25 Solved Examples Across the Chapter

The problems below run in rough order of difficulty and cover every tool in the chapter: conditional probability by counting and by formula (Examples 1 to 5), the multiplication theorem for chained draws (6 to 10), independence and its complement tricks (11 to 15), total probability (16 to 18), and the full range of Bayes' theorem applications (19 to 25), ending with the alternating-game and system-reliability classics.

For every problem, try covering the solution and attempting it first. The final answer matters less than the setup line — naming the events, the priors and the conditionals correctly is where marks are won and lost.

Examples 1 to 7 — Conditional Probability and Chained Draws

Example 1 — Both conditionals from the same data

If $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E|F)$ and $P(F|E)$.

Step 1 — apply the definition each way: $$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}, \qquad P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$$

Answer: $P(E|F) = \dfrac{2}{3}$, $P(F|E) = \dfrac{1}{3}$ — the same intersection, two different denominators, two different answers.

Example 2 — Conditioning on the first two tosses

A coin is tossed three times. Find the probability of a head on the third toss, given that the first two tosses both showed heads.

Step 1 — reduce the sample space: given "first two are heads", only $\{HHH, HHT\}$ survive, each equally likely.

Step 2 — count favourable outcomes: the third toss is a head only in $HHH$.

Answer: $P = \dfrac{1}{2}$. The first two tosses tell us nothing about the third — a preview of independence.

Example 3 — Dice with a condition on one die

A die is thrown twice. Find the probability that the sum is 8, given that the second throw shows 4.

Step 1 — reduced sample space: second throw 4 leaves $(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)$ — six outcomes.

Step 2 — favourable: sum 8 requires the first throw to be 4, i.e. only $(4,4)$.

Answer: $P(\text{sum } 8 \mid \text{second is } 4) = \dfrac{1}{6}$.

Example 4 — Sum condition the other way

A die is thrown twice. Find the probability that the sum exceeds 9, given that the first throw shows 5.

Step 1 — reduced sample space: $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$.

Step 2 — favourable: sum greater than 9 needs the second throw to be 5 or 6: $(5,5), (5,6)$.

Answer: $P = \dfrac{2}{6} = \dfrac{1}{3}$.

Example 5 — Conditioning inside the deck

A card drawn from a deck of 52 is known to be a face card (jack, queen or king). Find the probability that it is a black queen.

Step 1 — reduced sample space: there are $12$ face cards.

Step 2 — favourable: black queens are the queen of spades and the queen of clubs — $2$ cards.

Answer: $P(\text{black queen} \mid \text{face card}) = \dfrac{2}{12} = \dfrac{1}{6}$.

Example 6 — First black, then white

An urn contains 10 black and 5 white balls. Two balls are drawn one after the other without replacement. Find the probability that the first is black and the second is white.

Step 1 — first draw: $P(B_1) = \frac{10}{15} = \frac{2}{3}$.

Step 2 — second draw given the first: 14 balls remain, 5 of them white: $P(W_2|B_1) = \frac{5}{14}$.

Step 3 — multiplication theorem: $$P(B_1 \cap W_2) = \frac{2}{3} \times \frac{5}{14} = \frac{5}{21}$$

Answer: $\dfrac{5}{21}$.

Example 7 — Two face cards in a row

Two cards are drawn without replacement from a deck of 52. Find the probability that both are face cards.

Step 1 — chain the draws: $$P(F_1 \cap F_2) = P(F_1)\,P(F_2|F_1) = \frac{12}{52} \times \frac{11}{51}$$

Step 2 — simplify: $\frac{12}{52} = \frac{3}{13}$, so $P = \frac{3}{13} \times \frac{11}{51} = \frac{33}{663} = \frac{11}{221}$.

Answer: $\dfrac{11}{221}$.

Examples 8 to 13 — Longer Chains and Independence

Example 8 — Both red without replacement

A bag contains 5 white and 8 red balls. Two balls are drawn successively without replacement. Find the probability that both are red.

Step 1 — chain: $P(R_1) = \frac{8}{13}$, then $P(R_2|R_1) = \frac{7}{12}$.

Step 2 — multiply: $$P(\text{both red}) = \frac{8}{13} \times \frac{7}{12} = \frac{56}{156} = \frac{14}{39}$$

Answer: $\dfrac{14}{39}$.

Example 9 — Three good bulbs

A box holds 15 bulbs of which 5 are defective. Three bulbs are drawn one by one without replacement. Find the probability that none is defective.

