03 Probability

A progressive guide to modeling uncertainty and solving probability problems with events, counting methods, conditional probability, independence, and Bayes’ theorem.

Experiments and models

is a mathematical language for uncertainty. It assigns each a value from 00 to 11: values near 00 indicate that an is unlikely, and values near 11 indicate that it is likely. The same can be expressed as a percentage; for example, 0.25=25%0.25=25\%.

A experiment is a process with an uncertain result. A single possible result is an outcome. The is the complete set of possible outcomes and is denoted by SS. For a fair six-sided die,

S={1,2,3,4,5,6}.S=\{1,2,3,4,5,6\}.

An is a subset of the . If AA is the that an even number is rolled, then

A={2,4,6}.A=\{2,4,6\}.

When all outcomes are equally likely, the of an is

P(A)=number of outcomes in Anumber of outcomes in S.P(A)=\frac{\text{number of outcomes in }A}{\text{number of outcomes in }S}.

For the even-number , P(A)=36=12P(A)=\frac{3}{6}=\frac{1}{2}. This favorable-outcomes ratio should be used only when the outcomes are equally likely.

Takeaway: Start every problem by identifying the uncertain experiment, its outcomes, and the of interest.

rules and complements

A valid model follows three basic rules:

  1. 0≤P(A)≤10\le P(A)\le 1 for every AA.

  2. P(S)=1P(S)=1, because some outcome in the must occur.

  3. If events cannot occur together, their probabilities add:

P(A∪B)=P(A)+P(B).P(A\cup B)=P(A)+P(B).

The of AA, written AcA^c or A′A', contains every outcome that is not in AA. The rule is

P(Ac)=1−P(A).P(A^c)=1-P(A).

For example, if the that a package arrives late is 0.080.08, then the that it does not arrive late is

P(Ac)=1−0.08=0.92.P(A^c)=1-0.08=0.92.

This rule is often the quickest method for “not” questions.

Takeaway: Check that every is between 00 and 11, and use the rule when the desired is the opposite of a known .

Combining events

The intersection of AA and BB, written A∩BA\cap B, means that both events occur. The of AA and BB, written A∪BA\cup B, means that at least one occurs. The general addition rule is

P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

The intersection is subtracted because outcomes in both events would otherwise be counted twice.

For a six-sided die, let A={2,4,6}A=\{2,4,6\} be the of rolling an even number, and let B={4,5,6}B=\{4,5,6\} be the of rolling a number greater than 33. Then A∩B={4,6}A\cap B=\{4,6\}, so

P(A∪B)=36+36−26=46=23.P(A\cup B)=\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}=\frac{2}{3}.

Mutually exclusive events cannot occur together. For these events, P(A∩B)=0P(A\cap B)=0, so the addition rule becomes

P(A∪B)=P(A)+P(B).P(A\cup B)=P(A)+P(B).

Mutual exclusivity is not the same as independence. Two nonzero- mutually exclusive events cannot be , because the occurrence of one makes the other impossible.

Takeaway: For “or,” use the addition rule and subtract the overlap unless the events are mutually exclusive.

Counting outcomes

Counting methods determine how many outcomes are possible without listing them all.

The multiplication principle states that if a process has mm choices for its first step and nn choices for its second step, then it has

m×nm\times n

possible outcomes. For a password with one letter followed by two digits, when repetition is allowed, the number of passwords is

26×10×10=2,600.26\times 10\times 10=2{,}600.

A is used when order matters. The number of ways to select and arrange rr objects from nn distinct objects is

nPr=n!(n−r)!.{}_nP_r=\frac{n!}{(n-r)!}.

For example, choosing a president and vice president from 1010 people gives

10P2=10×9=90.{}_{10}P_2=10\times 9=90.

A is used when order does not matter. The number of ways to choose rr objects from nn distinct objects is

nCr=(nr)=n!r!(n−r)!.{}_nC_r=\binom{n}{r}=\frac{n!}{r!(n-r)!}.

A three-person committee chosen from 1010 people can be formed in

(103)=10!3!7!=120\binom{10}{3}=\frac{10!}{3!7!}=120

ways.

Decision rule: Use a for ordered roles or rankings. Use a for an unordered group or selection.

restricts attention to outcomes in which a specified condition has already occurred. The notation P(A∣B)P(A\mid B) means the of AA given BB. When P(B)>0P(B)>0,

P(A∣B)=P(A∩B)P(B).P(A\mid B)=\frac{P(A\cap B)}{P(B)}.

