3.1 Terminology
SLO 3
Describe and apply probability concepts and distributions.
Before you can apply a probability concept you have to be able to state one. This section supplies the operating parts - experiment, sample space, event - and the notation for OR, AND, complement and conditional probability that every later rule in the chapter is written in.
Learning Objectives
By the end of this section, you will be able to:
- describe an experiment, list its sample space, and identify an event as a subset of that sample space;
- calculate the probability of an event when every outcome is equally likely, and explain what that number means over the long run;
- read and write the notation for OR, AND, and complement events, and find each one from a listed sample space;
- calculate a conditional probability and explain how conditioning shrinks the sample space;
- translate a sentence about a real situation into probability symbols, and translate symbols back into a sentence.
Probability is how we measure certainty. It is a number that tells you how sure you can be about the result of something whose result is not settled in advance — a coin toss, a card draw, a shipment of parts that may or may not contain a defective one.
Most of the work in a probability problem is not arithmetic. It is reading. Before you can compute anything you have to decide what is being repeated, what counts as a result, and which results you care about. This section gives those three things their proper names and their notation, so that the rest of the chapter has something precise to build on.
3.1.1 Experiments, Outcomes, and Sample Spaces
An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, the experiment is called a chance experiment.
Flipping one fair coin twice is an experiment. So is rolling a die, drawing a card, or selecting one student at random from a roster. The word does not mean a laboratory — it means a procedure you could repeat, whose result you cannot call in advance.
An outcome is a result of an experiment.
The sample space of an experiment is the set of all possible outcomes. The uppercase letter \(S\) denotes the sample space.
Everything that can happen has to appear in \(S\) somewhere, exactly once. If you can describe a result that isn't in your list, your list isn't the sample space yet.
Definition 3.1.3 — The sample space of two coin flips: every possible result appears in S exactly once.
There are three standard ways to display a sample space: list the possible outcomes, draw a tree diagram, or draw a Venn diagram. Sections 3.4 and 3.5 develop the last two. For now we list.
If you flip one fair coin, \(S = \{H, T\}\), where \(H\) is heads and \(T\) is tails. Two outcomes, and the set of them is the sample space.
An event is any combination of outcomes — that is, any subset of the sample space. Uppercase letters such as \(A\) and \(B\) represent events. The probability of an event \(A\) is written \(P(A)\).
Definition 3.1.4 — An event is a subset of the sample space: the outcomes that answer yes are the event.
The word combination is doing real work here. An event does not have to be a single outcome. If the experiment is flipping one fair coin, the event \(A\) might be "getting at most one head", which for a single flip is both outcomes at once. An event is a question you ask about the result, and the answer is yes for some outcomes and no for the others.
You spin a spinner with four equal wedges labelled 1, 2, 3, and 4.
a. Write the sample space \(S\).
b. Let \(A\) be the event that the spinner lands on an odd number. Write \(A\) as a set.
c. Let \(B\) be the event that the spinner lands on a number greater than 4. Write \(B\) as a set.
Solution
Part a — list every possible result. The spinner can stop on any of the four labels, so
$$S = \{1, 2, 3, 4\}$$Part b — keep the outcomes that answer yes. The odd labels are 1 and 3, so \(A = \{1, 3\}\).
Part c — an event can be empty. No label on this spinner is greater than 4, so \(B = \{\ \}\), the empty set. That is still a legitimate event; it just never happens.
Answer: \(S = \{1, 2, 3, 4\}\), \(A = \{1, 3\}\), \(B = \{\ \}\).
3.1.2 Probability and the Law of Large Numbers
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive.
Definition 3.1.5 — Every probability is a position on one line from 0 to 1.
Read that definition twice, because it is a definition about the long run and not about any one trial. \(P(A) = 0\) means the event \(A\) can never happen. \(P(A) = 1\) means \(A\) always happens. \(P(A) = 0.5\) means \(A\) is equally likely to occur or not to occur.
Here is what "long-term relative frequency" looks like in practice. Flip one fair coin repeatedly — 20 times, then 2,000, then 20,000 — and keep track of the fraction of flips that came up heads. That fraction wanders early and settles late. It approaches 0.5, which is the probability of heads.
Equally likely means that each outcome of an experiment occurs with equal probability.
Definition 3.1.6 — A fair die: six faces, six identical probabilities.
Toss a fair, six-sided die and each face — 1, 2, 3, 4, 5, or 6 — is as likely to come up as any other. Toss a fair coin and \(H\) and \(T\) are equally likely. Guess at random on a true/false question and you are equally likely to be right or wrong.
When every outcome in the sample space is equally likely, computing a probability becomes counting:
$$P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}$$Toss a fair dime and a fair nickel together. The sample space is \(S = \{HH, HT, TH, TT\}\), where the first letter is the dime and the second is the nickel — four outcomes. Let \(A\) be the event of getting exactly one head. Two outcomes meet that condition, \(HT\) and \(TH\), so
$$P(A) = \frac{2}{4} = 0.5$$Now roll one fair six-sided die and let \(E\) be the event of rolling a number that is at least five. Two outcomes qualify, 5 and 6, so \(P(E) = \frac{2}{6}\).
Roll that die only a handful of times and you should not be surprised if your results look nothing like \(\frac{2}{6}\). Roll it a very large number of times and you would expect about \(\frac{2}{6}\) of the rolls to land on "at least five" — not exactly \(\frac{2}{6}\), but closer to it the longer you go. That pull toward the theoretical value has a name.
The law of large numbers states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability.
A single flip tells you nothing about whether a coin is fair. An insurance company cannot predict which house will burn, but it can predict what fraction will, and it prices policies off that fraction. Probability is a statement about many trials, not about the next one.
Definition 3.1.7 — The running fraction of heads settles onto 0.5 as the flips pile up.
The outcomes do not arrange themselves into any pattern or order to make this happen. Each trial is as unpredictable as the last. What changes is that the accumulated fraction has more and more trials underneath it, so any one strange run matters less. (The word empirical is often used in place of observed — an empirical probability is one you got by watching, a theoretical probability is one you got by counting.)
A fair, eight-sided die has faces numbered 1 through 8. Let \(A\) be the event of rolling a number less than 3.
a. Find \(P(A)\).
b. Marisol rolls the die 16 times and gets a number less than 3 on only one roll. She concludes the die is unfair. Is her conclusion supported?
Solution
Part a — count, then divide. Every face is equally likely, so we count outcomes. The numbers less than 3 are 1 and 2, so \(A = \{1, 2\}\) has 2 outcomes out of 8:
$$P(A) = \frac{2}{8} = \frac{1}{4} = 0.25$$Part b — 16 rolls is not the long run. We would expect about \((0.25)(16) = 4\) qualifying rolls, and Marisol got 1. That is a gap, but 16 trials is far too few to conclude anything. The law of large numbers promises the relative frequency approaches 0.25 as the number of rolls grows large — it promises nothing about 16 rolls. She would need many more trials before the data could support her claim of unfairness.
Answer: \(P(A) = 0.25\); no, her 16 rolls are too small a sample to judge the die.
Watch the law of large numbers happen
Open rāSHio and choose File → Random Numbers… to generate a column of a few hundred whole numbers from 1 to 2, treating 1 as heads. Read the running share of heads down the column and you are watching exactly what this section describes: the fraction lurches over the first handful of draws and then tightens onto 0.5. Generate a short column and a long one and compare — the long one is not more random, it is just further along.
