Probability · Lesson 5 of 5

Chapter Summary and Practice

“Connect the chapter’s probability methods and use them to solve mixed problems with explicit assumptions.”

Learning Objectives

• Connect conditional, joint, independent, total, and reverse probabilities. • Choose an appropriate sample space, partition, or sequence before calculating. • Reconstruct the principal formulas and explain their conditions. • Solve mixed counting, reliability, repeated-trial, and Bayes problems. • Check consistency, interpret results, and recognise common reasoning errors.

One probability model, several useful viewpoints

The methods in this chapter all organise the same underlying probability weights. Conditioning focuses on the part consistent with known information. Multiplication follows a required sequence, total probability combines disjoint routes, and Bayes’ rule compares those routes after an observation. Choosing the right viewpoint makes a complicated story manageable.

Begin with the experiment rather than with a formula. Identify what is randomly selected, which elementary outcomes are equally likely, and what information is given. Name the events. Then decide whether you need a restricted fraction, a joint event, a union, a forward mixture, or a reverse conditional probability.

Question or idea	Useful relationship	What to check
An event after known information	$P (E ∣ F) = P (E \cap F) / P (F)$	The condition has positive probability.
Two required events	$P (E \cap F) = P (E) P (F ∣ E)$	Update the second probability when the first changes the experiment.
Independence	$P (E \cap F) = P (E) P (F)$	Establish the product relationship; do not assume it.
A conditional complement	$P (E^{'} ∣ F) = 1 - P (E ∣ F)$	Keep the same condition.
At least one independent event	$1 - P (E^{'}) P (F^{'})$	Its complement is neither.
All routes to evidence	$P (A) = \sum_{i} P (E_{i}) P (A ∣ E_{i})$	Use a disjoint exhaustive partition.
A case given evidence	$P (E_{i} ∣ A) = P (E_{i}) P (A ∣ E_{i}) / P (A)$	Include every evidence source in the denominator.
A numerical description of an outcome	A random variable maps each outcome to a real number.	Different outcomes may have the same value.

Reconstruct the rules instead of guessing them

A formula is easier to remember when its pieces have meanings. In a conditional fraction, the numerator is the part satisfying both requirements and the denominator is everything consistent with the condition. Multiplying back by that denominator gives the joint probability rule.

Independence is the case in which the conditional factor is unchanged. A partition then lets us split any target event into disjoint intersections and add their joint probabilities. Dividing one such intersection by the total target probability reverses the conditioning and gives Bayes’ rule. This chain explains why these formulas belong to one chapter.

The central connectionLaTeX

P (A \cap E_{i}) P (A) P (E_{i} ∣ A) = P (E_{i}) P (A ∣ E_{i}), = j \sum P (E_{j}) P (A ∣ E_{j}), = \frac{P ( E _{i} ) P ( A ∣ E _{i} )}{P ( A )}

The cases

E_{i}

form a positive-probability partition and

P (A) > 0

. The same joint contribution appears in all three lines.

Remember also the conditional addition rule: add the two conditional probabilities and subtract their conditional overlap. For three mutually independent events, all three pairwise products and the three-way product must hold. Returning a ball and remixing preserves the composition; removing it changes the next conditional probability. These distinctions matter more than surface similarities between stories.

Example — Similar family questions with different conditions

Problem
Under the equally likely ordered boy/girl model, compare the probability of two boys given at least one boy with the probability of two girls given that the elder child is a girl.

1.For the first question, $bb, b g, g b$ remain. Only $bb$ satisfies the target, giving $1/3$ .
2.For the second, $g b, g g$ remain when the elder child is written first. Only $g g$ satisfies the target, giving $1/2$ .
3.The difference comes from the information, not from a change in the underlying four-outcome model. List the conditioned outcomes whenever a verbal description is ambiguous.

Example — Replacement changes a joint probability

Problem
A bag contains five red and three blue balls. Find the probability of two red draws with replacement and without replacement.

1.With replacement and remixing, the red probability is $5/8$ on each draw. Independence gives $(5/8)^{2} = 25/64$ .
2.Without replacement, a first red leaves four red among seven balls. Multiplication gives $(5/8) (4/7) = 5/14$ .
3.Both calculations use multiplication. What changes is whether the second factor remains $5/8$ or becomes the conditional value $4/7$ .

Example — Forward and reverse in the same model

Problem
Two sources have priors $0.7, 0.3$ and evidence likelihoods $0.1, 0.4$ . Find the evidence probability and then the probability of the second source given the evidence.

1.The joint contributions are $0.7 (0.1) = 0.07$ and $0.3 (0.4) = 0.12$ .
2.The evidence probability is the sum, $0.19$ . This is a forward total-probability calculation.
3.The reverse answer is the second contribution divided by that total: $0.12/0.19 = 12/19$ . The posterior exceeds the prior because the evidence is more likely under the second source.

