6.1: The Probability of Calculus

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Inductive arguments, recall, are arguments whose premises support their conclusions insofar as they make them more probable. The more probable the conclusion in light of the premises, the stronger the argument; the less probable, the weaker. As we saw in the last chapter, it is often impossible to say with any precision exactly how probable the conclusion of a given inductive argument is in light of its premises; often, we can only make relative judgments, noting that one argument is stronger than another, because the conclusion is more probable, without being able to specify just how much more probable it is.

Sometimes, however, it is possible to specify precisely how probable the conclusion of an inductive argument is in light of its premises. To do that, we must learn something about how to calculate probabilities; we must learn the basics of the probability calculus. This is the branch of mathematics dealing with probability computations. (Don’t freak out about the word ‘calculus’. We’re not doing derivatives and integrals here; we’re using that word in a generic sense, as in ‘a system for performing calculations’, or something like that. Also, don’t get freaked out about ‘mathematics’. This is really simple, fifth-grade stuff: adding and multiplying fractions and decimals.) We will cover its most fundamental rules and learn to perform simple calculations. After that preliminary work, we use the tools provided by the probability calculus to think about how to make decisions in the face of uncertainty, and how to adjust our beliefs in the light of evidence. We will consider the question of what it means to be rational when engaging in these kinds of reasoning activities.

Finally, we will turn to an examination of inductive arguments involving statistics. Such arguments are of course pervasive in public discourse. Building on what we learned about probabilities, we will cover some of the most fundamental statistical concepts. This will allow us to understand various forms of statistical reasoning—from different methods of hypothesis testing to sampling techniques. In addition, even a rudimentary understanding of basic statistical concepts and reasoning methods will put us in a good position to recognize the myriad ways in which statistics are misunderstood, misused, and deployed with the intent to manipulate and deceive. As Mark Twain said, “There are three kinds of lies: lies, damned lies, and statistics.” (Twain attributes the remark to British Prime Minister Benjamin Disraeli, though it’s not really clear who said it first.) Advertisers, politicians, pundits—everybody in the persuasion business—trot out statistical claims to bolster their arguments, and more often than not they are either deliberately or mistakenly committing some sort of fallacy. We will end with a survey of these sorts of errors.

But first, we examine the probability calculus. Our study of how to compute probabilities will divide neatly into two sections, corresponding to the two basic types of probability calculations one can make. There are, on the one hand, probabilities of multiple events all occurring—or, equivalently, multiple propositions all being true; call these conjunctive occurrences. We will first learn how to calculate the probabilities of conjunctive occurrences—that this event and this other event and some other event and so on will occur. On the other hand, there are probabilities that at least one of a set of alternative events will occur—or, equivalently, that at least one of a set of propositions will be true; call these disjunctive occurrences. In the second half of our examination of the probability calculus we will learn how to calculate the probabilities of disjoint occurrences— that this event or this other event or some other event or... will occur.

Conjunctive Occurrences

Recall from our study of sentential logic that conjunctions are, roughly, ‘and’-sentences. We can think of calculating the probability of conjunctive occurrences as calculating the probability that a particular conjunction is true. If you roll two dice and want to know your chances of getting “snake eyes” (a pair of ones), you’re looking for the probability that you’ll get a one on the first die and a one on the second.

Such calculations can be simple or slightly more complex. What distinguishes the two cases is whether or not the events involved are independent. Events are independent when the occurrence of one has no effect on the probability that any of the others will occur. Consider the dice mentioned above. We considered two events: one on die #1, and one on die #2. Those events are independent. If I get a one on die #1, that doesn’t affect my chances of getting a one on the second die; there’s no mysterious interaction between the two dice, such that what happens with one can affect what happens with the other. They’re independent. (If you think otherwise, you’re committing what’s known as the Gambler’s Fallacy. It’s surprisingly common. Go to a casino and you’ll see people committing it. Head to the roulette wheel, for example, where people can bet on whether the ball lands in a red or a black space. After a run of say, five reds in a row, somebody will commit the fallacy: “Red is hot! I’m betting on it again.” This person believes that the results of the previous spins somehow affect the probability of the outcome of the next one. But they don’t. Notice that an equally compelling (and fallacious) case can be made for black: “Five reds in a row? Black is due. I’m betting on black.”) On the other hand, consider picking two cards from a standard deck (and keeping them after they’re drawn). (A standard deck has 52 playing cards, equally divided among four suits (hearts, diamonds, clubs, and spades) with 13 different cards in each suit: Ace (A), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K).) Here are two events: the first card is a heart, the second card is a heart. Those events are not independent. Getting a heart on the first draw affects your chances of getting a second heart (it makes the second heart less likely).

