Conditional probability — P(A - B) = P(A∩B) - P(A): Exercises

Before we start, one picture to fix the whole idea in your head: conditioning shrinks the universe.

Figure — Conditional probability — P(A - B) = P(A∩B) - P(A)

The big peach rectangle is the whole sample space. Once you are told $B$ happened, you are trapped inside the violet blob $B$ . The only part of $A$ that still counts is the magenta overlap $A \cap B$ . That is all $P (A ∣ B)$ measures.

Level 1 — Recognition

Here you only need to name the pieces and read the formula correctly.

Recall Solution L1.1

WHAT: The formula is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . WHY: $B$ is the given event, so it is the denominator (the new universe). $P (A ∣ B) = \frac{0.12}{0.30} = 0.4$ Check: $0.4 < 1$ , as any probability must be. ✔

Recall Solution L1.2

Answer: undefined. You may never divide by $P (B) = 0$ . Conditioning on an impossible event has no meaning — there is no shrunken universe to sit inside. The formula requires $P (B) > 0$ .

Recall Solution L1.3

The event right of the bar is $> 3$ , so we divide by $P (> 3)$ . ( $> 3 = {4, 5, 6}$ , so $P (> 3) = \frac{3}{6} = \frac{1}{2}$ .) The bar's buddy is the bottom.

Level 2 — Application

Now plug real counts into the formula.

Recall Solution L2.1

Step 1 — overlap. $A \cap B = {3, 6}$ , so $P (A \cap B) = \frac{2}{6}$ . Step 2 — condition. $P (B) = \frac{4}{6}$ . Step 3 — divide. $P (A ∣ B) = \frac{2/6}{4/6} = \frac{2}{4} = \frac{1}{2}$ Sanity check inside the new universe: of ${3, 4, 5, 6}$ , the multiples of $3$ are ${3, 6}$ — that is $2$ out of $4$ . ✔

Recall Solution L2.2

There are $26$ red cards. Use the multiplication rule $P (R_{1} \cap R_{2}) = P (R_{1}) P (R_{2} ∣ R_{1})$ . Step 1: $P (R_{1}) = \frac{26}{52} = \frac{1}{2}$ . Step 2: After one red is gone, $25$ reds remain in $51$ cards, so $P (R_{2} ∣ R_{1}) = \frac{25}{51}$ . Step 3: $P (both red) = \frac{26}{52} \cdot \frac{25}{51} = \frac{25}{102}$

Recall Solution L2.3

WHAT we condition on: the first draw was magenta, so one magenta is gone. New universe: $7$ marbles remain — $2$ magenta, $5$ orange. $P (2nd orange ∣ 1st magenta) = \frac{5}{7}$ Only the count of what is left matters, not what colour we removed's own future.

Level 3 — Analysis

Here you must choose the right tool: complement, multiplication, or independence check.

Recall Solution L3.1

The correct complement law keeps the condition fixed: $P (A ∣ B) + P (A^{c} ∣ B) = 1 ⟹ P (A^{c} ∣ B) = 1 - 0.7 = 0.3$ Why this and not $P (A ∣ B^{c})$ : inside the fixed universe $B$ , the events $A$ and $A^{c}$ split it completely, so their conditional probabilities add to $1$ . Changing the condition to $B^{c}$ moves to a different universe entirely.

Recall Solution L3.2

Tool choice: use the independence test $P (A \cap B) = ? P (A) P (B)$ (see Independence of events). $P (A) P (B) = 0.4 \times 0.5 = 0.20 = P (A \cap B)$ They are independent. Equivalently $P (A ∣ B) = \frac{0.2}{0.5} = 0.4 = P (A)$ — knowing $B$ changed nothing.

Recall Solution L3.3

Tool choice: "of bike-owners, $25%$ ..." is a conditional $P (S ∣ K) = 0.25$ . Use multiplication: $P (K \cap S) = P (K) P (S ∣ K) = 0.60 \times 0.25 = 0.15$ So $15%$ of the whole class own both.

Level 4 — Synthesis

Combine multiplication, total probability, and reversing a condition.