Step 1 — three-event multiplication theorem: $$P(G_1 \cap G_2 \cap G_3) = P(G_1)\,P(G_2|G_1)\,P(G_3|G_1 \cap G_2)$$

Step 2 — substitute the shrinking counts: $$= \frac{10}{15} \times \frac{9}{14} \times \frac{8}{13} = \frac{2}{3} \times \frac{9}{14} \times \frac{8}{13} = \frac{24}{91}$$

Answer: $\dfrac{24}{91}$.

Example 10 — First ace exactly on the second draw

Cards are drawn one by one without replacement from a deck of 52. Find the probability that the first ace appears exactly on the second draw.

Step 1 — translate: the first card must be a non-ace AND the second an ace.

Step 2 — chain: $$P = P(\text{non-ace first}) \times P(\text{ace second} \mid \text{non-ace first}) = \frac{48}{52} \times \frac{4}{51} = \frac{12}{13} \times \frac{4}{51} = \frac{48}{663} = \frac{16}{221}$$

Answer: $\dfrac{16}{221}$.

Example 11 — Independence check plus both conditionals

Events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B) = 0.8$ and $P(A \cap B) = 0.32$. Are they independent? Find $P(A|B)$ and $P(B|A)$.

Step 1 — product test: $P(A)\,P(B) = 0.4 \times 0.8 = 0.32 = P(A \cap B)$ — independent.

Step 2 — the conditionals collapse to the plain probabilities (that is what independence means): $$P(A|B) = \frac{0.32}{0.8} = 0.4 = P(A), \qquad P(B|A) = \frac{0.32}{0.4} = 0.8 = P(B)$$

Answer: independent; $P(A|B) = 0.4$, $P(B|A) = 0.8$.

Example 12 — Independence inside a single die throw

A die is thrown. Let $A$ be the event "the number is at most 4" and $B$ the event "the number is even". Are $A$ and $B$ independent?

Step 1 — the three numbers: $A = \{1,2,3,4\}$, $B = \{2,4,6\}$, $A \cap B = \{2,4\}$: $$P(A) = \frac{4}{6} = \frac{2}{3}, \qquad P(B) = \frac{1}{2}, \qquad P(A \cap B) = \frac{2}{6} = \frac{1}{3}$$

Step 2 — test: $P(A)\,P(B) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} = P(A \cap B)$.

Answer: independent. Exactly half of $A$'s four outcomes are even, matching the overall proportion of evens — the counting reason the test passes.

Example 13 — Two shooters fire at a target

$A$ hits a target with probability $\frac{3}{4}$ and $B$ hits it with probability $\frac{1}{2}$; they fire independently, one shot each. Find the probability that (i) the target is hit, (ii) both hit it, (iii) exactly one hits it.

Step 1 — (ii) is a plain product: $P(A \cap B) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$.

Step 2 — (i) via the at-least-one formula: $$P(\text{hit}) = 1 - P(A')\,P(B') = 1 - \frac{1}{4} \times \frac{1}{2} = 1 - \frac{1}{8} = \frac{7}{8}$$

Step 3 — (iii) two disjoint ways: $$P(\text{exactly one}) = P(A)P(B') + P(A')P(B) = \frac{3}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}$$

Answer: (i) $\dfrac{7}{8}$, (ii) $\dfrac{3}{8}$, (iii) $\dfrac{1}{2}$.

Examples 14 to 19 — Independence Applications and Total Probability

Example 14 — Two candidates, one post

The probabilities that $A$ and $B$ are selected for a post are $\frac{1}{7}$ and $\frac{1}{5}$ respectively, independently (selections are for different organisations). Find the probability that (i) both are selected, (ii) exactly one is selected, (iii) none is selected.

Step 1 — (i): $P(A \cap B) = \frac{1}{7} \times \frac{1}{5} = \frac{1}{35}$.

Step 2 — (ii): $$P(\text{exactly one}) = P(A)P(B') + P(A')P(B) = \frac{1}{7} \cdot \frac{4}{5} + \frac{6}{7} \cdot \frac{1}{5} = \frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$$

Step 3 — (iii): $P(A' \cap B') = \frac{6}{7} \times \frac{4}{5} = \frac{24}{35}$.

Answer: (i) $\dfrac{1}{35}$, (ii) $\dfrac{2}{7}$, (iii) $\dfrac{24}{35}$. Check: $\frac{1}{35} + \frac{10}{35} + \frac{24}{35} = 1$ ✓ — the three cases exhaust all possibilities.

Example 15 — At least one of three

Three students attempt a problem independently, with probabilities of success $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{1}{4}$. Find the probability that the problem is solved.

Step 1 — the complement of "solved" is "all three fail": the failure probabilities are $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$.

Step 2 — multiply (complements of independent events are independent): $$P(\text{all fail}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}$$

Step 3 — subtract: $P(\text{solved}) = 1 - \frac{1}{4} = \frac{3}{4}$.