When events are independent, things are simple. We calculate the probability of their conjunctive occurrence by multiplying the probabilities of their individual occurrences. This is the Simple Product Rule:

P(a • b • c • ...) = P(a) x P(b) x P(c) x ...

This rule is abstract; it covers all cases of the conjunctive occurrence of independent events. ‘a’, ‘b’, and ‘c’ refer to events; the ellipses indicate that there may be any number of them. When we write ‘P’ followed by something in parentheses, that’s just the probability of the thing in parentheses coming to pass. On the left-hand side of the equation, we have a bunch of events with dots in between them. The dot means the same thing it did in SL: it’s short for and. So this equation just tells us that to compute the probability of a and b and c (and however many others there are) occurring, we just multiply together the individual probabilities of those events occurring on their own.

Go back to the dice above. We roll two dice. What’s the probability of getting a pair of ones? The events—one on die #1, one on die #2—are independent, so we can use the Simple Product Rule and just multiply together their individual probabilities.

What are those probabilities? We express probabilities as numbers between 0 and 1. An event with a probability of 0 definitely won’t happen (a proposition with a probability of 0 is certainly false); an event with a probability of 1 definitely will happen (a proposition with a probability of 1 is certainly true). Everything else is a number in between: closer to 1 is more probable; closer to 0, less. So, how probable is it for a rolled die to show a one? There are six possible outcomes when you roll a die; each one is equally likely. When that’s the case, the probability of the particular outcome is just 1 divided by the number of possibilities. The probability of rolling a one is 1/6.

So, we calculate the probability of rolling “snake eyes” as follows:

P(one on die #1 • one on die #2) = P(one on die #1) x P(one on die #2)
= 1/6 x 1/6
= .0278

If you roll two dice a whole bunch of times, you’ll get a pair of one a little less than 3% of the time.

We noted earlier that if you draw two cards from a deck, two possible outcomes—first card is a heart, second card is a heart —are not independent. So we couldn’t calculate the probability of getting two spades using the Simple Product Rule. We could only do that if we made the two events independent—if we stipulated that after drawing the first card, you put it (randomly) back into the deck, so you’re picking at random from a full deck of cards each time. In that case, you’ve got a 1/4 chance of picking a heart each time, so the probability of picking two in a row would be 1/4 x 1/4—and the probability of picking three in a row would be 1/4 x 1/4 x 1/4, and so on.

Of course the more interesting question—and the more practical one, if you’re a card player looking for an edge—is the original one: what’s the probability of, say, drawing three hearts assuming, as is the case in all real-life card games, that you keep the cards as you draw them? As we noted, these events— heart on the first card, heart on the second card, heart on the third card— are not independent, because each time you succeed in drawing a heart, that affects your chances(negatively) of drawing another one. Let’s think about this effect in the current case. The probability of drawing the first heart from a well-shuffled, complete deck is simple: 1/4. It’s the subsequent hearts that are complicated. How much of an effect does success at drawing that first heart have on the probability of drawing the second one? Well, if we’ve already drawn one heart, the deck from which we’re attempting to draw the second is different from the original, full deck: specifically, it’s short the one card already drawn—so there are only 51 total—and it’s got fewer hearts now—12 instead of the original 13. 12 out of the remaining 51 cards are hearts, then. So the probability of drawing a second heart, assuming the first one has already been picked, is 12/51. If we succeed in drawing the second heart, what are our chances at drawing a third? Again, in this case, the deck is different: we’re now down to 50 total cards, only 11 of which are hearts. So the probability of getting the third heart is 11/50.

It’s these fractions—1/4, 12/51, and 11/50—that we must multiply together to determine the probability of drawing three straight hearts while keeping the cards. The result is (approximately) .013—a lower probability than that of picking 3 straight hearts when the cards are not kept, but replaced after each selection: 1/4 x 1/4 x 1/4 = .016 (approximately). This is as it should be: it’s harder to draw three straight hearts when the cards are kept, because each success diminishes the probability of drawing another heart. The events are not independent.