Figure 3.1.1 — Generating a column of random whole numbers in rāSHio: File → Random Numbers….
3.1.3 When the Outcomes Are Not Equally Likely
A coin, die, or other chance device is biased when its outcomes are not equally likely — that is, when at least one outcome occurs with a different probability than another.
Bias does not change what can happen, only how often. \(S = \{1, 2, 3, 4, 5, 6\}\) either way. What breaks is the shortcut of dividing by 6 — you now need the individual probabilities, which is what later sections give you tools for.
Definition 3.1.8 — The Belgian euro coin came out 56/44 over 250 trials, not the 50/50 a fair coin predicts.
Two math professors in Europe had their statistics students test the Belgian one-euro coin. In 250 trials a head came up 56% of the time and a tail 44% of the time. The data seem to show the coin is not fair — though 250 trials is not many, and more repetitions would be needed to draw a firm conclusion about the bias.
Dice can be biased too. Look at the dice in a game you have at home: the spots on each face are usually small holes carved out and then painted so the spots are visible. Those holes remove material, and different faces have different numbers of holes, so the faces may not weigh the same. Whether your dice are actually biased is hard to say, but the possibility is real. Casinos make a great deal of money on outcomes from rolling dice, so casino dice are made differently to eliminate it: the faces are flat, and the holes are completely filled with paint of the same density as the body of the die, so each face is equally likely to come up.
The counting formula above has a condition attached: every outcome in the sample space must be equally likely. In many real situations that condition fails.
Devon is building a carnival spinner with three wedges, colored red, blue, and green. On their spinner the red wedge takes up half the circle, and the blue and green wedges split the rest evenly.
a. Write the sample space.
b. Devon claims \(P(\text{red}) = \frac{1}{3}\). Explain why their answer is wrong.
c. Find \(P(\text{red})\), \(P(\text{blue})\), and \(P(\text{green})\).
Solution
Part a — the sample space is about what can happen, not how often. There are three possible results:
$$S = \{\text{red}, \text{blue}, \text{green}\}$$Part b — the counting shortcut needs equally likely outcomes. Devon divided by 3, which assumes all three colors are equally likely, and they are not: red covers half the circle while blue and green cover a quarter each. Because the three outcomes are not equally likely, \(\frac{\text{number in } A}{\text{number in } S}\) does not apply to their spinner.
Part c — use area instead of a count. The spinner is equally likely to stop at any point on the circle, so probability follows area:
$$P(\text{red}) = \frac{1}{2}, \qquad P(\text{blue}) = \frac{1}{4}, \qquad P(\text{green}) = \frac{1}{4}$$As a check, these sum to 1, which they must.
Answer: \(S = \{\text{red}, \text{blue}, \text{green}\}\); the counting shortcut fails because the wedges are unequal; \(P(\text{red}) = \frac{1}{2}\), \(P(\text{blue}) = P(\text{green}) = \frac{1}{4}\).
3.1.4 OR, AND, and the Complement
An outcome is in the event \(A\) OR \(B\) if the outcome is in \(A\), or is in \(B\), or is in both \(A\) and \(B\). The event \(A\) OR \(B\) is also written as \(A\) union \(B\), with notation \(A \cup B\).
Definition 3.1.9 — A OR B covers both circles, and the overlap is shaded once, not twice.
Everyday English uses "or" to mean one or the other but not both. Mathematics does not. \(A\) OR \(B\) includes the overlap.
For example, let \(A = \{1, 2, 3, 4, 5\}\) and \(B = \{4, 5, 6, 7, 8\}\). Then
$$A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}$$Notice that 4 and 5 are not listed twice. A set records what is in it, not how many ways an element got there.
An outcome is in the event \(A\) AND \(B\) if the outcome is in both \(A\) and \(B\) at the same time. The event \(A\) AND \(B\) is also written as \(A\) intersection \(B\), with notation \(A \cap B\).
Definition 3.1.10 — A AND B is only the lens: the outcomes in both sets at the same time.
Using the same two sets, \(A \cap B = \{4, 5\}\) — only the elements that appear in both lists.
The complement of event \(A\) is denoted \(A'\), read "\(A\) prime". \(A'\) consists of all outcomes that are NOT in \(A\). Because every outcome is either in \(A\) or in \(A'\) and never both,
$$P(A) + P(A') = 1$$
Definition 3.1.11 — A and its complement tile the whole sample space, so their probabilities add to 1.
For example, let \(S = \{1, 2, 3, 4, 5, 6\}\) and let \(A = \{1, 2, 3, 4\}\). Then \(A' = \{5, 6\}\), and
$$P(A) = \frac{4}{6}, \qquad P(A') = \frac{2}{6}, \qquad P(A) + P(A') = \frac{4}{6} + \frac{2}{6} = 1$$"At least one" problems are usually miserable to count directly and easy to count backwards. The complement of "at least one head" is "no heads at all", which is a single outcome. Subtract from 1 and you are done.
Once you can name single events you need a way to combine them. Three combinations cover almost everything: one event or the other, both events at once, and everything the event leaves out.
Let \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), \(A = \{2, 4, 6, 8, 10\}\), and \(B = \{1, 2, 3, 4, 5\}\).
a. Find \(A\) OR \(B\) and \(P(A \text{ OR } B)\).
b. Find \(A\) AND \(B\) and \(P(A \text{ AND } B)\).
c. Find \(B'\) and \(P(B')\).
Solution
Part a — union keeps everything, once. Combine both lists and drop repeats:
$$A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\}$$That is 8 outcomes out of 10, so \(P(A \text{ OR } B) = \frac{8}{10} = 0.8\).
Part b — intersection keeps only the overlap. The values in both lists are 2 and 4:
$$A \cap B = \{2, 4\}$$That is 2 outcomes out of 10, so \(P(A \text{ AND } B) = \frac{2}{10} = 0.2\).
Part c — the complement is everything left over. \(B' = \{6, 7, 8, 9, 10\}\), which is 5 outcomes out of 10, so \(P(B') = \frac{5}{10} = 0.5\). Check: \(P(B) + P(B') = 0.5 + 0.5 = 1\).
Answer: \(A \cup B = \{1,2,3,4,5,6,8,10\}\) with probability 0.8; \(A \cap B = \{2,4\}\) with probability 0.2; \(B' = \{6,7,8,9,10\}\) with probability 0.5.
3.1.5 Conditional Probability
The conditional probability of \(A\) given \(B\) is written \(P(A \mid B)\). It is the probability that event \(A\) will occur given that event \(B\) has already occurred. Conditioning reduces the sample space: we calculate the probability of \(A\) from the reduced sample space \(B\). The formula is
$$P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)}$$where \(P(B)\) is greater than zero.
A screening test that is positive does not tell you the probability of having the disease — it tells you the probability given a positive test. Confusing \(P(A \mid B)\) with \(P(A)\) is one of the most expensive mistakes in medicine and law, and it is the whole reason this notation exists.
Definition 3.1.12 — Conditioning on B discards the outcomes outside B, so the denominator drops from 6 to 3.
The phrase "reduces the sample space" is the part worth holding onto. Once you know \(B\) happened, outcomes outside \(B\) are off the table. You are no longer working in \(S\); you are working inside \(B\), and asking what fraction of \(B\) is also \(A\).
