Independent events — multiplication rule: Exercises

Recall the one tool this whole page tests: $P (A \cap B) = P (A) P (B) — valid only when A, B are ==independent==.$ Everything else (the Complement Rule trick, the General Multiplication Rule, the Binomial Distribution) is built on top of this. If a symbol below feels unfamiliar, revisit the parent note.

A tiny glossary so no notation surprises you:

$P (A)$ ::: the probability (a number between $0$ and $1$ ) that event $A$ happens.
$A \cap B$ ::: " $A$ and $B$ " — both happen. The $\cap$ symbol is the overlap of two events.
$A \cup B$ ::: " $A$ or $B$ " (or both) — at least one of them happens. The $\cup$ symbol is the combined region of two events.
$B^{c}$ ::: "not $B$ " — the complement, everything where $B$ does not happen. See Complement Rule.
$P (A ∣ B)$ ::: "probability of $A$ given $B$ already happened" — see Conditional Probability.
$\prod$ ::: a big "multiply all of these together" sign, like $\sum$ but for products.

Level 1 — Recognition

Goal: decide whether events are independent, and apply the raw rule once.

Recall Solution L1·Q1

Independence check. The coin cannot feel what the die does — there is no physical link. So knowing the die landed on $6$ leaves the coin's chance of heads at $\frac{1}{2}$ . That is exactly the meaning $P (heads ∣ 6) = P (heads)$ , so independent. ✓ Apply the rule. $P (heads \cap 6) = P (heads) P (6) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} \approx 0.0833.$

Recall Solution L1·Q2

$P (A) = \frac{4}{52} = \frac{1}{13}$ (four Kings).
$P (B) = \frac{13}{52} = \frac{1}{4}$ (thirteen Hearts).
$A \cap B =$ "King of Hearts", one card: $P (A \cap B) = \frac{1}{52}$ . Test: $P (A) P (B) = \frac{1}{13} \cdot \frac{1}{4} = \frac{1}{52}$ . This equals $P (A \cap B)$ , so $A$ and $B$ are independent. ✓ (Notice: independence here is a coincidence of the deck's balanced structure, not a physical "no-talking" story — the definition still decides it.)

Level 2 — Application

Goal: chain the rule across repeated trials and use the "at least one" complement.

Recall Solution L2·Q1

Replacing puts the bag back exactly as it was, so the second draw's odds are untouched → independent. $P (R_{1}) = \frac{4}{10}, P (R_{2}) = \frac{4}{10} .$ $P (R_{1} \cap R_{2}) = \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} = 0.16.$

Recall Solution L2·Q2

"At least one miss" is the opposite of "no misses" = "all three made". Use the Complement Rule. $P (all made) = 0. 8^{3} = 0.512.$ $P (at least one miss) = 1 - 0.512 = 0.488.$ Why the complement? Counting "1 or 2 or 3 misses" is three messy cases; "all made" is a single clean product.

Recall Solution L2·Q3

Each fails to trigger with probability $1 - 0.7 = 0.3$ . All fail together (independence of failures): $P (none trigger) = 0. 3^{3} = 0.027.$ $P (at least one triggers) = 1 - 0.027 = 0.973.$

Level 3 — Analysis

Goal: tell independence from dependence, and see how the answer shifts.

The figure below shows the same bag drawn twice. On the left, the marble is replaced, so the bag for draw 2 is identical to the start — the green label reads $P (R_{2}) = 4/10 = 0.40$ . On the right, one red has been removed (shown as a dashed red outline marked "removed"), so draw 2 sees only $3$ red of $9$ — the orange label reads $P (R_{2} ∣ R_{1}) = 3/9 = 0.33$ . The shrinking red count is exactly why the without-replacement answer comes out smaller.

Figure — Independent events — multiplication rule

Recall Solution L3·Q1

The first draw changes the bag, so the second draw is conditional — not independent. Use the General Multiplication Rule: $P (R_{1} \cap R_{2}) = P (R_{1}) P (R_{2} ∣ R_{1}) .$

$P (R_{1}) = \frac{4}{10}$ .
After removing one red: $3$ red of $9$ left, so $P (R_{2} ∣ R_{1}) = \frac{3}{9} = \frac{1}{3}$ . $P (R_{1} \cap R_{2}) = \frac{4}{10} \cdot \frac{1}{3} = \frac{4}{30} = \frac{2}{15} \approx 0.1333.$ Comparison: with replacement gave $0.16$ ; without gives $0.133$ . Dependence lowered the chance — this matches the figure's right panel, where removing a red thins the red count for draw 2.