In general, when events are not independent, we have to make the same move that we made in the three-hearts case. Rather than considering the stand-alone probability of a second and third heart— as we could in the case where the events were independent—we had to consider the probability of those events assuming that other events had already occurred. We had to ask what the probability was of drawing a second heart, given that the first one had already been drawn; then we asked after the probability of drawing the third heart, given that the first two had been drawn.

We call such probabilities—the likelihood of an event occurring assuming that others have occurred—conditional probabilities. When events are not independent, the Simple Product Rule does not apply; instead, we must use the General Product Rule:

P(a • b • c • ...) = P(a) x P(b | a) x P(c | a • b) x ...

The term ‘P(b | a)’ stands for the conditional probability of b occurring, provided a already has. The term ‘P(c | a • b)’ stands for the conditional probability of c occurring, provided a and b already have. If there were a fourth event, d, we would have a this term on the right-hand side of the equation: ‘P(d | a • b • c)’. And so on.

Let’s reinforce our understanding of how to compute the probabilities of conjunctive occurrences with a sample problem:

There is an urn filled with marbles of various colors. Specifically, it contains 20 red marbles, 30 blue marbles, and 50 white marbles. If we select 4 marbles from the earn at random, what’s the probability that all four will be blue, (a) if we replace each marble after drawing it, and (b) if we keep each marble after drawing it?

Let’s let ‘B1’ stand for the event of picking a blue marble on the first selection; and we’ll let ‘B2’, ‘B3’, and ‘B4’ stand for the events of picking blue on the second, third, and fourth selections, respectively. We want the probability of all of these events occurring:

P(B1 • B2 • B3 • B4) = ?

(a) If we replace each marble after drawing it, then the events are independent: selecting blue on one drawing doesn’t affect our chances of selecting blue on any other; for each selection, the urn has the same composition of marbles. Since the events are independent in this case, we can use the Simple Product Rule to calculate the probability:

P(B1 • B2 • B3 • B4) = P(B1) x P(B2) x P(B3) x P(B4)

And since there are 100 total marbles in the urn, and 30 of them are blue, on each selection we have a 30/100 (= .3) probability of picking a blue marble.

P(B1 • B2 • B3 • B4) = .3 x .3 x .3 x .3 = .0081

(b) If we don’t replace the marbles after drawing them, then the events are not independent: each successful selection of a blue marble affects our chances (negatively) of drawing another blue marble. When events are not independent, we need to use the General Product Rule:

P(B1 • B2 • B3 • B4) = P(B1) x P(B2 | B1) x P(B3 | B1 • B2) x P(B4 | B1 • B2 • B3)

On the first selection, we have the full urn, so P(B1) = 30/100. But for the second term in our product, we have the conditional probability P(B2 | B1); we want to know the chances of selecting a second blue marble on the assumption that the first one has already been selected. In that situation, there are only 99 total marbles left, and 29 of them are blue. For the third term in our product, we have the conditional probability P(B3 | B1 • B2); we want to know the chances of drawing a third blue marble on the assumption that the first and second ones have been selected. In that situation, there are only 98 total marbles left, and 28 of them are blue. And for the final term—P(B4 | B1 • B2 • B3)—we want the probability of a fourth blue marble, assuming three have already been picked; there are 27 left out of a total of 97.

P(B1 • B2 • B3 • B4) = 30/100 x 29/99 x 28/98 x 27/97 = .007 (approximately)

Disjunctive Occurrences

Conjunctions are (roughly) ‘and’-sentences. Disjunctions are (roughly) ‘or’-sentences. So we can think of calculating the probability of disjunctive occurrences as calculating the probability that a particular disjunction is true. If, for example, you roll a die and you want to know the probability that it will come up with an odd number showing, you’re looking for the probability that you’ll roll a one or you’ll roll a three or you’ll roll a five.

As was the case with conjunctive occurrences, such calculations can be simple or slightly more complex. What distinguishes the two cases is whether or not the events involved are mutually exclusive. Events are mutually exclusive when at most one of them can occur—when the occurrence of one precludes the occurrence of any of the others. Consider the die mentioned above. We considered three events: it comes up showing one, it comes up showing three, and it comes up showing five. Those events are mutually exclusive; at most one of them can occur. If I roll a one, that means I can’t roll a three or a five; if I roll a three, that means I can’t roll a one or a five; and so on. (At most one of them can occur; notice, it’s possible that none of them occur.) On the other hand, consider the dice example from earlier: rolling two dice, with the events under consideration rolling a one on die #1 and rolling a one on die #2. These events are not mutually exclusive. It’s not the case that at most one of them could happen; they could both happen—we could roll snake eyes.