Suppose we toss one fair, six-sided die, so \(S = \{1, 2, 3, 4, 5, 6\}\). Let \(A\) be the event that the face is 2 or 3, and let \(B\) be the event that the face is even, so \(B = \{2, 4, 6\}\).
To find \(P(A \mid B)\) by counting, work inside \(B\). Of the three outcomes in \(B\), only one of them — the 2 — is also 2 or 3. So \(P(A \mid B) = \frac{1}{3}\). Notice we divided by 3, the size of \(B\), not by 6, the size of \(S\).
The formula gives the same answer. Remember that \(S\) has six outcomes:
$$P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}$$The numerator counts the outcomes that are both 2-or-3 and even, over 6. The denominator counts the outcomes that are even, over 6. The sixes cancel, which is exactly the "reduced sample space" idea showing up in the algebra.
Sometimes you learn something before you have to make a judgement. Conditional probability is how we build that extra information into the number.
A fair, six-sided die is rolled. Let \(A\) be the event that the roll is greater than 3, and let \(B\) be the event that the roll is odd.
a. Find \(P(A)\).
b. Find \(P(A \mid B)\).
c. Explain in a sentence why the two answers differ.
Solution
Part a — work in the full sample space. \(A = \{4, 5, 6\}\), which is 3 outcomes out of 6:
$$P(A) = \frac{3}{6} = \frac{1}{2}$$Part b — shrink the sample space to \(B\). \(B = \{1, 3, 5\}\), three outcomes. Of those, only 5 is greater than 3, so
$$P(A \mid B) = \frac{1}{3}$$Using the formula instead: \(A \text{ AND } B = \{5\}\), so \(P(A \text{ AND } B) = \frac{1}{6}\) and \(P(B) = \frac{3}{6}\), giving \(\frac{1/6}{3/6} = \frac{1}{3}\).
Part c — knowing the roll is odd is bad news for "greater than 3". Two of the three high faces (4 and 6) are even, so learning the roll is odd removes most of the ways \(A\) could have happened. The probability drops from \(\frac{1}{2}\) to \(\frac{1}{3}\).
Answer: \(P(A) = \frac{1}{2}\), \(P(A \mid B) = \frac{1}{3}\); conditioning on "odd" eliminates 4 and 6, two of the three outcomes that made \(A\) true.
3.1.6 Understanding Terminology and Symbols
Read each problem carefully and think about what the events actually are. Understanding the wording is the first important step in solving a probability problem, and it is the step most often rushed. Reread the problem as many times as you need to. Identify the event of interest clearly. Then check whether the wording states a condition that would make the probability conditional, and if it does, identify exactly what that condition is.
A few phrasings show up constantly and are worth memorizing as translations. "Of all the women, the probability that a student has short hair" means \(P(\text{short hair} \mid \text{woman})\) — the phrase "of all the ..." names the reduced sample space. "The probability that a student is a woman and has short hair" means \(P(\text{woman AND short hair})\), computed over everyone. Those two sentences describe different numbers, and the only difference in the English is a handful of small words.
It also helps to know that a probability model is a choice, not a discovery. It can be useful to model a process as random even when it is not truly random — nobody thinks a coin's flight is genuinely unpredictable given enough physics, but treating it as random gives answers that match what actually happens.
One more piece of housekeeping. As you get more comfortable with the notation, it gets abbreviated. If it is already clear that the process is "rolling a die", we write \(P(\text{rolling a } 1)\) as simply \(P(1)\). The shorter form means exactly the same thing.
The sample space \(S\) is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (for example, \((1, 4)\)).
a. \(S =\) ?
Let event \(A\) be that the sum is even and event \(B\) be that the first number is prime.
b. \(A =\) ?, \(B =\) ?
c. \(P(A) =\) ?, \(P(B) =\) ?
d. \(A\) AND \(B =\) ?, \(A\) OR \(B =\) ?
e. \(P(A \text{ AND } B) =\) ?, \(P(A \text{ OR } B) =\) ?
f. \(A' =\) ?, \(P(A') =\) ?
g. \(P(A) + P(A') =\) ?
h. \(P(A \mid B) =\) ?, \(P(B \mid A) =\) ?; are the probabilities equal?
Solution
Part a — pair every first value with every second value. Three choices for the first number and four for the second give \(3 \times 4 = 12\) ordered pairs:
$$S = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$$Part b — check each pair against each condition. Add the two entries of each pair and keep the even sums; separately, keep the pairs whose first number is 2 or 3 (the primes in the range 1 to 3 — note that 1 is not prime):
$$A = \{(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)\}$$ $$B = \{(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$$Part c — count over 12. \(A\) has 6 pairs and \(B\) has 8:
$$P(A) = \frac{6}{12} = \frac{1}{2}, \qquad P(B) = \frac{8}{12} = \frac{2}{3}$$Part d — intersection, then union. The pairs in both lists are those with a prime first number and an even sum:
$$A \text{ AND } B = \{(2,2), (2,4), (3,1), (3,3)\}$$ $$A \text{ OR } B = \{(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}$$Part e — count those. The intersection has 4 pairs; the union has 10:
$$P(A \text{ AND } B) = \frac{4}{12} = \frac{1}{3}, \qquad P(A \text{ OR } B) = \frac{10}{12} = \frac{5}{6}$$Part f — the complement is the odd sums.
$$A' = \{(1,2), (1,4), (2,1), (2,3), (3,2), (3,4)\}, \qquad P(A') = \frac{6}{12} = \frac{1}{2}$$Part g — check the total. \(P(A) + P(A') = \frac{1}{2} + \frac{1}{2} = 1\).
Part h — condition each way. Inside \(B\) there are 8 pairs, 4 of which have an even sum, so \(P(A \mid B) = \frac{4}{8} = \frac{1}{2}\). Inside \(A\) there are 6 pairs, 4 of which have a prime first number, so \(P(B \mid A) = \frac{4}{6} = \frac{2}{3}\). They are not equal.
Answer: \(P(A) = \frac{1}{2}\), \(P(B) = \frac{2}{3}\), \(P(A \text{ AND } B) = \frac{1}{3}\), \(P(A \text{ OR } B) = \frac{5}{6}\), \(P(A') = \frac{1}{2}\), \(P(A) + P(A') = 1\), \(P(A \mid B) = \frac{1}{2}\), \(P(B \mid A) = \frac{2}{3}\); not equal.
The sample space \(S\) is the whole numbers starting at one and less than 20. Let event \(A\) be the even numbers and event \(B\) be the numbers greater than 13.
a. \(S =\) ?
b. \(A =\) ?, \(B =\) ?
c. \(P(A) =\) ?, \(P(B) =\) ?
d. \(A\) AND \(B =\) ?, \(A\) OR \(B =\) ?
e. \(P(A \text{ AND } B) =\) ?, \(P(A \text{ OR } B) =\) ?
f. \(A' =\) ?, \(P(A') =\) ?
g. \(P(A) + P(A') =\) ?
h. \(P(A \mid B) =\) ?, \(P(B \mid A) =\) ?; are the probabilities equal?
Solution
Part a — list the sample space. The whole numbers from 1 up to but not including 20:
$$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\}$$That is 19 outcomes, so 19 is the denominator for every unconditional probability below.
