Question

Determine whether you think the given pair of events is independent.

You frequent casinos. You have lost money gambling.

Answer 1

mutually exclusive are independent their intersection is empty

When we were working out the probability of the ball landing in a black or red pocket, we were dealing with two separate events, the ball landing in a black pocket and the ball landing in a red pocket. These two events are mutually exclusive because it’s impossible for the ball to land in a pocket that’s both black and red.

If two events are mutually exclusive, only one of the two can occur.

What about the black and even events? This time the events aren’t mutually exclusive. It’s possible that the ball could land in a pocket that’s both black andeven. The two events intersect.

If two events intersect, it’s possible they can occur simultaneously.

BRAIN POWER

What sort of effect do you think this intersection could have had on the probability?

Problems at the intersection

Calculating the probability of getting a black or even went wrong because we included black and even pockets twice. Here’s what happened.

First of all, we found the probability of getting a black pocket and the probability of getting an even number.

When we added the two probabilities together, we counted the probability of getting a black and even pocket twice.

To get the correct answer, we need to subtract the probability of getting both black and even. This gives us

P(Black or Even) = P(Black) + P(Even) – P(Black and Even)

We can now substitute in the values we just calculated to find P(Black or Even):

P(Black or Even) = 18/38 + 18/38 – 10/38 = 26/38 = 0.684

Some more notation

There’s a more general way of writing this using some more math shorthand.

First of all, we can use the notation A ∩ B to refer to the intersection between A and B. You can think of this symbol as meaning “and.” It takes the common elements of events.

A ∪ B, on the other hand, means the union of A and B. It includes all of the elements in A and also those in B. You can think of it as meaning “or.”

If A ∪ B =1, then A and B are said to be exhaustive. Between them, they make up the whole of S. They exhaust all possibilities.

SHARPEN YOUR PENCIL

On the previous page, we found that

(P(Black or Even) = P(Black) + P(Even) – P(Black and Even)

Write this equation for A and B using ∩ and ∪ notation.

SHARPEN YOUR PENCIL SOLUTION

On the previous page, we found that

(P(Black or Even) = P(Black) + P(Even) – P(Black and Even)

Write this equation for A and B using ∩ and ∪ notation.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

NOTE

P(A or B)

P(A and B)

It’s not actually that different.

Mutually exclusive events have no elements in common with each other. If you have two mutually exclusive events, the probability of getting A and B is actually 0—so P(A ∩ B) = 0. Let’s revisit our black-or-red example. In this bet, getting a red pocket on the roulette wheel and getting a black pocket are mutually exclusive events, as a pocket can’t be both red and black. This means that P(Black ∩ Red) = 0, so that part of the equation just disappears.

WATCH IT!

There’s a difference between exclusive and exhaustive.

If events A and B are exclusive, then

P(A ∩ B) = 0

If events A and B are exhaustive, then

P(A ∪ B) = 1

BE THE PROBABILITY

Your job is to play like you’re the probability and shade in the area that represents each of the following probabilities on the Venn diagrams.

BE THE PROBABILITY SOLUTION

Your job was to play like you’re the probability and shade in the area that represents each of the probabilities on the Venn diagrams.

EXERCISE

50 sports enthusiasts at the Head First Health Club are asked whether they play baseball, football, or basketball. 10 only play baseball. 12 only play football. 18 only play basketball. 6 play baseball and basketball but not football. 4 play football and basketball but not baseball.

Draw a Venn diagram for this probability space. How many enthusiasts play baseball in total? How many play basketball? How many play football?

Are any sports’ rosters mutually exclusive? Which sports are exhaustive (fill up the possibility space)?

VITAL STATISTICS: A OR B

To find the probability of getting event A or B, use

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∪ means OR

∩ means AND

EXERCISE SOLUTION

50 sports enthusiasts at the Head First Health Club are asked whether they play baseball, football or basketball. 10 only play baseball. 12 only play football. 18 only play basketball. 6 play baseball and basketball but not football. 4 play football and basketball but not baseball.

Draw a Venn diagram for this probability space. How many enthusiasts play baseball in total? How many play basketball? How many play football?

Are any sports’ rosters mutually exclusive? Which sports are exhaustive (fill up the possibility space)?

By adding up the values in each circle in the Venn diagram, we can determine that there are 16 total baseball players, 28 total basketball players, and 16 total football players.