When events are mutually exclusive, things are simple. We calculate the probability of their disjunctive occurrence by adding the probabilities of their individual occurrences. This is the Simple Addition Rule:

P(a ∨ b ∨ c ∨ ...) = P(a) + P(b) + P(c) + ...

This rule exactly parallels the Simple Product Rule from above. We replace that rule’s dots with wedges, to reflect the fact that we’re calculating the probability of disjunctive rather than conjunctive occurrences. And we replace the multiplication signs with additions signs on the right- hand side of the equation to reflect the fact that in such cases we add rather than multiply the individual probabilities.

Go back to the die above. We roll it, and we want to know the probability of getting an odd number. There are three mutually exclusive events—rolling a one, rolling a three, and rolling a five—and we want their disjunctive probability; that’s P(one ∨ three ∨ five). Each individual event has a probability of 1/6, so we calculate the disjunctive occurrence with the Simple Addition Rule thus:

P(one ∨ three ∨ five) = P(one) + P(three) + P(five)
= 1/6 +1/6 +1/6 = 3/6 = 1/2

This is a fine result, because it’s the result we knew was coming. Think about it: we wanted to know the probability of rolling an odd number; half of the numbers are odd, and half are even; so the answer better be 1/2. And it is.

Now, when events are not mutually exclusive, the Simple Addition Rule cannot be used; its results lead us astray. Consider a very simple example: flip a coin twice; what’s the probability that you’ll get heads at least once? That’s a disjunctive occurrence: we’re looking for the probability that you’ll get heads on the first toss or heads on the second toss. But these two events—heads on toss #1, heads on toss #2—are not mutually exclusive. It’s not the case that at most one can occur; you could get heads on both tosses. So in this case, the Simple Addition Rule will give us screwy results. The probability of tossing heads is 1/2, so we get this:

If we use the Simple Addition Rule in this case, we get the result that the probability of throwing heads at least once is 1; that is, it’s absolutely certain to occur. Talk about screwy! We’re not guaranteed to gets heads at least once; we could toss tails twice in a row.

In cases such as this, where we want to calculate the probability of the disjunctive occurrence of events that are not mutually exclusive, we must do so indirectly, using the following universal truth:

P(success) = 1 - P(failure)

This formula holds for any event or combination of events whatsoever. It says that the probability of any occurrence (singular, conjunctive, disjunctive, whatever) is equal to 1 minus the probability that it does not occur. ‘Success’ = it happens; ‘failure’ = it doesn’t. Here’s how we arrive at the formula. For any occurrence, there are two possibilities: either it will come to pass or it will not; success or failure. It’s absolutely certain that at least one of these two will happen; that is, P(success ∨ failure) = 1. Success and failure are (obviously) mutually exclusive outcomes (they can’t both happen). So we can express P(success ∨ failure) using the Simple Addition Rule: P(success ∨ failure) = P(success) + P(failure). And as we’ve already noted, P(success ∨ failure) = 1, so P(success) + P(failure) = 1. Subtracting P(failure) from each side of the equation gives us our universal formula: P(success) = 1 - P(failure).

Let’s see how this formula works in practice. We’ll go back to the case of flipping a coin twice. What’s the probability of getting at least one head? Well, the probability of succeeding in getting at least one head is just 1 minus the probability of failing. What does failure look like in this case? No heads; two tails in a row. That is, tails on the first toss and tails on the second toss. See that ‘and’ in there? (I italicized it.) This was originally a disjunctive-occurrence calculation; now we’ve got a conjunctive occurrence calculation. We’re looking for the probability of tails on the first toss and tails on the second toss:

P(tails on toss #1 • tails on toss #2) = ?

We know how to do problems like this. For conjunctive occurrences, we need first to ask whether the events are independent. In this case, they clearly are. Getting tails on the first toss doesn’t affect my chances of getting tails on the second. That means we can use the Simple Product Rule:

P(tails on toss #1 • tails on toss #2) = P(tails on toss #1) x P(tails on toss #2)
=1/2x1/2 =1/4

Back to our universally true formula: P(success) = 1 – P(failure). The probability of failing to toss at least one head is 1/4. The probability of succeeding in throwing at least one head, then, is just 1 – 1/4 = 3/4. (This makes good sense. If you throw a coin twice, there are four distinct ways things could go: (1) you throw heads twice; (2) you throw heads the first time, tails the second; (3) you throw tails the first time, heads the second; (4) you throw tails twice. In three out of those four scenarios (all but the last), you’ve thrown at least one head.)

