅓.
Now, this rule is only a special case of a more general principle. In general, for all events, and not just mutually exclusive ones, the following is true:
General Rule for P(A or B)
Let A, B be two events. Then, P(A or B) = P(A) + P(B) - P(A and B).
I.e. the probability that A or B occurs is equal to the probability that A occurs plus the probability that B occurs minus the probability that A and B occur.
And again, if you like to see why such rules are true, click here. Here is an example of using this rule:
You are wondering whether to go to the cafe. You would go if you knew that Bertrand or Simone was going. There is a 45% chance that Bertrand will go. There is a 20% chance Simone will go, and there is a 15% chance that both Bertrand and Simone go. What is the likelihood that Bertrand or Simone will go to the cafe?
Let B = "Bertrand goes to the cafe" and let S = "Simone goes to the cafe." Then, the question gives us that P(B) = .45, P(S) = .2, and P(B and S) = .15. Following our rule, P(A or B) = P(A) + P(B) - P(A and B) = .45 + .2 - .15 = .5. Thus, there is a 1/2 chance that Bertrand or Simone will go.
In our next post, we will look at a strategy that can help us solve some tricky questions: instead of finding the probability of some event, try finding the probability that it does not occur.
In a bag, there are 5 red marbles, 2 blue marbles, and 1 pink marble. I will pick one marble from the bag, set it aside, and then pick another marble from the bag. Is the event of my drawing a blue marble on the first draw mutually exclusive with drawing a pink marble second? Is drawing a pink marble first mutually exclusive with drawing a pink marble second?
Drawing a blue marble first and a pink marble second are not mutually exclusive; it is possible to do both. But, drawing a pink marble first is mutually exclusive with drawing a pink marble second. After all, since there is only one pink marble and we do not replace the marbles after we draw them, once you draw the first pink marble, you've drawn the only one there is! You can't draw a second pink marble.
Jane is worried that her new neighbor both (i) likes bad music, and (ii) is willing to blare his preferred kind of music at all hours. She estimates the probability that her neighbor likes bad music at .4, and the probability of his constantly blaring music at .3. And she estimates the probability that (i) or (ii) is true at .6. What probability should she assign to the worst possible outcome: her neighbor both likes bad music and is willing to blare music constantly?
This is a different way in which we can apply the rule we just learned. We have been thinking of our rule as a way to calculate P(A or B). But, if you are given P(A or B), P(A), and P(B), we can also use it as a way of calculating P(A and B). Let's plug in the numbers our question gives: .4 + .3 - P(A and B) = .6. Then, subtracting .6 and multiplying by -1 on both sides, we get: P(A and B) = .1.
Intuitively, this may seem a little odd, but we must remind ourselves that the formulas we use are true, and that we are allowed to manipulate them algebraically however we wish.
There are two events, A and B. The probability of just A occurring is r. The probability of just B occurring is s. The probability of neither A or B occurring is t. What is the probability that both A and B occur?
Now, there are four possibilities in total: (1) A and B, (2) A and not-B, (3) not-A and B, and (4) not-A and not-B. We know that one of these four must happen since they capture all the logical possibilities. Furthermore, the possibilities are all mutually exclusive since it is impossible for any two of them to happen. For example, suppose (2) and (4) were both true. Then A and not-A would be true, which would be impossible - either A is true or it is not! Thus, we can apply our above rule: P(Possibility 1 or 2 or 3 or 4 occurs) = P(Possibility 1) + P(Possibility 2) + P(Possibility 3) + P(Possibility 4) = 1. The question gives us the probability of (ii), (iii), and (iv). It asks us about the probability of (i). Plugging in the values we get from the question, we get: P(Possibility 1) + r + s + t = 1, and thus, P(Possibility 1) = 1 - r - s - t.