Probability and conditional information

Define the sample space and the event

An experiment generates data. An observation records numerical or categorical information. A sample space S is the set of all possible outcomes; it may be finite or infinite. Each outcome is a sample point, element or member of S. Outcomes are equally likely when they have equal chances. An event E is a subset of S.

Relative to a universal set ε, $A'$ is the complement of A; $A\cup B$ contains elements in A or B or both; $A\cap B$ contains elements in both. The empty set is $\varnothing$ or $\{\}$, and n(A) is the number of elements in A. In a probability problem the universal outcome set is S.

For a fair coin, H and T denote a head and a tail. Three tosses give the finite sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$. Tossing repeatedly until a tail gives the infinite stopping-string space $S=\{T,HT,HHT,HHHT,\ldots\}$.

Event	Three tosses	Stop at the first tail
A tail is observed once	$\{HHT,HTH,THH\}$	All of S
Tails are observed at least twice	$\{HTT,THT,TTH,TTT\}$	$\varnothing$
Heads are observed three times	$\{HHH\}$	$\{HHHT\}$

The stopping strings have probabilities 1/2,1/4,1/8,…, not equal weights. This model records completed stopping strings; endless heads has probability zero under independent fair tosses. The source phrase “drawn at random means all outcomes are equally likely” describes its intended uniform finite-selection model. Randomness by itself does not guarantee uniform probabilities.

Count only when elementary outcomes are equally likely

For a finite, nonempty sample space of equally likely outcomes,

$$P(E)=\frac{\text{number of outcomes in E}}{\text{total number of outcomes}}=\frac{n(E)}{n(S)}.$$

The source’s function note views probability as a map from events, which are subsets of the outcome space, to numbers in [0, 1].

The seven printed results and their reasoning are:

1. If $E=\varnothing$, then $n(E)=0$, so $P(E)=0/n(S)=0$.
2. If E=S, then n(E)=n(S), so $P(E)=n(S)/n(S)=1$.
3. Since $E\subseteq S$, $0\le n(E)\le n(S)$. Divide by n(S)>0 to obtain $0\le P(E)\le1$.
4. E and E′ partition S: $n(E)+n(E')=n(S)$. Divide by n(S) to get $P(E)+P(E')=1$, hence $P(E')=1-P(E)$.
5. Adding P(A) and P(B) counts the overlap twice. Subtract one copy: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
6. Intersection is symmetric: $P(A\cap B)=P(B\cap A)$.
7. Union is symmetric: $P(A\cup B)=P(B\cup A)$.

Recognise mutually exclusive events

Mutually exclusive events have no common outcomes: $A\cap B=\varnothing$, so $P(A\cap B)=0$. For one die roll, getting 1 and getting 2 are mutually exclusive. Getting 1 and getting an odd number are not: outcome 1 is shared. The notes acknowledge OpenIntro Statistics, second edition for this explanation.

Substitute the zero overlap into the addition rule:

$$P(A\cup B)=P(A)+P(B)-0=P(A)+P(B).$$

For pairwise mutually exclusive $A_1,\ldots,A_n$, $P(A_1\cup\cdots\cup A_n)=P(A_1)+\cdots+P(A_n)$.

The source tests mutual exclusivity through either zero intersection probability or equality in the addition formula. These are equivalent to an empty intersection in the finite positive-probability elementary-outcome model used here. In general probability spaces, zero probability need not mean an empty set; the set definition remains no shared outcomes.

Change the reference set when information is given

John rolls a fair die and Peter guesses its result. A specified number has probability 1/6. Given that the result is even, the possible outcomes reduce from {1,2,3,4,5,6} to {2,4,6}, and each has conditional probability 1/3.

Mary’s class has 25 students: 15 boys and 10 girls. Selecting one student uniformly gives Mary probability 1/25. Given that a girl was selected, the reference set is the 10 girls and Mary’s probability becomes 1/10. This concrete example applies the same change of denominator.