Counting arrangements in independent repetitions

Some mixed problems ask how many successes occur in a fixed number of trials. The chapter’s multiplication and addition rules are enough to construct the needed expression. Count the arrangements of successes first, then assign the probability of each arrangement.

If each of $n$ independent trials has success probability $p$ , a particular arrangement with $r$ successes and $n - r$ failures has probability $p^{r} (1 - p)^{n - r}$ . There are $(r n)$ ways to select the success positions. These arrangements are disjoint, so their probabilities add. This is the counting pattern needed for the chapter’s mixed repeated-trial question; no mean or variance formulas are needed.

A fixed number of successesLaTeX

P (exactly r successes) = (r n) p^{r} (1 - p)^{n - r}

Here

(r n) = n! / [r! (n - r)!]

counts the choices of success positions. Trials must have the same success probability and be mutually independent.

Example — At most six right-handed people

Problem
Model ten randomly sampled people as independent, each right-handed with probability $0.9$ . Find the probability that at most six are right-handed.

1.Let $X$ count right-handed people. For a fixed count $r$ , choose the $r$ positions and multiply their success/failure probabilities, giving $P (X = r) = (r 10) (0.9)^{r} (0.1)^{10 - r}$ .
2.At most six includes the seven disjoint counts $r = 0, 1, 2, 3, 4, 5, 6$ . Add their probabilities: $P (X \leq 6) = \sum_{r = 0}^{6} (r 10) (0.9)^{r} (0.1)^{10 - r}$ .
3.The sum is $0.0127951984$ , about $1.28%$ . A single arrangement with six right-handed people is only one part of the event; omitting its arrangement count or omitting counts below six would undercount.

Bring probability into another mathematical setting

An event can be defined by an algebraic condition, a calendar pattern, or a system failure. Translate that condition into outcomes before calculating. Cross-topic wording does not require a new probability rule.

Example — A positive determinant

Problem
Each entry of a two-by-two determinant is independently $0$ or $1$ , with probability $1/2$ for each. Find the probability that the determinant is positive.

1.Write the determinant as $a d - b c$ . Both products are either $0$ or $1$ . Positivity therefore requires $a d = 1$ and $b c = 0$ .
2.The first condition means $a = d = 1$ , with probability $1/4$ . The second excludes only $b = c = 1$ , so it has probability $3/4$ .
3.These conditions involve separate independent entries, so their joint probability is $(1/4) (3/4) = 3/16$ . Equivalently, three of the sixteen equally likely entry arrangements work.

Example — Two subsystem failures

Problem
Subsystem A fails with probability $0.2$ , B fails alone with probability $0.15$ , and both fail with probability $0.15$ . Find the probability A has failed given B has failed, and the probability A fails alone.

1.Let $A, B$ now name failure events. B fails through disjoint possibilities B alone or both, so $P (B) = 0.15 + 0.15 = 0.30$ .
2.Conditioning gives $P (A ∣ B) = P (A \cap B) / P (B) = 0.15/0.30 = 0.5$ .
3.A alone is the part of A outside B: $P (A \cap B^{'}) = P (A) - P (A \cap B) = 0.20 - 0.15 = 0.05$ . No independence assumption was supplied or needed.

Check the model as well as the arithmetic

A numerical answer between zero and one is necessary, but it does not guarantee a correct model. Verify that the listed routes cover the evidence, that joint probabilities do not exceed their marginals, and that every denominator matches its condition. Test whether the answer moves in a sensible direction after evidence.

Probability one and set containment

From $P (B ∣ A) = 1$ with $P (A) > 0$ , we obtain $P (A ∖ B) = 0$ . In finite models where every elementary outcome has positive probability, this implies $A \subseteq B$ . In a general probability model, zero-probability outcomes can remain in $A ∖ B$ , so literal containment is a stronger claim. The corresponding positive-denominator condition also matters when a union equation leads to a conditional probability of one.

Four common shortcuts to reject

Do not equate mutually exclusive with independent. Do not reverse a conditional probability without accounting for base rates. Do not count unequal-weight leaves as equally likely. Do not infer equal marginals merely because both reverse conditional probabilities are zero. Each shortcut discards information required by the probability model.

A historical connection

Early probability problems grew from games of chance and questions about counting favourable outcomes, including work by Cardano and Galileo. The work associated with Pascal and Fermat connected algebraic and counting approaches; Huygens developed early treatments of such problems, Bernoulli studied repeated-trial patterns, and Kolmogorov later provided the axiomatic framework. The chapter follows that same movement from concrete outcomes to general probability rules.

Example — Why one sum can be more likely than another

Problem
Three fair independent dice are rolled. Compare the probabilities of a sum of $9$ and a sum of $10$ , the counting issue described in the historical note.