So, generally speaking, when we’re calculating the probability of disjunctive occurrences and the events are not mutually exclusive, we need to do so indirectly, by calculating the probability of the failure of any of the disjunctive occurrences to come to pass and subtracting that from 1. This has the effect of turning a disjunctive occurrence calculation into a conjunctive occurrence calculation: the failure of a disjunction is a conjunction of failures. This is a familiar point from our study of SL in Chapter 4. Failure of a disjunction is a negated disjunction; negated disjunctions are equivalent to conjunctions of negations. This is one of DeMorgan’s Laws:

~ (p ∨ q) ≡ ~ p • ~ q

Let’s reinforce our understanding of how to compute probabilities with another sample problem. This problem will involve both conjunctive and disjunctive occurrences.

There is an urn filled with marbles of various colors. Specifically, it contains 20 red marbles, 30 blue marbles, and 50 white marbles. If we select 4 marbles from the urn at random, what’s the probability that all four will be the same color, (a) if we replace each marble after drawing it, and (b) if we keep each marble after drawing it? Also, what’s the probability that at least one of our four selections will be red, (c) if we replace each marble after drawing it, and (d) if we keep each marble after drawing it?

This problem splits into two: on the one hand, in (a) and (b), we’re looking for the probability of drawing four marbles of the same color; on the other hand, in (c) and (d), we want the probability that at least one of the four will be red. We’ll take these two questions up in turn.

First, the probability that all four will be the same color. We dealt with a narrower version of this question earlier when we calculated the probability that all four selections would be blue. But the present question is broader: we want to know the probability that they’ll all be the same color, not just one color (like blue) in particular, but any of the three possibilities—red, white, or blue. There are three ways we could succeed in selecting four marbles of the same color: all four red, all four white, or all four blue. We want the probability that one of these will happen, and that’s a disjunctive occurrence:

P(all 4 red ∨ all 4 white ∨ all 4 blue) = ?

When we are calculating the probability of disjunctive occurrences, our first step is to ask whether the events involved are mutually exclusive. In this case, they clearly are. At most, one of the three events—all four red, all four white, all four blue—will happen (and probably none of them will); we can’t draw four marbles and have them all be red and all be white, for example. Since the events are mutually exclusive, we can use the Simple Addition Rule to calculate the probability of their disjunctive occurrence:

P(all 4 red ∨ all 4 white ∨ all 4 blue) = P(all 4 red) + P(all 4 white) + P(all 4 blue)

So we need to calculate the probabilities for each individual color—that all will be red, all white, and all blue—and add those together. Again, this is the kind of calculation we did earlier, in our first practice problem, when we calculated the probability of all four marbles being blue. We just have to do the same for red and white. These are calculations of the probabilities of conjunctive occurrences:

P(R1 • R2 • R3 • R4) = ? P(W1 • W2 • W3 • W4) = ?

(a) If we replace the marbles after drawing them, the events are independent, and so we can use the Simple Product Rule to do our calculations:

P(R1 • R2 • R3 • R4) = P(R1) x P(R2) x P(R3) x P(R4) P(W1 • W2 • W3 • W4) = P(W1) x P(W2) x P(W3) x P(W4)

Since 20 of the 100 marbles are red, the probability of each of the individual red selections is .2; since 50 of the marbles are white, the probability for each white selection is .5.

P(R1 • R2 • R3 • R4) = .2 x .2 x .2 x .2 = .0016 P(W1 • W2 • W3 • W4) = .5 x .5 x .5 x .5 = .0625

In our earlier sample problem, we calculated the probability of picking four blue marbles: .0081. Putting these together, the probability of picking four marbles of the same color:

P(all 4 red ∨ all 4 white ∨ all 4 blue) = P(all 4 red) + P(all 4 white) + P(all 4 blue) = .0016 + .0625 + .0081 = .0722

(b) If we don’t replace the marbles after each selection, the events are not independent, and so we must use the General Product Rule to do our calculations. The probability of selecting four red marbles is this:

P(R1 • R2 • R3 • R4) = P(R1) x P(R2 | R1) x P(R3 | R1 • R2) x P(R4 | R1 • R2 • R3)

We start with 20 out of 100 red marbles, so P(R1) = 20/100. On the second selection, we’re assuming the first red marble has been drawn already, so there are only 19 red marbles left out of a total of 99; P(R2 | R1) = 19/99. For the third selection, assuming that two red marbles have been drawn, we have P(R3 | R1 • R2) = 18/98. And on the fourth selection, we have P(R4 | R1 • R2 • R3) = 17/97.