Recall Solution L3·Q2

Independence test: independence would require $P (jam ∣ rain) = P (jam) = 0.3$ . But we're told it's $0.6 \neq = 0.3$ . So dependent — rain raises jam odds. Use the general rule (the conditional value, not $0.3$ ): $P (rain \cap jam) = P (rain) P (jam ∣ rain) = 0.4 \cdot 0.6 = 0.24.$ (Multiplying $0.4 \cdot 0.3 = 0.12$ would be the blind-independence error — nearly half the true value.)

Level 4 — Synthesis

Goal: combine independence with complements, systems, and multiple stages.

The figure below plots reliability against the number of parallel copies $n$ of a component that fails with probability $0.2$ . The blue curve is $P (unit works) = 1 - 0. 2^{n}$ ; each point is labelled with its value. The dotted gray line marks a single copy's reliability ( $0.8$ ); the dashed red line marks the $0.99$ target. Notice how the blue curve rockets past the target with just a few copies — that steep climb is the pedagogical point about redundancy.

Recall Solution L4·Q1

"System works" = "pump works and filter works". Independent parts multiply: $P (works) = 0.95 \cdot 0.9 = 0.855.$ $P (fails) = 1 - 0.855 = 0.145.$

Recall Solution L4·Q2

"At least one works" = opposite of "all fail". Each fails with prob $0.2$ : $P (all fail) = 0. 2^{3} = 0.008.$ $P (unit works) = 1 - 0.008 = 0.992.$ Design insight: one component was $0.8$ reliable; three in parallel jump to $0.992$ . That is the whole point of redundancy — the blue curve at $n = 3$ in the figure above sits just above the dashed $0.99$ target.

Recall Solution L4·Q3

"Exactly one" splits into two disjoint cases: (A solves, B fails) OR (A fails, B solves). Within each case, the two independent facts multiply; across the two cases we add (they can't both be true at once — Mutually Exclusive Events). $P (A only) = 0.6 \cdot (1 - 0.5) = 0.6 \cdot 0.5 = 0.30.$ $P (B only) = (1 - 0.6) \cdot 0.5 = 0.4 \cdot 0.5 = 0.20.$ $P (exactly one) = 0.30 + 0.20 = 0.50.$

Level 5 — Mastery

Goal: reason backwards, prove structure, and connect to bigger machinery.

Recall Solution L5·Q1

$P (at least one pass) = 1 - (1 - p)^{n} = 1 - 0. 5^{n} .$ Require $1 - 0. 5^{n} \geq 0.99 \Rightarrow 0. 5^{n} \leq 0.01.$ Test values: $0. 5^{6} = 0.015625$ (too big), $0. 5^{7} = 0.0078125 \leq 0.01$ ✓. So the smallest $n$ is $7$ .

Recall Solution L5·Q2

We must show $P (A \cap B^{c}) = P (A) P (B^{c})$ . Split $A$ into the part inside $B$ and the part outside $B$ (these are disjoint and together make all of $A$ ): $P (A) = P (A \cap B) + P (A \cap B^{c}) .$ So $P (A \cap B^{c}) = P (A) - P (A \cap B)$ . Use independence $P (A \cap B) = P (A) P (B)$ : $P (A \cap B^{c}) = P (A) - P (A) P (B) = P (A) (1 - P (B)) = P (A) P (B^{c}) .$ That is exactly the independence condition for $A$ and $B^{c}$ . $■$ (This is why "the machine works" and "each part fails" can both be handled with the same product logic.)

Recall Solution L5·Q3

First recover $P (A \cap B)$ from the addition rule (stated at the top of this page) $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Recall $A \cup B$ means " $A$ or $B$ (or both)": $0.7 = 0.5 + 0.4 - P (A \cap B) \Rightarrow P (A \cap B) = 0.2.$ Independence would need $P (A) P (B) = 0.5 \cdot 0.4 = 0.20$ . Since $P (A \cap B) = 0.20$ equals $0.20$ , the events are independent. ✓

Recall Feynman recap: the whole page in one breath

If two things don't talk to each other, the chance of both is the product of their chances — that's the one move. When they do talk (a marble removed, rain feeding traffic), swap the second factor for its given-value. "At least one" is always cleanest as "one minus (all fail)". And "exactly one" means multiply inside each case, add across the cases.

Connections

Independent events — multiplication rule — the parent rule every problem above tests.
Conditional Probability — needed the moment independence breaks (L3, L5·Q2).
General Multiplication Rule — the without-replacement / rain-jam solutions.
Complement Rule — powers every "at least one" answer.
Mutually Exclusive Events — why disjoint cases add in L4·Q3.
Binomial Distribution — L5·Q1 is a first taste of "n independent trials".
Bayes' Theorem — the natural next step once conditionals stack up.

Exercises — Independent events — multiplication rule