The baseball and football events are mutually exclusive. Nobody plays both baseball and football, so P(Baseball ∩ Football) = 0

The events for baseball, football, and basketball are exhaustive. Together, they fill the entire possibility space, so P(Baseball ∪ Football ∪ Basketball) = 1

THERE ARE NO DUMB QUESTIONS

Q:	Q: Are A and AI mutually exclusive or exhaustive?
A:	A: Actually they’re both. A and AI can have no common elements, so they are mutually exclusive. Together, they make up the entire possibility space so they’re exhaustive too.
Q:	Q: Isn’t P(A ∩ B) + P(A ∩ BI) just a complicated way of saying P(A)?
A:	A: Yes it is. It can sometimes be useful to think of different ways of forming the same probability, though. You don’t always have access to all the information you’d like, so being able to think laterally about probabilities is a definite advantage.
Q:	Q: Is there a limit on how many events can intersect?
A:	A: No. When you’re referring to the intersection between several events, use more ∩’s. As an example, the intersection of events A, B, and C is A ∩ B ∩ C. Finding probabilities for multiple intersections can sometimes be tricky. We suggest that if you’re in doubt, draw a Venn diagram and take a good, hard look at which probabilities need to be added together and which need to be subtracted.

Another unlucky spin...

We know that the probability of the ball landing on black or even is 0.684, but, unfortunately, the ball landed on 23, which is red and odd.

...but it’s time for another bet

Even with the odds in our favor, we’ve been unlucky with the outcomes at the roulette table. The croupier decides to take pity on us and offers a little inside information. After she spins the roulette wheel, she’ll give us a clue about where the ball landed, and we’ll work out the probability based on what she tells us.

Should we take this bet?

How does the probability of getting even given that we know the ball landed in a black pocket compare to our last bet that the ball would land on black or even. Let’s figure it out.

Conditions apply

The croupier says the ball has landed in a black pocket. What’s the probability that the pocket is also even?

This is a slightly different problem

We don’t want to find the probability of getting a pocket that is both black and even, out of all possible pockets. Instead, we want to find the probability that the pocket is even, given that we already know it’s black.

In other words, we want to find out how many pockets are even out of all the black ones. Out of the 18 black pockets, 10 of them are even, so

It turns out that even with some inside information, our odds are actually lower than before. The probability of even given black is actually less than the probability of black or even.

However, a probability of 0.556 is still better than 50% odds, so this is still a pretty good bet. Let’s go for it.

Find conditional probabilities

So how can we generalize this sort of problem? First of all, we need some more notation to represent conditional probabilities, which measure the probability of one event occurring relative to another occurring.

If we want to express the probability of one event happening given another one has already happened, we use the “|” symbol to mean “given.” Instead of saying “the probability of event A occurring given event B,” we can shorten it to say

P(A | B)

NOTE

The probability of A give that we know B has happened.

So now we need a general way of calculating P(A | B). What we’re interested in is the number of outcomes where both A and B occur, divided by all the B outcomes. Looking at the Venn diagram, we get:

We can rewrite this equation to give us a way of finding P(A ∩ B)

P(A ∩ B) = P(A | B) × P(B)

It doesn’t end there. Another way of writing P(A ∩ B) is P(B ∩ A). This means that we can rewrite the formula as

P(B ∩ A) = P(B | A) × P(A)

In other words, just flip around the A and the B.

Venn diagrams aren’t always the best way of visualizing conditional probability.

Don’t worry, there’s another sort of diagram you can use—a probability tree.

You can visualize conditional probabilities with a probability tree

It’s not always easy to visualize conditional probabilities with Venn diagrams, but there’s another sort of diagram that really comes in handy in this situation—the probability tree. Here’s a probability tree for our problem with the roulette wheel, showing the probabilities for getting different colored and odd or even pockets.

The first set of branches shows the probability of each outcome, so the probability of getting a black is 18/38, or 0.474. The second set of branches shows the probability of outcomes given the outcome of the branch it is linked to. The probability of getting an odd pocket given we know it’s black is 8/18, or 0.444.

Trees also help you calculate conditional probabilities

Probability trees don’t just help you visualize probabilities; they can help you to calculate them, too.

Let’s take a general look at how you can do this. Here’s another probability tree, this time with a different number of branches. It shows two levels of events: A and A^I and B and B^I. A^I refers to every possibility not covered by A, and B^I refers to every possibility not covered by B.

You can find probabilities involving intersections by multiplying the probabilities of linked branches together. As an example, suppose you want to find P(A ∩ B). You can find this by multiplying P(B) and P(A | B) together. In other words, you multiply the probability on the first level B branch with the probability on the second level A branch.

Using probability trees gives you the same results you saw earlier, and it’s up to you whether you use them or not. Probability trees can be time-consuming to draw, but they offer you a way of visualizing conditional probabilities.

PROBABILITY MAGNETS

Duncan’s Donuts are looking into the probabilities of their customers buying donuts and coffee. They drew up a probability tree to show the probabilities, but in a sudden gust of wind, they all fell off. Your task is to pin the probabilities back on the tree. Here are some clues to help you.