Recall Solution L4.1

Tool: the Law of total probability splits $D$ across the partition ${M_{1}, M_{2}}$ : $P (D) = P (M_{1}) P (D ∣ M_{1}) + P (M_{2}) P (D ∣ M_{2})$ $= 0.70 \times 0.02 + 0.30 \times 0.05 = 0.014 + 0.015 = 0.029$ So $2.9%$ of all bolts are defective. Read the tree above: each path multiplies down, then we add the two "defective" leaves.

Recall Solution L4.2

Tool: Bayes' Theorem flips the condition correctly: $P (M_{2} ∣ D) = \frac{P ( M _{2} ) P ( D ∣ M _{2} )}{P ( D )} = \frac{0.30 \times 0.05}{0.029} = \frac{0.015}{0.029} = \frac{15}{29} \approx 0.517$ Even though $M_{2}$ makes only $30%$ of bolts, it accounts for over half of the defects — because its defect rate is higher.

Recall Solution L4.3

Step 1 — total probability of $+$ : $P (+) = 0.02 \times 0.95 + 0.98 \times 0.10 = 0.019 + 0.098 = 0.117$ Step 2 — Bayes: $P (sick ∣ +) = \frac{0.02 \times 0.95}{0.117} = \frac{0.019}{0.117} \approx 0.1624$ Why so low? The disease is rare, so most positives are false positives from the huge healthy group. Conditioning correctly restricts you to everyone who tested $+$ , then finds the sick fraction — about $16%$ , not $95%$ .

Level 5 — Mastery

Model a full scenario from scratch.

Recall Solution L5.1

Each urn is chosen with probability $\frac{1}{3}$ , and each urn holds $5$ balls. $P (M ∣ U_{1}) = \frac{2}{5}, P (M ∣ U_{2}) = \frac{4}{5}, P (M ∣ U_{3}) = \frac{1}{5}$ (a) Law of total probability: $P (M) = \frac{1}{3} (\frac{2}{5} + \frac{4}{5} + \frac{1}{5}) = \frac{1}{3} \cdot \frac{7}{5} = \frac{7}{15}$ (b) Bayes: $P (U_{2} ∣ M) = \frac{\frac{1}{3} \cdot \frac{4}{5}}{\frac{7}{15}} = \frac{4/15}{7/15} = \frac{4}{7}$ Urn $2$ is the richest in magenta, so seeing magenta makes it the most likely source — over half.

Recall Solution L5.2

Strategy: "at least one" is easiest via its complement "no ace." Chain the conditional no-ace draws (each draw shrinks the deck). $P (no ace) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50}$ $= \frac{48 \cdot 47 \cdot 46}{52 \cdot 51 \cdot 50} = \frac{103776}{132600} = \frac{4324}{5525}$ $P (at least one ace) = 1 - \frac{4324}{5525} = \frac{1201}{5525} \approx 0.2174$

Recall Solution L5.3

Sample space: ${H H, H T, T H, T T}$ , each with probability $\frac{1}{4}$ (see Sample space and events). $A = {H H, H T}$ so $P (A) = \frac{1}{2}$ . $B = {H T, T H}$ so $P (B) = \frac{1}{2}$ . $A \cap B = {H T}$ so $P (A \cap B) = \frac{1}{4}$ . Independence test: $P (A) P (B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = P (A \cap B)$ — independent! Conditional: $P (A ∣ B) = \frac{1/4}{1/2} = \frac{1}{2} = P (A)$ . Knowing "exactly one head" gives no clue about which flip it was.

Recall Feynman recap: the whole ladder in one breath

Every problem here is the same move: someone tells you something, so your box of possibilities shrinks. L1 just names the pieces. L2 counts inside the smaller box. L3 picks whether to complement, multiply, or test independence. L4 walks down a tree, adds up paths (total probability), and reverses arrows (Bayes). L5 does all of it at once. Denominator is always the thing you were told.

Connections

Multiplication rule of probability — powers every "without replacement" chain (L2, L5).
Law of total probability — sums conditionals over a partition (L4, L5).
Bayes' Theorem — reverses the condition (L4.2, L4.3, L5.1b).
Independence of events — the arithmetic test (L3.2, L5.3).
Sample space and events — the universe we shrink.
Tree diagrams — visual chaining of the L4/L5 steps.