Part b — write each event as a subset.
$$A = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}, \qquad B = \{14, 15, 16, 17, 18, 19\}$$Part c — count and divide. \(A\) has 9 outcomes and \(B\) has 6:
$$P(A) = \frac{9}{19}, \qquad P(B) = \frac{6}{19}$$Part d — intersection, then union. The values that are both even and greater than 13 are 14, 16, and 18. The union is everything in either list, each written once:
$$A \text{ AND } B = \{14, 16, 18\}$$ $$A \text{ OR } B = \{2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19\}$$Part e — count those two sets. The intersection has 3 outcomes and the union has 12:
$$P(A \text{ AND } B) = \frac{3}{19}, \qquad P(A \text{ OR } B) = \frac{12}{19}$$Part f — the complement is the odd numbers.
$$A' = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, \qquad P(A') = \frac{10}{19}$$Part g — the two must fill the sample space.
$$P(A) + P(A') = \frac{9}{19} + \frac{10}{19} = 1$$Part h — condition each way. For \(P(A \mid B)\), work inside \(B\), which has 6 outcomes; 3 of them (14, 16, 18) are even, so \(P(A \mid B) = \frac{3}{6} = \frac{1}{2}\). For \(P(B \mid A)\), work inside \(A\), which has 9 outcomes; 3 of them are greater than 13, so \(P(B \mid A) = \frac{3}{9} = \frac{1}{3}\).
No, the probabilities are not equal. The numerator is the same event both times, but the reduced sample space is different — \(B\) has 6 outcomes and \(A\) has 9.
Answer: \(P(A) = \frac{9}{19}\), \(P(B) = \frac{6}{19}\), \(P(A \text{ AND } B) = \frac{3}{19}\), \(P(A \text{ OR } B) = \frac{12}{19}\), \(P(A') = \frac{10}{19}\), \(P(A) + P(A') = 1\), \(P(A \mid B) = \frac{1}{2}\), \(P(B \mid A) = \frac{1}{3}\); not equal.
Check these probabilities in rāSHio
Open rāSHio and choose Calculators → Probability to check the answers above without redoing the arithmetic. Enter the size of the sample space and the counts you found for each event, and the panel returns the AND together with both conditional values. The point is not to skip the counting — that is the skill this section teaches — but to check the overlap before you use it in \(P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B)\), where subtracting it is the step most often missed.
Figure 3.1.2 — Checking the intersection and the conditional probabilities in rāSHio: Calculators → Probability.
A number is randomly selected from 1 to 10. Describe the sample space \(S\), identify each of the following events with a subset of \(S\), and compute its probability. An outcome is the number selected.
a. Event \(T\) = the outcome is 5.
b. Event \(A\) = the number is more than 6.
c. Event \(B\) = the number is odd.
d. The complement of \(A\).
e. \(A\) GIVEN \(B\).
f. \(B\) GIVEN \(A\).
g. \(A\) OR \(B\).
h. \(A\) OR \(B'\).
i. Event \(C\) = the outcome is a composite number.
j. Event \(D\) = the number is a multiple of 3.
Solution
Set up the sample space. \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\), ten equally likely outcomes.
Part a. \(T = \{5\}\), so \(P(T) = \frac{1}{10}\).
Part b. \(A = \{7, 8, 9, 10\}\), so \(P(A) = \frac{4}{10} = \frac{2}{5}\).
Part c. \(B = \{1, 3, 5, 7, 9\}\), so \(P(B) = \frac{5}{10} = \frac{1}{2}\).
Part d. \(A' = \{1, 2, 3, 4, 5, 6\}\), so \(P(A') = \frac{6}{10} = \frac{3}{5}\).
Part e — reduce to \(B\). Inside \(B = \{1, 3, 5, 7, 9\}\), the values more than 6 are 7 and 9, so \(A \mid B = \{7, 9\}\) and \(P(A \mid B) = \frac{2}{5}\).
Part f — reduce to \(A\). Inside \(A = \{7, 8, 9, 10\}\), the odd values are 7 and 9, so \(B \mid A = \{7, 9\}\) and \(P(B \mid A) = \frac{2}{4} = \frac{1}{2}\).
Part g — union. \(A \text{ OR } B = \{1, 3, 5, 7, 8, 9, 10\}\), so \(P(A \text{ OR } B) = \frac{7}{10}\).
Part h — union with a complement. \(B' = \{2, 4, 6, 8, 10\}\), so \(A \text{ OR } B' = \{2, 4, 6, 7, 8, 9, 10\}\) and \(P(A \text{ OR } B') = \frac{7}{10}\). Two different events can carry the same probability.
Part i — composite means more than two factors. The composite numbers from 1 to 10 are 4, 6, 8, 9, and 10 (1 is neither prime nor composite), so \(C = \{4, 6, 8, 9, 10\}\) and \(P(C) = \frac{5}{10} = \frac{1}{2}\).
Part j. \(D = \{3, 6, 9\}\), so \(P(D) = \frac{3}{10}\).
Answer: \(P(T) = \frac{1}{10}\), \(P(A) = \frac{2}{5}\), \(P(B) = \frac{1}{2}\), \(P(A') = \frac{3}{5}\), \(P(A \mid B) = \frac{2}{5}\), \(P(B \mid A) = \frac{1}{2}\), \(P(A \text{ OR } B) = \frac{7}{10}\), \(P(A \text{ OR } B') = \frac{7}{10}\), \(P(C) = \frac{1}{2}\), \(P(D) = \frac{3}{10}\).
A fair, six-sided die is rolled. Describe the sample space \(S\), identify each of the following events with a subset of \(S\), and compute its probability. An outcome is the number of dots that show up.
a. Event \(T\) = the outcome is two.
b. Event \(A\) = the outcome is an even number.
c. Event \(B\) = the outcome is less than four.
d. The complement of \(A\).
e. \(A\) GIVEN \(B\).
f. \(B\) GIVEN \(A\).
g. \(A\) AND \(B\).
h. \(A\) OR \(B\).
i. \(A\) OR \(B'\).
j. Event \(N\) = the outcome is a prime number.
k. Event \(I\) = the outcome is seven.
Solution
Set up the sample space first. \(S = \{1, 2, 3, 4, 5, 6\}\), six equally likely outcomes.
Part a. \(T = \{2\}\), so \(P(T) = \frac{1}{6}\).
Part b. \(A = \{2, 4, 6\}\), so \(P(A) = \frac{3}{6} = \frac{1}{2}\).
Part c. \(B = \{1, 2, 3\}\), so \(P(B) = \frac{3}{6} = \frac{1}{2}\).
Part d. \(A' = \{1, 3, 5\}\), so \(P(A') = \frac{3}{6} = \frac{1}{2}\).
Part e — reduce to \(B\). Inside \(B = \{1, 2, 3\}\), the only even value is 2, so \(A \mid B = \{2\}\) and \(P(A \mid B) = \frac{1}{3}\).
Part f — reduce to \(A\). Inside \(A = \{2, 4, 6\}\), the only value less than four is 2, so \(B \mid A = \{2\}\) and \(P(B \mid A) = \frac{1}{3}\).
Part g — both at once. \(A \text{ AND } B = \{2\}\), so \(P(A \text{ AND } B) = \frac{1}{6}\).