1.The $216$ ordered die triples are equally likely. The possible sums themselves are not equally likely, because different sums can arise from different numbers of triples.
2.Fix the first die at $1, 2, 3, 4, 5, 6$ . For total $9$ , the required remaining sums are $8, 7, 6, 5, 4, 3$ , with $5, 6, 5, 4, 3, 2$ ordered pairs respectively: $25$ triples.
3.For total $10$ , the remaining sums are $9, 8, 7, 6, 5, 4$ , with $4, 5, 6, 5, 4, 3$ pairs respectively: $27$ triples. Thus the probabilities are $25/216$ and $27/216 = 1/8$ . Counting weighted outcomes, rather than treating named event values equally, resolves the comparison.

Check your understanding

These questions mix methods deliberately. State what is known and what is being requested before choosing the calculation. A method is justified by the event structure, not by the topic named in the story.

Quick check

If $P (A \cap B) = 0.15$ and $P (B) = 0.5$ , what is $P (A ∣ B)$ ?

Quick check

Which expression is valid for a general two-stage joint event?

Quick check

Two positive-probability events are mutually exclusive. What follows?

Quick check

Independent events have probabilities $0.2$ and $0.5$ . What is the probability of at least one?

Quick check

To calculate total probability, why use a partition?

Quick check

A source contributes 0.02 to evidence whose total probability is 0.05. What is its posterior?

Quick check

A random variable counts heads in two tosses. How many outcomes can have the same value one?

Quick check

The determinant with independent binary entries is positive with what probability?

Quick check

For a random leap year with uniformly likely starting weekdays, what is the probability of 53 Tuesdays?

Quick check

If $P (A ∣ B) > P (A)$ with the conditional probability defined, which relation follows?

Practice Problems

For $P (A) > 0$ , find $P (B ∣ A)$ if (a) $A \subseteq B$ , (b) $A \cap B = \emptyset$ .
For two children with equally likely ordered outcomes, find (a) probability both are boys given at least one boy; (b) probability both are girls given the elder child is a girl.
A population contains equally many men and women. Grey hair occurs in $5%$ of men and $0.25%$ of women. Given a grey-haired person selected uniformly from the population, find the probability that the person is male.
Assume ten independently sampled people each have right-handed probability $0.9$ . Express and evaluate the probability that at most six are right-handed.
A leap year has $366$ days. Assuming uniformly likely starting weekdays, find the probability it has $53$ Tuesdays.
Four equally likely boxes A, B, C, D contain respectively (red, white, black) counts $(1, 6, 3), (6, 2, 2), (8, 1, 1), (0, 6, 4)$ . A uniformly drawn ball from the chosen box is red. Find the posterior probabilities of A, B, C, and D.
In a hypothetical model, baseline heart-attack probability is $0.4$ . One intervention reduces this risk by $30%$ of its baseline value; another reduces it by $25%$ of its baseline value. Each is chosen with probability $1/2$ . Given a heart attack after the intervention, find the probability that the first was chosen.
Each entry of a two-by-two determinant is independently and uniformly $0$ or $1$ . Find the probability the determinant is positive, explaining your outcome count.
A fails with probability $0.2$ , B alone fails with probability $0.15$ , and both fail with probability $0.15$ . Find $P (A ∣ B)$ and the probability of A alone.
Bag I has three red and four black balls; bag II has four red and five black. Transfer a random ball from I to II, then draw a random ball from II. Given a red final draw, find the probability that the transferred ball was black.
Given $P (A) > 0$ and $P (B ∣ A) = 1$ , state what follows about $P (A ∖ B)$ . When can this be strengthened to literal set containment?
If $P (A ∣ B) > P (A)$ , prove $P (B ∣ A) > P (B)$ . Explain why the needed denominators are positive.
Suppose $P (A) + P (B) - P (A \cap B) = P (A)$ . If $P (B) > 0$ , find $P (A ∣ B)$ . What can happen if $P (B) = 0$ ?

(a) $A \cap B = A$ , so the ratio is $1$ . (b) The intersection is empty, giving $0$ .

Key Takeaways

• Identify the experiment, known information, requested event, and probability assumptions before choosing a formula. • Conditional probability restricts and rescales; multiplication builds a joint event; independence removes dependence from its factors. • A partition lets total probability combine all routes to an observation, and Bayes’ rule compares those routes afterward. • Complements simplify at-least-one events, and counting arrangements combines with independence in repeated trials. • Random variables give numerical descriptions of the same underlying outcomes. • Check positive denominators, consistent event probabilities, all alternative cases, and the meaning of the final result.

Previous · Lesson 4

Reasoning Backward from Evidence

End of chapter

One probability model, several useful viewpoints

Reconstruct the rules instead of guessing them

Counting arrangements in independent repetitions

Bring probability into another mathematical setting

Check the model as well as the arithmetic

Check your understanding

Quiz

Practice Problems

Key Takeaways