P(R1 • R2 • R3 • R4) = 20/100 x 19/99 x 18/98 x 17/97 = .0012 (approximately)

The same considerations apply to our calculation of drawing four white marbles, except that we start with 50 of those on the first draw:

P(W1 • W2 • W3 • W4) = 50/100 x 49/99 x 48/98 x 47/97 = .0587 (approximately)

In our earlier sample problem, we calculated the probability of picking four blue marbles as .007. Putting these together, the probability of picking four marbles of the same color:

P(all 4 red ∨ all 4 white ∨ all 4 blue) = P(all 4 red) + P(all 4 white) + P(all 4 blue) = .0012 + .0587 + .007 = .0669 (approximately)

As we would expect, there’s a slightly lower probability of selecting four marbles of the same color when we don’t replace them after each selection.

We turn now to the second half of the problem, in which we are asked to calculate the probability that at least one of the four marbles selected will be red. The phrase ‘at least one’ is a clue: this is a disjunctive occurrence problem. We want to know the probability that the first marble will be red or the second will be red or the third or the fourth:

P(R1 ∨ R2 ∨ R3 ∨ R4) = ?

When our task is to calculate the probability of disjunctive occurrences, the first step is to ask whether the events are mutually exclusive. In this case, they are not. It’s not the case that at most one of our selections will be a red marble; we could pick two or three or even four (we calculated the probability of picking four just a minute ago). That means that wecan’t use the Simple Addition Rule to make this calculation. Instead, we must calculate the probability indirectly, relying on the fact that P(success) = 1 - P(failure). We must subtract the probability that we don’t select any red marbles from 1:

P(R1 ∨ R2 ∨ R3 ∨ R4) = 1 - P(no red marbles)

As is always the case, the failure of a disjunctive occurrence is just a conjunction of individual failures. Not getting any red marbles is failing to get a red marble on the first draw and failing to get one on the second draw and failing on the third and on the fourth:

P(R1 ∨ R2 ∨ R3 ∨ R4) = 1 - P(~ R1 • ~ R2 • ~ R3 • ~ R4)

In this formulation, ‘~ R1’ stands for the eventuality of not drawing a red marble on the first selection, and the other terms for not getting red on the subsequent selections. Again, we’re just borrowing symbols from SL.

Now we’ve got a conjunctive occurrence problem to solve, and so the question to ask is whether the events ~ R1, ~ R2, and so on are independent or not. And the answer is that it depends on whether we replace the marbles after drawing them or not.

(c) If we replace the marbles after each selection, then failure to pick red on one selection has no effect on the probability of failing to select red subsequently. It’s the same urn— with 20 red marbles out of 100—for every pick. In that case, we can use the Simple Product Rule for our calculation:

P(R1 ∨ R2 ∨ R3 ∨ R4) = 1 - [P(~ R1) x P(~ R2) x P(~ R3) x P(~ R4)]

Since there are 20 red marbles, there are 80 non-red marbles, so the probability of picking a color other than red on any given selection is .8.

P(R1 ∨ R2 ∨ R3 ∨ R4) = 1 - (.8 x .8 x .8 x .8) = 1 - .4096 = .5904

(d) If we don’t replace the marbles after each selection, then the events are not independent, and we must use the General Product Rule for our calculation. The quantity that we subtract from 1 will be this:

P(~ R1) x P(~ R2 | ~ R1) x P(~ R3 | ~ R1 • ~ R2) x P(~ R4 | ~ R1 • ~ R2 • ~ R3) = ?

On the first selection, our chances of picking a non-red marble are 80/100. On the second selection, assuming we chose a non-red marble the first time, our chances are 79/99. And on the third and fourth selections, the probabilities are 78/98 and 77/97, respectively. Multiplying all these together, we get .4033 (approximately), and so our calculation of the probability of getting at least one red marble looks like this:

P(R1 ∨ R2 ∨ R3 ∨ R4) = 1 - .4033 = .5967 (approximately)

We have a slightly better chance at getting a red marble if we don’t replace them, since each selection of a non-red marble makes the urn’s composition a little more red-heavy.