P(Donuts) = 3/4

P(Coffee | DonutsI) = 1/3

P(Donuts ∩ Coffee) = 9/20

HANDY HINTS FOR WORKING WITH TREES

1. Work out the levels

Try and work out the different levels of probability that you need. As an example, if you’re given a probability for P(A | B), you’ll probably need the first level to cover B, and the second level A.

2. Fill in what you know

If you’re given a series of probabilities, put them onto the tree in the relevant position.

3. Remember that each set of branches sums to 1

If you add together the probabilities for all of the branches that fork off from a common point, the sum should equal 1. Remember that P(A) = 1 – P(A^I).

4. Remember your formula

You should be able to find most other probabilities by using

PROBABILITY MAGNETS SOLUTION

Duncan’s Donuts are looking into the probabilities of their customers buying Donuts and Coffee. They drew up a probability tree to show the probabilities, but in a sudden gust of wind they all fell off. Your task is to pin the probabilities back on the tree. Here are some clues to help you.

P(Donuts) = 3/4

P(Coffee | DonutsI) = 1/3

P(Donuts ∩ Coffee) = 9/20

EXERCISE

We haven’t quite finished with Duncan’s Donuts! Now that you’ve completed the probability tree, you need to use it to work out some probabilities.

1. P(Donuts^I)

2. P(Donuts^I ∩ Coffee)

3. P(Coffee^I | Donuts)

4. P(Coffee)

   NOTE

   Hint: How many ways are there of getting coffee? (You can get coffee with or without donuts.)

5. P(Donuts | Coffee)

   NOTE

   Hint: maybe some of your other answers can help you.

EXERCISE SOLUTION

Your job was to use the completed probability tree to work out some probabilities.

1. P(Donuts^I)

   1/4

   NOTE

   We can read this one off the tree. We were given P(Donuts) = 3/4, so P(DonutsI) must be 1/4.

2. P(Donuts^I ∩ Coffee)

   1/12

   NOTE

   We can find this by multiplying together P(Donuts^I) and P(Coffee | Donuts^I). We’ve just found P(Donuts^I) = 1/4, and looking at the tree, P(Coffee | Donuts^I) = 1/3. Multiplying these together gives 1/12.

3. P(Coffee^I | Donuts)

   2/5

   NOTE

   We can read this off the tree.

4. P(Coffee)

   8/15

   NOTE

   This probability is tricky, so don’t worry if you didn’t get it.

   To get P(Coffee), we need to add together P(Coffee ∩ Donuts) and P(Coffee ∩ Donuts^I). This gives us 1/12 + 9/20 = 8/15.

5. P(Donuts | Coffee)

   27/32

   NOTE

   You’ll only be able to do this if you found P(Coffee).

   P(Donuts | Coffee) = P(Donuts ∩ Coffee) / P(Coffee). This gives us (9/20) / (8 / 15) = 27/32.

VITAL STATISTICS: CONDITIONS

THERE ARE NO DUMB QUESTIONS

Q:	Q: I still don’t get the difference between P(A ∩ B) and P(A | B).
A:	A: P(A ∩ B) is the probability of getting both A and B. With this probability, you can make no assumptions about whether one of the events has already occurred. You have to find the probability of both events happening without making any assumptions. P(A | B) is the probability of event A given event B. In other words, you make the assumption that event B has occurred, and you work out the probability of getting A under this assumption.
Q:	Q: So does that mean that P(A | B) is just the same as P(A)?
A:	A: No, they refer to different probabilities. When you calculate P(A | B), you have to assume that event B has already happened. When you work out P(A), you can make no such assumption.
Q:	Q: Is P(A | B) the same as P(B | A)? They look similar.
A:	A: It’s quite a common mistake, but they are very different probabilities. P(A | B) is the probability of getting event A given event B has already happened. P(B | A) is the probability of getting event B given event A occurred. You’re actually finding the probability of a different event under a different set of assumptions.
Q:	Q: Are probability trees better than Venn diagrams?
A:	A: Both diagrams give you a way of visualizing probabilities, and both have their uses. Venn diagrams are useful for showing basic probabilities and relationships, while probability trees are useful if you’re working with conditional probabilities. It all depends what type of problem you need to solve.
Q:	Q: Is there a limit to how many sets of branches you can have on a probability tree?
A:	A: In theory there’s no limit. In practice you may find that a very large probability tree can become unwieldy, but you may still find it easier to draw a large probability tree than work through complex probabilities without it.
Q:	Q: If A and B are mutually exclusive, what is P(A | B)?
A:	A: If A and B are mutually exclusive, then P(A ∩ B) = 0 and P(A | B) = 0. This makes sense because if A and B are mutually exclusive, it’s impossible for both events to occur. If we assume that event B has occurred, then it’s impossible for event A to happen, so P(A | B) = 0.