Part h — either one. \(A \text{ OR } B = \{1, 2, 3, 4, 6\}\), so \(P(A \text{ OR } B) = \frac{5}{6}\).
Part i — union with a complement. \(B' = \{4, 5, 6\}\), so \(A \text{ OR } B' = \{2, 4, 5, 6\}\) and \(P(A \text{ OR } B') = \frac{4}{6} = \frac{2}{3}\).
Part j. The primes in \(S\) are 2, 3, and 5, so \(N = \{2, 3, 5\}\) and \(P(N) = \frac{3}{6} = \frac{1}{2}\).
Part k — an impossible event. A six-sided die does not have seven dots, so \(I = \{\ \}\) and \(P(I) = 0\).
Answer: \(P(T) = \frac{1}{6}\), \(P(A) = \frac{1}{2}\), \(P(B) = \frac{1}{2}\), \(P(A') = \frac{1}{2}\), \(P(A \mid B) = \frac{1}{3}\), \(P(B \mid A) = \frac{1}{3}\), \(P(A \text{ AND } B) = \frac{1}{6}\), \(P(A \text{ OR } B) = \frac{5}{6}\), \(P(A \text{ OR } B') = \frac{2}{3}\), \(P(N) = \frac{1}{2}\), \(P(I) = 0\).
Table 3.1.1 describes the distribution of a random sample of 100 individuals, organized by gender and whether they prefer tea or coffee.
| Tea | Coffee | |
|---|---|---|
| Men | 22 | 26 |
| Women | 16 | 36 |
Let \(M\) be the event that the subject is a man, \(W\) that the subject is a woman, \(T\) that the subject likes tea, and \(C\) that the subject likes coffee. Compute the following probabilities:
a. \(P(M)\)
b. \(P(W)\)
c. \(P(T)\)
d. \(P(C)\)
e. \(P(M \text{ AND } T)\)
f. \(P(W \text{ AND } C)\)
g. \(P(M \text{ OR } W)\)
h. \(P(M \text{ OR } T)\)
i. \(P(W \text{ OR } C)\)
j. \(P(M')\)
k. \(P(T \mid M)\)
l. \(P(W \mid C)\)
m. \(P(C \mid W)\)
Solution
Totals first. Men: \(22 + 26 = 48\). Women: \(16 + 36 = 52\). Tea: \(22 + 16 = 38\). Coffee: \(26 + 36 = 62\). Grand total 100.
Parts a-d — the margins over 100. \(P(M) = 0.48\), \(P(W) = 0.52\), \(P(T) = 0.38\), \(P(C) = 0.62\).
Parts e-f — single cells. \(P(M \text{ AND } T) = \frac{22}{100} = 0.22\) and \(P(W \text{ AND } C) = \frac{36}{100} = 0.36\).
Part g — the two genders fill the table. \(P(M \text{ OR } W) = 1\).
Parts h-i — union, minus the overlap.
$$P(M \text{ OR } T) = \frac{48 + 38 - 22}{100} = \frac{64}{100} = 0.64$$ $$P(W \text{ OR } C) = \frac{52 + 62 - 36}{100} = \frac{78}{100} = 0.78$$Part j — complement. \(P(M') = 1 - 0.48 = 0.52\), which is \(P(W)\).
Parts k-m — reduce the sample space each time.
$$P(T \mid M) = \frac{22}{48} = 0.4583$$ $$P(W \mid C) = \frac{36}{62} = 0.5806$$ $$P(C \mid W) = \frac{36}{52} = 0.6923$$Answer: 0.48, 0.52, 0.38, 0.62, 0.22, 0.36, 1, 0.64, 0.78, 0.52, 0.4583, 0.5806, 0.6923 (conditional probabilities rounded to four decimal places).
Table 3.1.2 describes the distribution of a random sample of 100 individuals, organized by sex assigned at birth and whether they are right- or left-handed.
| Right-handed | Left-handed | |
|---|---|---|
| Males | 43 | 9 |
| Females | 44 | 4 |
Let \(M\) be the event that the subject is male, \(F\) that the subject is female, \(R\) that the subject is right-handed, and \(L\) that the subject is left-handed. Compute the following probabilities:
a. \(P(M)\)
b. \(P(F)\)
c. \(P(R)\)
d. \(P(L)\)
e. \(P(M \text{ AND } R)\)
f. \(P(F \text{ AND } L)\)
g. \(P(M \text{ OR } F)\)
h. \(P(M \text{ OR } R)\)
i. \(P(F \text{ OR } L)\)
j. \(P(M')\)
k. \(P(R \mid M)\)
l. \(P(F \mid L)\)
m. \(P(L \mid F)\)
Solution
Get the row and column totals first — every answer comes from them. Males: \(43 + 9 = 52\). Females: \(44 + 4 = 48\). Right-handed: \(43 + 44 = 87\). Left-handed: \(9 + 4 = 13\). Grand total 100, so every unconditional probability is just the count divided by 100.
Parts a-d — the four margins. \(P(M) = \frac{52}{100} = 0.52\), \(P(F) = \frac{48}{100} = 0.48\), \(P(R) = \frac{87}{100} = 0.87\), \(P(L) = \frac{13}{100} = 0.13\).
Parts e-f — an AND is a single cell. \(M \text{ AND } R\) is the male/right-handed cell, 43, so \(P(M \text{ AND } R) = 0.43\). \(F \text{ AND } L\) is the female/left-handed cell, 4, so \(P(F \text{ AND } L) = 0.04\).
Part g — the two sexes fill the table. Every subject is male or female, so \(P(M \text{ OR } F) = \frac{100}{100} = 1\).
Parts h-i — union, subtracting the overlap so it is not counted twice.
$$P(M \text{ OR } R) = \frac{52 + 87 - 43}{100} = \frac{96}{100} = 0.96$$ $$P(F \text{ OR } L) = \frac{48 + 13 - 4}{100} = \frac{57}{100} = 0.57$$Part j — the complement of male is female. \(P(M') = 1 - 0.52 = 0.48\).
Parts k-m — conditioning changes the denominator. For \(P(R \mid M)\), work inside the 52 males; 43 are right-handed:
$$P(R \mid M) = \frac{43}{52} = 0.8269$$For \(P(F \mid L)\), work inside the 13 left-handers; 4 are female:
$$P(F \mid L) = \frac{4}{13} = 0.3077$$For \(P(L \mid F)\), work inside the 48 females; 4 are left-handed:
$$P(L \mid F) = \frac{4}{48} = 0.0833$$Parts l and m use the same cell, 4, and get different answers because they divide by different reduced sample spaces.
Answer: 0.52, 0.48, 0.87, 0.13, 0.43, 0.04, 1, 0.96, 0.57, 0.48, 0.8269, 0.3077, 0.0833 (conditional probabilities rounded to four decimal places).
Problem Set 3.1
Problem 1. In a particular college class, there are men and women students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (You cannot find numerical answers here — you were not given enough information to find any probability values. Concentrate on understanding the symbols.) Let \(W\) be the event that a student is a woman, \(M\) that a student is a man, \(S\) that a student has short hair, and \(L\) that a student has long hair.
a) The probability that a student does not have long hair.
b) The probability that a student is a man or has short hair.
c) The probability that a student is a woman and has long hair.
d) The probability that a student is a man, given that the student has long hair.
e) The probability that a student has long hair, given that the student is a man.
f) Of all the women students, the probability that a student has short hair.
g) Of all students with long hair, the probability that a student is a woman.
h) The probability that a student is a woman or has long hair.
i) The probability that a randomly selected student is a man with short hair.
j) The probability that a student is a woman.
