13.2 Probability and Logic
This module discusses how to apply probabilities to individual cases, and now to compute the probability of a conjunction of events or a disjunction of events. It is important to pay attention to whether events are independent of one another, and whether events are mutually exclusive.
Table of Contents
- 13.2 Probability and Logic
13.2.1 Making and Applying Generalizations
What’s the probability that a yellow sock will be pulled out?
Review: Making Generalizations
In the previous sub-module, we observed three different ways of calculating probabilities for events. It will be useful to review them briefly.
First, there’s the theoretical probability of an event, assuming all the options are equally probable and the outcome is truly random: for instance, the probability that a dice will roll a four on one roll is 1 in 6, and the probability of drawing an Ace of Spades from a deck of 52 cards is 1 in 52.
Second, there is the statistical or frequency probability of an event, given our observations of occurrences of similar events in the past. For instance, if the hotel staff have observed that 1 in every 20 guests at their hotel in the past have been smokers, then the probability that the next guest who arrives will be a smoker is 1 in 20. If 4 out of 5 people polled support requiring dogs to be leashed and licensed, then the probability that a random person on the street will support leashing and licensing dogs is 4 in 5.
Third, there is the subjective or Bayesian probability of an event, our sense of subjective confidence about the event. For instance, if Barb is a barista, Barb might feel about 60% sure that the next customer who places an order is going to order a Latte with soy milk, based on her experience, even though she can’t precisely identify all of the factors which lead her to this conclusion. Her coworker Brad might instead feel about 40% sure, based on his different experience. Barb would be willing to pay 60 cents, and Brad 40 cents, on a bet with a return of $1 if next customer orders a Latte with soy milk.
Caution: Applying Generalizations
We have to be careful in how we apply probabilities about groups into probabilities about individual members of a group. For instance, suppose we have a statistic that 75% of all people released from prison are arrested for a crime again within 5 years of release. It follows that, all other things being equal, a randomly selected prisoner being released today has a 75% chance of returning within 5 years. This assumes we have no further information about the prisoner being released, however.
A probability about a group does not give us any information about a particular prisoner or their particular case. For instance, if Joe is a prisoner who is being released, we might have other information about Joe: that he’s kept in touch with his family, has a support network to return to, has gained some employable skills while in prison, has repaid the victim of the crime, and has resolved not to return to crime again. This information would reduce the probability below 75%. On the other hand, suppose that Joe has used his time in prison to gain skills in committing crimes more efficiently, has made closer connections with international criminal organizations, has learned to laugh at the misery of crime victims, and has already hatched a plan for a string of robberies as soon as he is released. In this case, the probability Joe will reoffend is much higher than 75%.
Suppose that 1% of the population has a particular disease. What is the probability that Eric has the disease? If we know nothing else about Eric, then it is 1%. But if we know that Eric is over 60 years old, and 10% of people over 60 have the disease, then suddenly the probability Eric has the disease increases to 10%. But, suppose that 100% of the people over 60 who have the disease are biologically female, and Eric is biologically male. Now, the probability Eric has the disease will drop to 0%. So, once again, we have to be careful in applying generalizations to individuals.
Suppose that Mary is over age 60 but Mary eats a vegan diet. Suppose that 10% of people over 60 have a disease, but only 2% of people who eat a vegan diet get the disease. What’s Mary’s chance of getting the disease? On the one hand, her eating a vegan diet makes it lower than the 10% of people over age 60, but her being over age 60 makes it higher than the 2% of people who eat a vegan diet. So, her chances are somewhere between 2 – 10%. To know precisely, we’d need to know how many people over age 60 eat a vegan diet, or how many people who eat a Vegan diet are over 60%, or, ideally, how many people over 60 who eat a vegan diet get the disease.
Caution: Comparing Probabilities
We also need to be very cautious when comparing probabilities. For instance, suppose that there is a lottery with 1 million tickets and each person can buy only one ticket. Suppose that ⅔ of the tickets are purchased by men, and ⅓ of the tickets are purchased by women. Suppose somebody asks: what is the probability that a woman will win the lottery? This question is ambiguous. The probability that the winner of the lottery will be a woman is ⅓. The probability that a given woman will win the lottery, however, is 1 in 1,000,000. Are men more likely than women to win the lottery? Again, the question is ambiguous. The probability that the winner of the lottery will be a man is ⅔, so it is twice as likely that a man will win the lottery than a woman. It does not follow that a given man is twice as likely to win the lottery than a given woman: the probability for a given man is still 1 in 1,000,000.