For P(B)>0,

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

In a finite equally likely model, the source proves this by dividing the favourable and reference counts by n(S):

$$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{n(A\cap B)/n(S)}{n(B)/n(S)}=\frac{P(A\cap B)}{P(B)}.$$

The source assumes both P(A)≠0 and P(B)≠0 in its introduction; only P(B)>0 is required to define P(A|B). The reverse conditional P(B|A) additionally needs P(A)>0.

If A and B are mutually exclusive and P(B)>0, then $P(A\mid B)=0/P(B)=0$. Rearranging gives $P(A\cap B)=P(A\mid B)P(B)$. Similarly $P(B\cap A)=P(B\mid A)P(A)$, so, when both conditionals are defined,

$$P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A).$$

The source deck illustration is retained using labelled suits and ranks. Its printed illustration credit is the card-image source; this is a provenance pointer, not additional course content.

Test independence using the probability model

Two events are independent when learning that one occurred does not change the probability of the other. The source example combines a coin toss and a die roll: knowing the coin is heads does not change the probability that the die is odd, assuming the two processes operate independently. This is a statement about probabilities, not a claim about causation.

Where the conditionals exist, independence gives $P(A\mid B)=P(A)$ and $P(B\mid A)=P(B)$. Substituting the first equality into the conditional formula gives $P(A\cap B)/P(B)=P(A)$, hence

$$P(A\cap B)=P(A)P(B).$$

The source lists three equivalent tests: either conditional equality when defined, or the product equality. The product test also works when an event has probability zero. For mutually independent $A_1,\ldots,A_n$, $P(A_1\cap\cdots\cap A_n)=P(A_1)\cdots P(A_n)$. Here independent means mutual independence; pairwise independence alone is not enough for that general product.

Complementing independent events preserves independence

If A and B are independent, so are A′ and B, A and B′, and A′ and B′. The source gives two full proofs and asks you to prove the remaining case in the activity below.

For A′ and B, start with the region in B outside A:

$$\begin{aligned} P(A'\cap B)&=P(A\cup B)-P(A)\\ &=P(A)+P(B)-P(A\cap B)-P(A)\\ &=P(B)-P(A)P(B)\\ &=P(B)[1-P(A)]=P(A')P(B). \end{aligned}$$

For both complements, use the complement of the union and factor:

$$\begin{aligned} P(A'\cap B')&=1-P(A\cup B)\\ &=1-[P(A)+P(B)-P(A\cap B)]\\ &=1-P(A)-P(B)+P(A)P(B)\\ &=[1-P(A)]-P(B)[1-P(A)]\\ &=[1-P(A)][1-P(B)]=P(A')P(B'). \end{aligned}$$

Try the analogous proof for A and B′ before opening its working. Example 3 then tests independence and mutual exclusivity separately, and Self-Practice 3 requires recovering all of A from its portions inside and outside B.

Choose a problem-solving tool

Approach	When it is useful
Table of outcomes	Two components and a small sample space.
Venn diagram	Multiple events, intersections and complements, such as $A'\cap B'$ and $(A\cap B')\cup B$.
Permutations and combinations	Selection without replacement or arrangement.
Tree diagram	A series of experiments or sequential events.

Example 4 constructs a complete 6×6 table for two throws, with the first throw across columns and the second down rows. Every ordered cell is equally likely; the displayed sums are not equally likely. Use the table for the intersection numerator, then choose the probability of the given event as denominator.

The source recommends the conditional probability formula.

Its alternative counting method first identifies the reduced reference space, then the favourable outcomes inside it. The Teacher guidance highlights these in that order.

Count ratios require equally likely elementary outcomes in the reference space; a reduced space being easy to list is a practical convenience.

The source repeats the same table for several methods; the native four-panel diagram retains all 36 cells and each distinct highlighted reference/favourable set.

Subtract sets before substituting probabilities

Venn diagrams make a region subtraction visible. In Example 5, the union is partitioned into B and A-only, so $P(B)=P(A\cup B)-P(A\cap B')$. This is the Teacher’s visual subtraction argument. Once B is known, independence replaces $P(A\cap B)$ by P(A)P(B) in the addition formula to find A.

For $A\cup B'$, identify the excluded region: B-only, or $B\cap A'$. The Teacher diagram splits B into its overlap with A and its part outside A. Subtract the overlap from B, then complement the result. Source circle areas are schematic and do not encode the numerical probabilities.