Problems 3.1.2 through 3.1.5: A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. One party favor is chosen from the box at random. Let \(H\) be the event of getting a hat, \(N\) the event of getting a noisemaker, \(F\) the event of getting a finger trap, and \(C\) the event of getting a bag of confetti.
Solution
Step 1 — read the four letters as a two-way split. Every student is a woman \(W\) or a man \(M\), and separately has short hair \(S\) or long hair \(L\). Because those two pairs are complements of each other, \(M = W'\) and \(S = L'\).
Step 2 — translate the key phrases. "Does not have" is a complement. "And" is an intersection, and both categories are named. "Given that" is a condition, and so is "of all the ..." — both name the reduced sample space, which becomes the second slot in the notation.
Step 3 — write each symbol.
a) \(P(L')\)
b) \(P(M \text{ OR } S)\)
c) \(P(W \text{ AND } L)\)
d) \(P(M \mid L)\)
e) \(P(L \mid M)\)
f) \(P(S \mid W)\) — "of all the women students" is the condition.
g) \(P(W \mid L)\) — "of all students with long hair" is the condition.
h) \(P(W \text{ OR } L)\)
i) \(P(M \text{ AND } S)\)
j) \(P(W)\)
Notice that parts d and e use the same two events in opposite order, and so do f and g. Order matters in a conditional probability.
Answer: a) \(P(L')\) b) \(P(M \text{ OR } S)\) c) \(P(W \text{ AND } L)\) d) \(P(M \mid L)\) e) \(P(L \mid M)\) f) \(P(S \mid W)\) g) \(P(W \mid L)\) h) \(P(W \text{ OR } L)\) i) \(P(M \text{ AND } S)\) j) \(P(W)\)
Problem 2. Find \(P(H)\).
Solution
Step 1 — find the size of the sample space. Add every favor in the box:
$$12 + 15 + 10 + 5 = 42$$Step 2 — count the favorable outcomes and divide. There are 12 hats.
$$P(H) = \frac{12}{42} = \frac{2}{7} \approx 0.29$$Answer: \(P(H) = \frac{2}{7} \approx 0.29\)
Problem 3. Find \(P(N)\).
Solution
Step 1 — the sample space is still all 42 favors.
Step 2 — count the noisemakers and divide. There are 15.
$$P(N) = \frac{15}{42} = \frac{5}{14} \approx 0.36$$Answer: \(P(N) = \frac{5}{14} \approx 0.36\)
Problem 4. Find \(P(F)\).
Solution
Step 1 — count the finger traps. There are 10 of them out of 42 favors.
Step 2 — divide and reduce.
$$P(F) = \frac{10}{42} = \frac{5}{21} \approx 0.24$$Answer: \(P(F) = \frac{5}{21} \approx 0.24\)
Problem 5. Find \(P(C)\).
Problems 3.1.6 through 3.1.11: A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. One jelly bean is chosen from the jar at random. Let \(B\) be the event of getting a blue jelly bean, \(G\) green, \(O\) orange, \(P\) purple, \(R\) red, and \(Y\) yellow.
Solution
Step 1 — count the bags of confetti. There are 5 out of 42.
Step 2 — divide.
$$P(C) = \frac{5}{42} \approx 0.12$$As a check, the four probabilities should sum to 1: \(\frac{12 + 15 + 10 + 5}{42} = \frac{42}{42} = 1\).
Answer: \(P(C) = \frac{5}{42} \approx 0.12\)
Problem 6. Find \(P(B)\).
Solution
Step 1 — the sample space is all 150 jelly beans.
Step 2 — count the blue beans and divide. There are 26.
$$P(B) = \frac{26}{150} = \frac{13}{75} \approx 0.17$$Answer: \(P(B) = \frac{13}{75} \approx 0.17\)
Problem 7. Find \(P(G)\).
Solution
Step 1 — count the green beans. There are 20 out of 150.
Step 2 — divide and reduce.
$$P(G) = \frac{20}{150} = \frac{2}{15} \approx 0.13$$Answer: \(P(G) = \frac{2}{15} \approx 0.13\)
Problem 8. Find \(P(Y)\).
Solution
Step 1 — count the yellow beans. There are 38 out of 150.
Step 2 — divide and reduce.
$$P(Y) = \frac{38}{150} = \frac{19}{75} \approx 0.25$$Answer: \(P(Y) = \frac{19}{75} \approx 0.25\)
Problem 9. Find \(P(R)\).
Solution
Step 1 — count the red beans. There are 22 out of 150.
Step 2 — divide and reduce.
$$P(R) = \frac{22}{150} = \frac{11}{75} \approx 0.15$$Answer: \(P(R) = \frac{11}{75} \approx 0.15\)
Problem 10. Find \(P(P)\).
Solution
Step 1 — count the purple beans. There are 28 out of 150.
Step 2 — divide and reduce.
$$P(P) = \frac{28}{150} = \frac{14}{75} \approx 0.19$$Answer: \(P(P) = \frac{14}{75} \approx 0.19\)
Problem 11. Find \(P(O)\).
Problems 3.1.12 through 3.1.17: There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (the Pacific Ocean region). One country is chosen at random. Let \(A\) be the event that a country is in Asia, \(E\) that it is in Europe, \(F\) that it is in Africa, \(N\) that it is in North America, \(O\) that it is in Oceania, and \(S\) that it is in South America.
Solution
Step 1 — the orange count is not given, so recover it by subtraction. The five stated colors account for
$$22 + 38 + 20 + 28 + 26 = 134$$so the orange beans number \(150 - 134 = 16\).
Step 2 — divide and reduce.
$$P(O) = \frac{16}{150} = \frac{8}{75} \approx 0.11$$Answer: \(P(O) = \frac{8}{75} \approx 0.11\)
Problem 12. Find \(P(A)\).
Solution
Step 1 — find the total number of countries. This total is the sample space for every problem in this group:
$$23 + 12 + 47 + 44 + 54 + 14 = 194$$Step 2 — count the countries in Asia and divide. There are 44.
$$P(A) = \frac{44}{194} = \frac{22}{97} \approx 0.23$$Answer: \(P(A) = \frac{22}{97} \approx 0.23\)
Problem 13. Find \(P(E)\).
Solution
Step 1 — the sample space is all 194 countries.
Step 2 — count the countries in Europe and divide. There are 47, and 47 is prime, so the fraction does not reduce.
$$P(E) = \frac{47}{194} \approx 0.24$$Answer: \(P(E) = \frac{47}{194} \approx 0.24\)
Problem 14. Find \(P(F)\).
Solution
Step 1 — count the countries in Africa. There are 54 out of 194.
Step 2 — divide and reduce by 2.
$$P(F) = \frac{54}{194} = \frac{27}{97} \approx 0.28$$Answer: \(P(F) = \frac{27}{97} \approx 0.28\)
Problem 15. Find \(P(N)\).
Solution
Step 1 — count the countries in North America. There are 23 out of 194.