Suppose you hear on the radio a statistic, sponsored by the Alaskan tourism board, that Arizonans are 80% more likely to suffer from heat stroke than Alaskans. This might make it sound like being in Arizona is a significant health risk. This is misleading, however. Suppose that the likelihood that an Alaskan will suffer from heat stroke is 0.001%. What is the likelihood that an Arizonan will suffer from heat stroke? If we’re not thinking clearly, we might conclude that the answer is 80.001%! Of course, that is incorrect. Instead, the likelihood would be 0.0018%, since 0.0018% is 80% higher than 0.001%. Being “80% more likely” amounts to an increase of only 0.0008% in overall risk. Similarly, claims that certain dietary changes increase or decrease the risk of certain diseases by a certain percentage need to be weighed in light of the overall likelihood of the disease.
13.2.2 Probability and Conjunctions
What’s the probability of a pair of aces?
The Probability of Independent Conjunctions
What is the probability that two independent events will both occur? For example, what is the probability that from a standard deck of 52 cards you randomly draw a queen and then, after returning it to the pile and reshuffling the deck, you draw another queen? (Reshuffling makes the draws independent: if we didn’t reshuffle, they would be dependent). Since we are asking what is the probability that these two events both occur, this is a matter of calculating the probability of a joint occurrence, a conjunction of events.
In the following, “a” and “b” will refer to independent events, and the locution “P(a)” stands for “the probability of a.” Here is how we calculate the probability of independent conjunctions:
P(a & b) = P(a) × P(b)
So, to apply this to my example of drawing two queens, we have to multiply the probability of drawing one queen, “P(a)” by the probability of drawing yet another queen, “P(b).” The probability of drawing a queen is 4 (for the 4 queens in a deck) divided by 52 (for 52 cards in a deck), or .077, the math is quite simple:
.077 × .077 = .0059
That is, there’s a less than 1% chance (.59% to be precise) of drawing two queens in this scenario. So, obviously, you’d not be wise to place a bet on that happening!
Suppose I wanted to know the probability of rolling a 12 when rolling two, six-sided dice. Since the only way to roll a 12 is when I roll a 6 on each die, I can compute the probability of rolling a 6 and then the independent probability of rolling another 6 on the other die. The probability of rolling a six on 1 die is just 1/6 = .166. Thus,
.166 × .166 = .028
Thus, you have a 2.8% chance of rolling a 12. We could have also calculated this using fractions instead of decimals:
1/6 × 1/6 = 1/36
The Probability of Dependent Conjunctions
Both of the examples above involve independent events. The events are “independent” because the fact that one event has happened does not increase or decrease the probability of the other event: the way in which one die rolls doesn’t change how the other die rolls. Suppose instead, however, that we don’t put the queen back in the deck and reshuffle the deck if we draw a queen on the first draw. Instead, we leave the queen on the table and draw from the remaining stack of 51 cards. Now, the probability of drawing two queens is represented by a more complex formula:
P(a & b) = P(a) × P(b|a)
The expression b|a means “b given a”, or “b if a occurs”, so P(b|a) means “what the probability of b would be assuming that a has occurred”. The probability of drawing the first queen was 4 divided by 52, or .077. But suppose we drew that first queen. There are now 3 queens remaining in the deck, and 51 cards remaining. So the probability of drawing a second queen is 3 divided by 51, or .059. So, the probability of drawing two queens in a row, without replacement, is:
.077 × .059 = .0045
Suppose you know there is an 80% chance that Mary will accept if John proposes to her, but only a 50% chance that John will propose. What is the probability of the conjunction: John proposes & Mary accepts?
.5 × .8 = .4
Garth Kemerling gives the following example:
- Suppose that a bag contains three red marbles, four blue marbles, and five white marbles. Then the probability of pulling out a white marble without looking is 5/12, the probability of pulling out a second is 4/11, and the probability of pulling out a third is 3/10. So the probability of pulling out three white marbles is 5/12 × 4/11 × 3/10, or 1/22. The chance of getting three red marbles, on the other hand, is 3/12 × 2/11 × 1/10, or only 1/220. (The Philosophy Pages, 2011)
Suppose that there are two students in a class of 10, Jorge and Lakshmi, and the teacher will pick at random two students to write their logic homework up on the board at the front. What’s the probability of both being chosen?
Well, first, let’s consider the probability of Jorge being chosen (1 in 10) followed by Lakshmi (1 in 9). That would be 1/10 x 1/9 = 1 in 90. But wait! We also have to consider the probability of Lakshmi being chosen (1 in 10) followed by Jorge (1 in 9). That also would be 1/10 x 1/9 = 1 in 90. Adding them together we get 2 in 90, which is 1 in 45. So the correct answer is 1 in 45
Another way to figure it is that there is a 2 in 10 chance of either of them being chosen first, and a 1 in 9 chance of the remaining person being chosen. So, 2/10 x 1/9 = 2 in 90 = 1 in 45.
The Conjunction Fallacy
Although some basic arithmetic is involved, the math isn’t too difficult when it comes to calculating probabilities for conjunctions. In spite of this, people tend to make a very common mistake when estimating the probability of a conjunction, known as the conjunction fallacy.