Exercises

1. Flip a coin 6 times; what’s the probability of getting heads every time?

2. Go into a racquetball court and use duct tape to divide the floor into four quadrants of equal area. Throw three super-balls in random directions against the walls as hard as you can. What’s the probability that all three balls come to rest in the same quadrant?

3. You’re at your grandma’s house for Christmas, and there’s a bowl of holiday-themed M&Ms— red and green ones only. There are 500 candies in the bowl, with equal number of each color. Pick one, note its color, then eat it. Pick another, note its color, and eat it. Pick a third, note its color, and eat it. What’s the probability that you ate three straight red M&Ms?

4. You and two of your friends enter a raffle. There is one prize: a complete set of Ultra Secret Rare Pokémon cards. There are 1000 total tickets sold; only one is the winner. You buy 20, and your friends each buy 10. What’s the probability that one of you wins those Pokémon cards?

5. You’re a 75% free-throw shooter. You get fouled attempting a 3-point shot, which means you get 3 free-throw attempts. What’s the probability that you make at least one of them?

6. Roll two dice; what’s the probability of rolling a seven? How about an eight?

7. In my county, 70% of people voted for Donald Trump. Pick three people at random. What’s the probability that at least one of them is a Trump voter?

8. You see these two boxes here on the table? Each of them has jelly beans inside. We’re going to play a little game, at the end of which you have to pick a random jelly bean and eat it. Here’s the deal with the jelly beans. You may not be aware of this, but food scientists are able to create jelly beans with pretty much any flavor you want—and many you don’t want. There is, in fact, such a thing as vomit-flavored jelly beans. (Really: http://mentalfloss.com/article/62593...-weird-flavors) Anyway, in one of my two boxes, there are 100 total jelly beans, 8 of which are vomit-flavored (the rest are normal fruit flavors). In the other box, I have 50 jelly beans, 7 of which are vomit-flavored. Remember, this all ends with you choosing a random jelly bean and eating it. But you have a choice between two methods of determining how it will go down: (a) You flip a coin, and the result of the flip determines which of the two boxes you choose a jelly bean from; (b) I dump all the jelly beans into the same box and you pick from that.Which option do you choose? Which one minimizes the probability that you’ll end up eating a vomit-flavored jelly bean? Or does it not make any difference?

9. For men entering college, the probability that they will finish a degree within four years is .329; for women, it’s .438. Consider two freshmen—Albert and Betty. What’s the probability that at least one of them will fail to complete college in at least four years? What’s the probability that exactly one of them will succeed in doing so?

10. I love Chex Mix. My favorite things in the mix are those little pumpernickel chips. But they’re relatively rare compared to the other ingredients. That’s OK, though, since my second-favorite are the Chex pieces themselves, and they’re pretty abundant. I don’t know what the exact ratios are, but let’s suppose that it’s 50% Chex cereal, 30% pretzels, 10% crunchy bread sticks, and 10% my beloved pumpernickel chips. Suppose I’ve got a big bowl of Chex Mix: 1,000 total pieces of food. If I eat three pieces from the bowl, (a) what’s the probability that at least one of them will be a pumpernickel chip? And (b) what’s the probability that either all three will be pumpernickel chips or all three will be my second-favorite—Chex pieces?

11. You’re playing draw poker. Here’s how the game works: a poker hand is a combination of five cards; some combinations are better than others; in draw poker, you’re dealt an initial hand, and then, after a round of wagering, you’re given a chance to discard some of your cards (up to three) and draw new ones, hoping to improve your hand; after another round of betting, you see who wins. In this particular hand, you’re initially dealt a 7 of hearts and the 4, 5, 6, and King of spades. This hand is quite weak on its own, but it’s very close to being quite strong, in two ways: it’s close to being a “flush”, which is five cards of the same suit (you have four spades); it’s also close to being a “straight”, which is five cards of consecutive rank (you have four in a row in the 4, 5, 6, and 7). A flush beats a straight, but in this situation that doesn’t matter; based on how the other players acted during the first round of betting, you’re convinced that either the straight or the flush will win the money in the end. The question is, which one should you go for? Should you discard the King, hoping to draw a 3 or an 8 to complete your straight? Or should you discard the 7 of hearts, hoping to draw a spade to complete your flush? What’s the probability for each? You should pick whichever one is higher. (Inspired by an exercise from Copi and Cohen, pp. 596 - 597)

This page titled 6.1: The Probability of Calculus is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Matthew Knachel via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.