Step 2 — divide. 23 is prime and does not divide 194, so the fraction stands.
$$P(N) = \frac{23}{194} \approx 0.12$$Answer: \(P(N) = \frac{23}{194} \approx 0.12\)
Problem 16. Find \(P(O)\).
Solution
Step 1 — count the countries in Oceania. There are 14 out of 194.
Step 2 — divide and reduce by 2.
$$P(O) = \frac{14}{194} = \frac{7}{97} \approx 0.07$$Answer: \(P(O) = \frac{7}{97} \approx 0.07\)
Problem 17. Find \(P(S)\).
Solution
Step 1 — count the countries in South America. There are 12 out of 194.
Step 2 — divide and reduce by 2.
$$P(S) = \frac{12}{194} = \frac{6}{97} \approx 0.06$$As a check, the six probabilities add to \(\frac{44 + 47 + 54 + 23 + 14 + 12}{194} = \frac{194}{194} = 1\).
Answer: \(P(S) = \frac{6}{97} \approx 0.06\)
Problem 18. What is the probability of drawing a red card in a standard deck of 52 cards?
Solution
Step 1 — count the red cards. A standard deck has four suits of 13 cards each. Two of those suits, hearts and diamonds, are red:
$$2 \times 13 = 26$$Step 2 — divide by the size of the deck.
$$P(\text{red}) = \frac{26}{52} = \frac{1}{2} = 0.5$$Answer: \(\frac{1}{2} = 0.5\)
Problem 19. What is the probability of drawing a club in a standard deck of 52 cards?
Solution
Step 1 — count the clubs. Clubs are one of the four suits, so there are 13 of them.
Step 2 — divide by 52.
$$P(\text{club}) = \frac{13}{52} = \frac{1}{4} = 0.25$$Answer: \(\frac{1}{4} = 0.25\)
Problem 20. What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?
Solution
Step 1 — list the sample space and the event. \(S = \{1, 2, 3, 4, 5, 6\}\), and the even faces are \(\{2, 4, 6\}\).
Step 2 — count and divide. Three of the six faces qualify.
$$P(\text{even}) = \frac{3}{6} = \frac{1}{2} = 0.5$$Answer: \(\frac{1}{2} = 0.5\)
Problem 21. What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?
Problems 3.1.22 and 3.1.23: Mai sees a game at a local fair. She has to throw a dart at a color wheel. Each section on the color wheel is equal in area. The wheel is divided into eight equal sections: four red, two yellow, one green, and one blue. Let \(B\) be the event of landing on blue, \(R\) the event of landing on red, \(G\) the event of landing on green, and \(Y\) the event of landing on yellow.
Solution
Step 1 — identify the primes on a die. A prime number has exactly two distinct factors, itself and 1. Among 1 through 6 the primes are 2, 3, and 5. Note that 1 is not prime, since it has only one factor.
Step 2 — count and divide. Three of the six faces qualify.
$$P(\text{prime}) = \frac{3}{6} = \frac{1}{2} = 0.5$$Answer: \(\frac{1}{2} = 0.5\)
Problem 22. If her dart lands on blue, she gets the biggest prize. Find \(P(B)\).
Solution
Step 1 — the sections are equal in area, so the outcomes are equally likely. That means we can count sections instead of measuring area. The wheel has 8 sections in all.
Step 2 — count the blue sections and divide. There is exactly one blue section, which is why it carries the biggest prize.
$$P(B) = \frac{1}{8} = 0.125$$Answer: \(P(B) = \frac{1}{8} = 0.125\)
Problem 23. If her dart lands on red, she doesn't get a prize. What is \(P(R)\)?
Problems 3.1.24 through 3.1.33: On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let \(I\) be the event that a player is an infielder, \(O\) that a player is an outfielder, \(H\) that a player is a great hitter, and \(N\) that a player is not a great hitter.
Solution
Step 1 — count the red sections. Four of the eight sections are red.
Step 2 — divide and reduce.
$$P(R) = \frac{4}{8} = \frac{1}{2} = 0.5$$Half the wheel wins Mai nothing, which is how the game makes money.
Answer: \(P(R) = \frac{1}{2} = 0.5\)
Problem 24. Write the symbols for the probability that a player is not an outfielder.
Solution
Step 1 — "not an outfielder" is a complement. The complement of \(O\) is written \(O'\).
Answer: \(P(O')\)
Problem 25. Write the symbols for the probability that a player is an outfielder or is a great hitter.
Solution
Step 1 — "or" is a union. Both categories are named without a condition, so this is an unconditional OR event.
Answer: \(P(O \text{ OR } H)\)
Problem 26. Write the symbols for the probability that a player is an infielder and is not a great hitter.
Solution
Step 1 — "and" is an intersection. The player must be both an infielder and not a great hitter at the same time.
Answer: \(P(I \text{ AND } N)\)
Problem 27. Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.
Solution
Step 1 — "given that" names the condition. The condition goes in the second slot, after the bar. Here the condition is being an infielder.
Answer: \(P(H \mid I)\)
Problem 28. Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.
Solution
Step 1 — the condition has switched. Now we are told the player is a great hitter, so \(H\) goes after the bar and \(I\) goes before it. Compare this with the previous problem: same two events, opposite order, different probability.
Answer: \(P(I \mid H)\)
Problem 29. Write the symbols for the probability that of all the outfielders, a player is not a great hitter.
Solution
Step 1 — "of all the outfielders" is a condition in disguise. It tells you the sample space has been reduced to outfielders, so \(O\) sits after the bar.
Answer: \(P(N \mid O)\)
Problem 30. Write the symbols for the probability that of all the great hitters, a player is an outfielder.
Solution
Step 1 — "of all the great hitters" is the condition. The reduced sample space is the great hitters, so \(H\) goes after the bar.
Answer: \(P(O \mid H)\)
Problem 31. Write the symbols for the probability that a player is an infielder or is not a great hitter.
Solution
Step 1 — "or" is a union, and "is not a great hitter" is the event \(N\). No condition is stated.
Answer: \(P(I \text{ OR } N)\)
Problem 32. Write the symbols for the probability that a player is an outfielder and is a great hitter.
Solution
Step 1 — "and" is an intersection. The player is an outfielder and a great hitter at the same time.
Answer: \(P(O \text{ AND } H)\)
Problem 33. Write the symbols for the probability that a player is an infielder.
Solution
Step 1 — a single event, no condition and no combination.
Answer: \(P(I)\)
Problem 34. What is the word for the set of all possible outcomes?
Solution
Step 1 — recall the vocabulary. The set of every possible outcome of an experiment is the sample space, written \(S\). An individual member of that set is an outcome, and any subset of it is an event.
Answer: The sample space.
Problem 35. What is conditional probability?
Solution
Step 1 — state what the notation means. A conditional probability \(P(A \mid B)\) is the likelihood that event \(A\) will occur given that event \(B\) has already occurred.
Step 2 — say why it differs from \(P(A)\). Conditioning reduces the sample space: instead of counting over all of \(S\), we count only within \(B\). That is why \(P(A \mid B)\) and \(P(A)\) are usually different numbers.
Answer: The likelihood that an event will occur given that another event has already occurred, computed on the reduced sample space of that second event.
Problem 36. A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. You randomly select one book. Let \(F\) be the event that the book is fiction and \(N\) the event that the book is nonfiction. What is the sample space?