The conjunction fallacy is best introduced with an example, made famous by Tversky & Kahneman (1983) in their paper “Extension versus intuitive reasoning: The conjunction fallacy in probability judgment”, published in Psychological Review.
Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations. Given this information about Linda, which of the following is more probable?
a. Linda is a bank teller.
b. Linda is a bank teller and is active in the feminist movement.
Most people who answer this question in psychological studies answer “b.” But that cannot be correct because it violates the basic rules of probability. In particular, notice that option b contains option a (i.e., Linda is a bank teller). But option b also contains more information—that Linda is also active in the feminist movement. The problem is that a conjunction of two independent events can never be more probable than either one of its conjuncts. Suppose we say it is very probable that Linda a bank teller (how boring, given the description of Linda which makes her sound interesting!). Let’s set the probability low, say .4. Then what is the probability of her being active in the feminist movement? Call that proposition (f). Let’s set that high, say .9. However, the probability that she is both a bank teller and active in the feminist movement (a & f) must be computed as the probability of a conjunction, like this:
P(a) × P(f) = P(a & f)
.4 × .9 = .36
So, the correct answer is “a”, because P(a) > P(a & f).
One explanation for why we fall into the conjunction fallacy is that what we are intuitively evaluating is the probability that someone’s testimony of a. or b. is knowledge, rather than the independent probability that a. or b. are true. Since testimony with more details is more likely to be knowledge, somebody who testifies that Linda is a bank teller and active in the feminist movement seems more likely to know Linda, since her being active in the feminist movement is consistent with what we know of her concern for discrimination and social justice; since saying Linda is a feminist fits with what we know of Linda, we’ll trust the person knows whom they’re talking about when they add that she is a bank teller. On the other hand, somebody who only tells us that Linda is a bank teller seems less likely to know Linda, since we had no other reason to think she was a bank teller, and they don’t seem to know anything else about her; perhaps it’s just a lucky guess, or maybe they’re thinking about a different person named “Linda”. In other words, we treat these as two dependent events, tied together because the same person testified to both, rather than independent events. The probability that Linda is a bank teller given that Linda is active in the feminist movement and the same person told us both, is perhaps much higher than the probability she is a bank teller: say it is .8 instead of .4. So, the probability would now be:
P(f) × P(a|f) = P(f & a)
.9 × .8 = .72
The conjunction of two dependent events can be greater than one of the conjuncts taken by itself, and P(f & a) > P(a).
13.2.3 Probability and Disjunctions
If you have 5, 6, 7 and 8, what’s the probability of drawing a 4 or a 9?
Probability of a Disjunction of Exclusive Events
Some events are mutually exclusive. For instance, if only one person can get the job after a job interview, then Donny getting the job and Barry getting the job are distinct events. The probability that one or the other of them will get the job is easy to calculate. Simply take the probability of one of them getting the job, and add the probability of the other getting the job.
Calculating the probability of a disjunction of two mutually exclusive events is simply a matter of figuring out the probability that either one event or another will occur. To calculate the probability of a disjunction we simply add the probability of the two events together:
P(A v B) = P(A) + P(B)
For example, suppose I wanted to calculate the probability of drawing randomly from a shuffled deck either a spade or a club. Since there a four suits (spades, clubs, diamonds, hearts) each with an equal number of cards, the probability of drawing a spade is 1⁄4 or .25. Likewise the probability of drawing a club is .25. No card is both a spade and a club. Thus, the probability of drawing either a spade or club is:
.25 + .25 = .50
So you have a 50% chance of drawing either a spade or a club.
Probability of a Disjunction of Non-Exclusive Events
Sometimes events are not mutually exclusive. For instance, suppose the probability that it rains is 8%, and the probability that it will be cloudy is 30%. What is the probability that it will either rain or be cloudy? We’d be mistaken to say 38%, because rain and clouds are not mutually exclusive: quite often it will both rain and be cloudy. So, we need to subtract out the overlap between the two events: the probability that both events will occur together.
Rain and clouds are dependent events. So, the probability of them both occurring is the probability of rain (8%) multiplied by the probability that if it is raining, then it is cloudy. It rarely rains when it isn’t cloudy, so let’s say the probability that it is cloudy if rainy is ⅞. Then, the probability probability of it both raining and being cloudy is 8% x ⅞ = 7%. So, get the probability that it would either rain or be cloudy, we’d subtract 38% minus 7% for a total of 31%.
The formula looks like this:
P(A v B) = P(A) + P(B) – P(A & B)
That is, the probability that one or the other or both of two events will occur is equal to the probability that the first will occur, plus the probability that the second will occur, minus the probability that they both occur. Here are some more examples:
- The probability of getting heads on one toss of a coin is .5 (or 1/2), and so is the probability of getting heads on a second toss of the same coin. Thus, the probability of getting heads at least once during two tosses of the coin is .5 + .5 – (.5 × .5), or .75 (3/4).