Answer: $\dfrac{3}{4}$.

Example 16 — Transfer a ball, then draw

Bag I contains 3 red and 4 black balls; Bag II contains 4 red and 5 black balls. One ball is transferred at random from Bag I to Bag II, and then a ball is drawn from Bag II. Find the probability that the drawn ball is red.

Step 1 — partition by the transferred ball: $P(R_t) = \frac{3}{7}$, $P(B_t) = \frac{4}{7}$.

Step 2 — conditionals for the draw (Bag II now has 10 balls): after a red transfer it holds 5 red, after a black transfer 4 red: $$P(\text{red}|R_t) = \frac{5}{10}, \qquad P(\text{red}|B_t) = \frac{4}{10}$$

Step 3 — total probability: $$P(\text{red}) = \frac{3}{7} \times \frac{5}{10} + \frac{4}{7} \times \frac{4}{10} = \frac{15}{70} + \frac{16}{70} = \frac{31}{70}$$

Answer: $\dfrac{31}{70}$.

Example 17 — The factory's overall defect rate

Machines A, B, C produce $25\%$, $35\%$ and $40\%$ of a factory's output with defect rates $5\%$, $4\%$ and $2\%$. What fraction of the factory's total output is defective?

Step 1 — total probability over the three machines: $$P(D) = 0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02$$

Step 2 — compute: $0.0125 + 0.0140 + 0.0080 = 0.0345$.

Answer: $0.0345$, i.e. $3.45\%$ of all output. This number is exactly the denominator of the Bayes computation that asks "which machine made this defective item?" — total probability and Bayes always travel together.

Example 18 — A two-stage experiment

A girl throws a die. If she gets 5 or 6 she tosses a coin three times; if she gets 1, 2, 3 or 4 she tosses it once. Find the probability that she obtains exactly one head.

Step 1 — partition by the die: $P(E_1) = P(5,6) = \frac{1}{3}$, $P(E_2) = P(1,2,3,4) = \frac{2}{3}$.

Step 2 — conditionals: three tosses give exactly one head in 3 of the 8 outcomes: $P(H|E_1) = \frac{3}{8}$; a single toss gives $P(H|E_2) = \frac{1}{2}$.

Step 3 — total probability: $$P(H) = \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2} = \frac{1}{8} + \frac{1}{3} = \frac{3 + 8}{24} = \frac{11}{24}$$

Answer: $\dfrac{11}{24}$.

Example 19 — Grey hair (unequal rates, equal priors)

Suppose $5\%$ of men and $0.25\%$ of women have grey hair, and there are equal numbers of men and women. A grey-haired person is selected at random. What is the probability that the person is male?

Step 1 — priors: $P(M) = P(W) = \frac{1}{2}$.

Step 2 — conditionals for $G$ = "grey hair": $P(G|M) = 0.05$, $P(G|W) = 0.0025$.

Step 3 — Bayes (the equal priors cancel): $$P(M|G) = \frac{0.05}{0.05 + 0.0025} = \frac{500}{525} = \frac{20}{21}$$

Answer: $\dfrac{20}{21}$. With equal priors, the posterior is just each group's rate divided by the total of the rates — grey hair is 20 times commoner among men, so the posterior is $\frac{20}{21}$.

Examples 20 to 25 — Bayes' Theorem at Full Strength

Example 20 — Four boxes of marbles

Boxes A, B, C, D contain marbles: A has 1 red, 6 white, 3 black; B has 6 red, 2 white, 2 black; C has 8 red, 1 white, 1 black; D has 0 red, 6 white, 4 black. A box is chosen at random and one marble drawn; it is red. Find the probability that it came from box A, from box B, and from box C.

Step 1 — priors and conditionals (each box has 10 marbles): $$P(\text{red}|A) = \frac{1}{10}, \quad P(\text{red}|B) = \frac{6}{10}, \quad P(\text{red}|C) = \frac{8}{10}, \quad P(\text{red}|D) = 0, \qquad \text{priors all } \frac{1}{4}$$

Step 2 — the equal priors cancel, so posteriors are proportional to $1 : 6 : 8 : 0$, total $15$: $$P(A|\text{red}) = \frac{1}{15}, \qquad P(B|\text{red}) = \frac{6}{15} = \frac{2}{5}, \qquad P(C|\text{red}) = \frac{8}{15}$$

Answer: $\dfrac{1}{15}$, $\dfrac{2}{5}$, $\dfrac{8}{15}$. (Box D gets posterior $0$ — it has no red marbles to offer, and the three nonzero posteriors add to 1.)

Example 21 — Meditation or medication?