Solution
Step 1 — decide what a single outcome is. One book is selected, so an outcome is one particular book. Every book has a unique title, so the eight fiction and four nonfiction books are twelve distinct outcomes.
Step 2 — list them using the numbering given. Writing \(F1\) for fiction book one and \(N1\) for nonfiction book one:
$$S = \{F1, F2, F3, F4, F5, F6, F7, F8, N1, N2, N3, N4\}$$Step 3 — check the size. That is \(8 + 4 = 12\) outcomes, one per book on the shelf.
Answer: \(S = \{F1, F2, F3, F4, F5, F6, F7, F8, N1, N2, N3, N4\}\), twelve equally likely outcomes.
Problem 37. What is the sum of the probabilities of an event and its complement?
Problems 3.1.38 and 3.1.39: You are rolling a fair, six-sided number cube. Let \(E\) be the event that it lands on an even number and \(M\) the event that it lands on a multiple of three.
Solution
Step 1 — an event and its complement cover the whole sample space. Every outcome is either in \(A\) or in \(A'\), and no outcome is in both.
Step 2 — so their probabilities fill the whole of 1.
$$P(A) + P(A') = 1$$Answer: 1
Problem 38. What does \(P(E \mid M)\) mean in words?
Solution
Step 1 — identify the two events. \(E\) is landing on an even number and \(M\) is landing on a multiple of three. The bar means "given".
Step 2 — put the condition into a sentence. The event after the bar has already happened, so we are told the roll was a multiple of three and asked how likely it is that it was also even.
Answer: \(P(E \mid M)\) is the probability that the cube lands on an even number, given that it has landed on a multiple of three.
Problem 39. What does \(P(E \text{ OR } M)\) mean in words?
Solution
Step 1 — OR is a union, and in mathematics it includes the overlap. So the event happens when the roll is even, or a multiple of three, or both.
Step 2 — write it as a sentence.
Answer: \(P(E \text{ OR } M)\) is the probability that the cube lands on an even number or on a multiple of three, including the rolls that are both.
Problem 40. A polling firm asked 1,045 residents of a city whether they approve of the mayor's actions in office. Table 3.1.3 gives the sample size and the percent who approve for each age group and for each gender group. Each grouping covers all 1,045 respondents.
| Group | Sample size | Percent who approve |
|---|---|---|
| Age 18-34 | 320 | 42 |
| Age 35-44 | 268 | 36 |
| Age 45-54 | 236 | 32 |
| Age 55 and over | 221 | 26 |
| Men | 501 | 40 |
| Women | 544 | 30 |
a) Define three events from the table.
b) Describe in words what the entry 40 means.
c) Describe in words the complement of the entry in part b.
d) Describe in words what the entry 30 means.
e) Out of the men and women, what percent are men?
f) Out of the women, what percent disapprove of the mayor?
g) Out of all the age groups, what percent approve of the mayor?
h) Find \(P(\text{Approve} \mid \text{Men})\).
i) Out of the age groups, what percent are more than 44 years old?
j) Find \(P(\text{Approve} \mid \text{Age} < 35)\).
Solution
Part a — name three events. Any three subsets of the 1,045 respondents work. For example: \(A\) = the respondent approves of the mayor; \(M\) = the respondent is a man; \(Y\) = the respondent is between 18 and 34 years old.
Part b — read the row the 40 sits in. The 40 is in the Men row of the "percent who approve" column, so it means that 40% of the 501 men surveyed approve of the mayor's actions.
Part c — the complement of that percentage. The men who do not approve are everyone else in that row:
$$100\% - 40\% = 60\%$$So 60% of the men surveyed do not approve.
Part d — same reading, Women row. 30% of the 544 women surveyed approve of the mayor.
Part e — men as a share of all respondents.
$$\frac{501}{1{,}045} \approx 0.479 = 47.9\%$$Part f — disapproval among women is the complement of approval. \(100\% - 30\% = 70\%\).
Part g — a weighted total across the four age groups. Multiply each group's size by its approval rate to get the number of approvers, add, then divide by 1,045:
$$(320)(0.42) + (268)(0.36) + (236)(0.32) + (221)(0.26) = 134.4 + 96.5 + 75.5 + 57.5 = 363.9$$ $$\frac{363.9}{1{,}045} \approx 0.348 = 34.8\%$$Do not average the four percentages directly — the groups are different sizes, so the larger groups have to count for more.
Part h — this is read straight off the table. The Men row gives the approval rate among men, which is exactly the conditional probability:
$$P(\text{Approve} \mid \text{Men}) = 0.40$$Part i — add the two oldest groups. More than 44 years old covers the 45-54 group and the 55-and-over group:
$$\frac{236 + 221}{1{,}045} = \frac{457}{1{,}045} \approx 0.437 = 43.7\%$$Part j — the 18-34 group is exactly "age under 35".
$$P(\text{Approve} \mid \text{Age} < 35) = 0.42$$Answer: b) 40% of the men approve. c) 60% of the men do not approve. d) 30% of the women approve. e) about 47.9%. f) 70%. g) about 34.8%. h) 0.40. i) about 43.7%. j) 0.42.
Problem 41. Explain what is wrong with the following statements. Use complete sentences.
a) If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend.
b) The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.
Solution
Part a — three things are wrong with adding the two chances.
First, a probability can never exceed 100%, so 130% is impossible on its face and should have stopped the writer immediately.
Second, rain on Saturday and rain on Sunday are not two outcomes of one experiment, so there is no rule that lets you add them. If you want the probability of rain at some point over the weekend, you need the union, and the union formula subtracts the overlap so it is not counted twice:
$$P(\text{Sat OR Sun}) = P(\text{Sat}) + P(\text{Sun}) - P(\text{Sat AND Sun})$$Third, the probability that it rains on both days is not given, and it cannot be recovered from the two individual chances alone — that would need the two days to be independent, which nobody has claimed.
Part b — one event contains the other. A home run is by definition a successful hit, so every home run is also a hit. That makes the home-run event a subset of the hit event, and a subset can never be more likely than the set that contains it. The player must have at least as many successful hits as home runs, so
$$P(\text{home run}) \le P(\text{hit})$$Answer: a) Probability cannot exceed 100%; the two chances belong to different days and cannot simply be added, and a correct union subtracts the overlap, which was never given. b) Every home run is a successful hit, so the home-run event is contained in the hit event and its probability can be at most as large, never greater.
Key Terms
experiment — a planned operation carried out under controlled conditions; a chance experiment is one whose result is not predetermined.
outcome — a single result of an experiment.
sample space — the set of all possible outcomes of an experiment, denoted \(S\).
event — any combination of outcomes; a subset of the sample space.
probability — the long-term relative frequency of an outcome, a number between 0 and 1 inclusive.
equally likely — the condition that every outcome of an experiment occurs with the same probability.
law of large numbers — as the number of repetitions increases, the observed relative frequency approaches the theoretical probability.
biased — describing a coin, die, or other device whose outcomes are not equally likely.
union (OR event) — \(A \cup B\), the event that the outcome is in \(A\), in \(B\), or in both.
intersection (AND event) — \(A \cap B\), the event that the outcome is in both \(A\) and \(B\) at the same time.
complement — \(A'\), the event that the outcome is not in \(A\); \(P(A) + P(A') = 1\).
conditional probability — \(P(A \mid B)\), the probability of \(A\) computed on the reduced sample space \(B\).