Rearranging produces the multiplication rule:

P(A∩B)=P(A∣B)P(B).P(A\cap B)=P(A\mid B)P(B).

It can also be written as

P(A∩B)=P(B∣A)P(A).P(A\cap B)=P(B\mid A)P(A).

Suppose a fair die is rolled. Let AA be the that the result is 22 or 33, and let B={2,4,6}B=\{2,4,6\} be the that the result is even. Once BB is known, only three outcomes remain possible, and just one of them belongs to AA. Therefore,

P(A∣B)=13.P(A\mid B)=\frac{1}{3}.

The same result follows from the formula:

P(A∣B)=P(A∩B)P(B)=1/63/6=13.P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/6}{3/6}=\frac{1}{3}.

In a two-way table, the condition determines the denominator. If a group contains 5252 people, a calculated within that group uses 5252, not the total number of people, as its denominator.

Takeaway: For “given” questions, restrict the to the condition and use the conditional as the denominator.

Independence and dependence

Two events are when knowing that one occurred does not change the of the other. Equivalent conditions, when the probabilities are defined, include

P(A∣B)=P(A)P(A\mid B)=P(A)

and

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

For events, the multiplication rule becomes

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

For example, tossing a fair coin and rolling a fair die are experiments. The of heads and a 66 is

P(heads and 6)=P(heads)P(6)=12⋅16=112.P(\text{heads and }6)=P(\text{heads})P(6)=\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}.

Independence differs from mutual exclusivity. On one die roll, rolling a 11 and rolling a 66 are mutually exclusive, because both cannot occur at once. They are not : knowing that a 11 occurred makes a 66 impossible.

Sampling with replacement is often because the population composition stays the same. Sampling without replacement is generally dependent because each selection changes the probabilities for later selections.

Takeaway: Independence means “no change in ,” while mutual exclusivity means “cannot occur together.”

and evidence

reverses the direction of a . When P(B)>0P(B)>0, it states

P(A∣B)=P(B∣A)P(A)P(B).P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}.

If AA and AcA^c partition the , the denominator can be expanded using the law of total :

P(B)=P(B∣A)P(A)+P(B∣Ac)P(Ac).P(B)=P(B\mid A)P(A)+P(B\mid A^c)P(A^c).

Combining these expressions gives

P(A∣B)=P(B∣A)P(A)P(B∣A)P(A)+P(B∣Ac)P(Ac).P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^c)P(A^c)}.

Consider a condition-testing example. Suppose P(D)=0.01P(D)=0.01, P(+∣D)=0.95P(+\mid D)=0.95, and P(+∣Dc)=0.05P(+\mid D^c)=0.05. First find the overall of a positive result:

P(+)=P(+∣D)P(D)+P(+∣Dc)P(Dc)P(+)=P(+\mid D)P(D)+P(+\mid D^c)P(D^c)
=(0.95)(0.01)+(0.05)(0.99)=0.059.=(0.95)(0.01)+(0.05)(0.99)=0.059.

Then

P(D∣+)=(0.95)(0.01)0.059≈0.161.P(D\mid +)=\frac{(0.95)(0.01)}{0.059}\approx 0.161.

Thus, a positive result implies approximately a 16.1%16.1\% of the condition in this population. The prior and the false-positive rate matter, so P(D∣+)P(D\mid +) is not generally equal to P(+∣D)P(+\mid D).

Takeaway: combines prior with new evidence and prevents the common mistake of reversing a without adjustment.

A practical problem-solving method

A reliable solution process keeps the definitions and denominators explicit.

  1. Define the experiment and identify what is uncertain.

  2. Specify the or the relevant population.

  3. Define the events with clear notation.

  4. Decide whether the outcomes are equally likely.

  5. Translate key words:

    • “not” suggests a ;

    • “or” suggests a and the addition rule;

    • “and” suggests an intersection and a multiplication rule;

    • “given” signals .

  6. Choose a method:

    • use permutations when order matters;

    • use combinations when order does not matter;

    • use when reversing a .

  7. Check that the result lies between 00 and 11 and fits the context.

For a final check, ask whether the denominator matches the population or restricted group being described. Also check whether overlap has been subtracted in an addition problem and whether independence has actually been established before multiplying probabilities directly.

Final takeaway: Model the situation first, select the rule that matches the wording, show the relevant denominator, and test whether the result is mathematically and contextually reasonable.