Keep the counting model consistent

For uniformly chosen finite outcomes, probability is favourable count divided by total count. With repeated letters, the numerator and denominator must describe the same distinct-arrangement model. SEPTEMBER has three identical Es; PROBABILITY has two Bs and two Is. Uniform shuffling of the physical copies gives every distinct arrangement the same number of labelled-copy realisations, justifying the ratio.

For three copies, not all together and all separated are different events. The Teacher’s SEPTEMBER warning is explicit: the former includes a pair together and a third copy elsewhere. For two copies, the complement of together is separated.

Read every tree branch relative to its preceding path

A probability tree traces sequential events. The first split is A or A′. Each of those nodes then splits into B or B′. The four complete path probabilities are:

$$P(A\cap B)=P(A)P(B\mid A),\quad P(A\cap B')=P(A)P(B'\mid A),$$ $$P(A'\cap B)=P(A')P(B\mid A'),\quad P(A'\cap B')=P(A')P(B'\mid A').$$

Each branch probability is conditional on the path so far; the first split is conditional on the initial sample space. Sibling alternatives from a reached node are mutually exclusive and exhaustive, and their probabilities sum to 1. This is the intended meaning of the source’s “nodes from the same branch” wording.

Multiply along a complete path; add disjoint complete paths for an event.

For a zero-probability preceding event, its ordinary conditional is undefined and that path contributes zero without needing a ratio there.

The source’s diagram note explains that Venn and tree diagrams present sample spaces, relationships between sets and event outcomes, helping calculate probabilities. Example 7 models two counters without replacement using both a sequential tree and combinations of distinct physical counters. The two methods use different but internally consistent elementary outcomes and agree on all three answers.

Add disjoint paths before conditioning on the result

Example 8 first divides customers by fuel type, then by whether they fill their tanks. To find a not-full tank, multiply along each of the three not-full paths and add them: the fuel categories are disjoint and exhaustive. To condition on a full tank, divide the desired fuel-and-full probability by the total probability of a full tank, including all fuel types.

Self-Practice 7 uses the same structure with machine identity and defect status. Production shares weight the three conditional defect rates. Independence of defects across different items does not mean that a machine’s identity and defect status are independent for the selected item.

Replacement keeps the composition fixed

In Example 9, replacement keeps the blue/white proportions unchanged between draws. For independent random draws, the joint probability factors into the two one-draw probabilities. The Teacher guidance therefore uses direct multiplication instead of a tree.

For an inclusive union, either add disjoint outcome cases or subtract the complement. “First blue or second white” includes BW, BB, WW and excludes only WB. This retains both source methods while making the overlap explicit.

Prefer a compact count when a full tree becomes large

Example 10 first uses a complete 27-path tree for three replacement draws. Its six all-different colour orders all have the same probability 0.03, so one path probability multiplied by 3! gives 0.18. The Teacher recommends this compact order count rather than teaching a lengthy tree construction. The complete native tree remains available for comparison.

Without replacement, repeated draw probabilities change with the remaining supplies.

The same example retains both source approaches: a probability for one colour order multiplied by its number of distinct orders, and an unordered selection count using distinct physical balls. The source’s 3^5=243 candidate-colour-string remark includes strings that would exhaust a colour and have zero probability.

It is not a uniform sample space.

In the six-path symbolic sum, the printed G1,W1,R2 indices are qualified as the intended G1,W2,R3 path.

Identify the repeat unit in a geometric progression

Example 11 restarts only after a complete round in which A, B and C all fail. The repeat probability is therefore (5/6)^3, not 5/6. The source case table counts A’s own attempts; each successive winning opportunity adds another full failed round. Sum the disjoint winning cases using an infinite geometric progression.

Example 12 counts at least one dart success in n throws, either by complementing all n misses or by summing first-hit cases 1 through n.

Once a hit occurs, later outcomes do not alter that event, as the Teacher notes emphasize.

Use the finite geometric sum or the complement expression, retain the strict threshold, reverse the inequality on dividing by a negative logarithm, and confirm the least integer with the source calculator table.