The chance of a heart attack is $40\%$. A meditation-and-yoga course reduces the risk by $30\%$; a certain drug reduces it by $25\%$. A patient chooses one of the two options with equal probability, and later suffers a heart attack. Find the probability that the patient followed the meditation-and-yoga course.

Step 1 — convert "risk reduced by" into conditionals: $$P(H|Y) = 0.40 \times (1 - 0.30) = 0.28, \qquad P(H|D) = 0.40 \times (1 - 0.25) = 0.30$$

Step 2 — priors: $P(Y) = P(D) = \frac{1}{2}$.

Step 3 — Bayes (priors cancel): $$P(Y|H) = \frac{0.28}{0.28 + 0.30} = \frac{28}{58} = \frac{14}{29}$$

Answer: $\dfrac{14}{29}$ — just under $\frac{1}{2}$, because the yoga group has the slightly lower attack rate.

Example 22 — Four boxes, unequal sizes

Coloured balls are distributed in four boxes. Box I holds 3 black of 18 balls, box II holds 2 black of 8, box III holds 1 black of 7, and box IV holds 4 black of 13. A box is selected at random and a ball drawn from it is black. Find the probability that it came from box III.

Step 1 — priors and conditionals: priors all $\frac{1}{4}$; $$P(A|E_1) = \frac{3}{18}, \quad P(A|E_2) = \frac{2}{8}, \quad P(A|E_3) = \frac{1}{7}, \quad P(A|E_4) = \frac{4}{13}$$

Step 2 — Bayes with the $\frac{1}{4}$'s cancelled: $$P(E_3|A) = \frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}} = \frac{156}{947} \approx 0.165$$

Answer: $\approx 0.165$. Unlike Example 20 the boxes have different sizes, so each conditional is a different-denominator fraction — work with exact fractions and combine at the end.

Example 23 — Transfer, draw, and reason backwards

Bag I contains 3 red and 4 black balls; Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II, which turns out red. Find the probability that the transferred ball was black.

Step 1 — hypotheses about the transfer: $P(B_t) = \frac{4}{7}$, $P(R_t) = \frac{3}{7}$.

Step 2 — conditionals for the red draw (Bag II then has 10 balls): $P(\text{red}|B_t) = \frac{4}{10}$, $P(\text{red}|R_t) = \frac{5}{10}$.

Step 3 — Bayes: $$P(B_t|\text{red}) = \frac{\frac{4}{7} \times \frac{4}{10}}{\frac{4}{7} \times \frac{4}{10} + \frac{3}{7} \times \frac{5}{10}} = \frac{16}{16 + 15} = \frac{16}{31}$$

Answer: $\dfrac{16}{31}$. This problem contains Example 16: its denominator $\frac{31}{70}$ is exactly the total probability computed there.

Example 24 — The alternating game

$A$ and $B$ throw a die alternately until one of them gets a six and wins. $A$ throws first. Find their respective probabilities of winning.

Step 1 — list the ways $A$ can win: on throw 1, or throw 3 (after two failures), or throw 5, … With $P(S) = \frac{1}{6}$, $P(F) = \frac{5}{6}$ and throws independent: $$P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \frac{1}{6} + \left(\frac{5}{6}\right)^4 \frac{1}{6} + \cdots$$

Step 2 — sum the infinite GP with first term $a = \frac{1}{6}$ and ratio $r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$: $$P(A \text{ wins}) = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{6}{11}$$

Step 3 — complement: $P(B \text{ wins}) = 1 - \frac{6}{11} = \frac{5}{11}$.

Answer: $A$: $\dfrac{6}{11}$, $B$: $\dfrac{5}{11}$. Going first is worth a $\frac{1}{11}$ edge.

Example 25 — System reliability

An electronic assembly has two subsystems A and B, with $P(\text{A fails}) = 0.2$, $P(\text{B fails alone}) = 0.15$ and $P(\text{A and B fail}) = 0.15$. Find (i) $P(\text{A fails} \mid \text{B has failed})$ and (ii) $P(\text{A fails alone})$.

Step 1 — reconstruct $P(\text{B fails})$: "B fails" splits into "B fails alone" and "both fail": $$P(B_f) = 0.15 + 0.15 = 0.30$$

Step 2 — (i) condition: $$P(A_f | B_f) = \frac{P(A_f \cap B_f)}{P(B_f)} = \frac{0.15}{0.30} = \frac{1}{2}$$

Step 3 — (ii) subtract the overlap: $$P(A_f \text{ alone}) = P(A_f) - P(A_f \cap B_f) = 0.2 - 0.15 = 0.05$$

Answer: (i) $\dfrac{1}{2}$, (ii) $0.05$. The phrase "fails alone" is the whole problem — translate it as "fails and the other does not" before touching any formula.