- What if we want to know the probability of getting heads five times in a row? Well, you can see, we got the distance between the probability and 1 by half with each additional coin toss:
- For the 3rd coin toss it will then be: .75 + 0.5 – (.75 x .5) = 1.25 – .375 = .875
- For the 4th coin toss we get: .875 + .5 – (.875 x .5) = .9375
- For the 5th toss: .9375 + .5 – (.9375 x .5) = .96875
- The chance of drawing one of the four aces from a standard deck of 52 cards is 4/52. Thus, the chance of drawing at least one ace in two draws is 4/52 + 4/52 – (4/52 × 3/51), or 33/221.
- Suppose that a bag contains three red marbles, four blue marbles, and five white marbles. Then the probability of pulling out a white marble without looking is 5/12. So the probability of pulling out at least one white marble in two tries is 5/12 + 5/12 – (5/12 × 4/11), or 15/22. (The chance of getting at least one red marble, on the other hand, is 3/12 + 3/12 – (3/12 × 2/11), or only 10/22.)
Probabilities Can’t Exceed 100%
Suppose we want to know the chances of flipping at least 1 head in 6 flips of a fair coin. You might reason as follows: There is a 50% chance I flip heads on the first flip, a 50% chance on the second, etc. Since I want to know the chance of flipping at least one head, then perhaps I should simply calculate the probability of the disjunction like this:
- .5 + .5 + .5 + .5 + .5 + .5 = 3 (or 300%)
However, this cannot be right, because the probability of any event is between 1 and 0 (including 0 and 1 for events that are impossible and absolutely certain). However, this way of calculating the probability leaves us with an event that is three times more than certain. And nothing is more than 100% certain—100% certainty is the limit. So something is wrong with the calculation.
Another way of seeing that something must be wrong with the calculation is that it isn’t impossible that I flip 6 tails in a row (and thus no heads). Since that is a real possibility (however improbable), it cannot be 100% certain that I flip at least one head. Here is the way to think about this problem. What is the probability that I flip all tails? That is, simply the probability of the conjunction of 6 events, each of which has the probability of .5 (or 50%):
- .5 × .5 × .5 × .5 × .5 × .5 = .015 (or 1.5%)
Then we simply use the rule for calculating the probability of a negation, since we want to know the chances that we don’t flip 6 tails in a row (i.e., we flip at least one head):
- 1 – .015 = .985
So the probability of flipping at least one head in 6 flips of the coin is 98.5%. (It would be exactly the same probability of flipping at least one tails in 6 flips.)
Remember to subtract the overlap for a disjunction of two non-exclusive events.
For instance, suppose you roll two six-sided dice once, or one die twice.
A map of the possibilities would look like the table below. There are 36 possible combinations of rolls.
The probability that one dice rolls a ‘5’ is 6 in 36, and the probability that the second dice rolls a ‘5’ is 6 in 36.
But the probability that either one or the other (that is, at least one) is a ‘5’ isn’t 12 in 36. It is 11 in 36.
That is because we have to subtract the overlapping case (1 in 36) where both are a ‘5’.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 1, 1 | 1, 2 | 1, 3 | 1, 4 | 1, 5 | 1, 6 |
| 2 | 2, 1 | 2, 2 | 2, 3 | 2, 4 | 2, 5 | 2, 6 |
| 3 | 3, 1 | 3, 2 | 3, 3 | 3, 4 | 3, 5 | 3, 6 |
| 4 | 4, 1 | 4, 2 | 4, 3 | 4, 4 | 4, 5 | 4, 6 |
| 5 | 5, 1 | 5, 2 | 5, 3 | 5, 4 | 5, 5 | 5, 6 |
| 6 | 6, 1 | 6, 2 | 6, 3 | 6, 4 | 6, 5 | 6, 6 |
Submodule 13.2 Quiz
Licenses and Attributions
Key Sources:
- Watson, Jeffrey (2019). Introduction to Logic. Licensed under: (CC BY-SA).
- Van Cleave, Matthew (2016), Introduction to Logic and Critical Thinking, pp 170-171, CC BY-SA 4.0
- Van Cleave, Matthew (2016), Introduction to Logic and Critical Thinking, pp 178, CC BY-SA 4.0
- Van Cleave, Matthew (2016). Introduction to Logic and Critical Thinking, pp 172-174. Copyright 2016. Licensed under: (CC BY-SA 4.0).
- Kemerling, Garth (2011). The Philosophy Pages, http://www.philosophypages.com. Copyright 2011. Licensed under: (CC BY-SA 3.0).
- Sock Drawer by Noricum, Shared under license CC BY-SA 2.0 via Flickr.
Next Page: 13.3 Conditional Probabilities