An experiment whose outcome cannot be predicted with certainty in advance, although the set of all possible outcomes is known, is called a random experiment (or a trial).
Tossing a fair coin: outcomes = {Head, Tail}. We don't know in advance which one will occur.
Drawing a card from a well-shuffled pack: outcomes = 52 cards. The drawn card cannot be predicted.
The set of all possible outcomes of a random experiment is called the sample space.
Tossing two coins: \(S = \{HH, HT, TH, TT\}\); \(|S|=4\).
Throwing a die: \(S = \{1,2,3,4,5,6\}\); \(|S|=6\).
An event is any subset of the sample space \(S\).
Throwing one die. \(A\) = "even number", \(B\) = "odd number". \(A \cap B = \varnothing\); they are mutually exclusive and also exhaustive.
Drawing one card from a pack: each of the 52 cards is an equally likely outcome.
An experiment is any activity that produces a result, called its outcome: tossing a coin, measuring the length of a table.
Probability is about random experiments only. A deterministic experiment has a single possible outcome, which happens with certainty, so there is nothing to measure.
A trial is one performance of the experiment. Tossing three coins once is one trial of the experiment “toss three coins”; drawing one card from a pack is one trial of “draw a card”. If three players A, B and C play a game, the sample space of “who wins” is \(S = \{A, B, C\}\).
Source note. The textbook defines exhaustive events as “the total number of possible outcomes”. That number is the count of exhaustive cases; the events themselves are exhaustive when their union is \(S\), as above. It writes the complement as \(E^1\); the usual mark is a prime, \(E'\).
The theory of probability began in the 17th century with problems of gambling on dice, in the correspondence of Blaise Pascal and Pierre de Fermat (1654). James (Jacob) Bernoulli's Ars Conjectandi (published in 1713, after his death) proved the first law of large numbers, and Abraham De Moivre's The Doctrine of Chances appeared in 1718. Bayes, Laplace, Chebyshev, Markov, Lyapunov, Lévy, von Mises, Fisher and Khinchin developed it further, and in 1933 Kolmogorov gave it the axioms of §2.3.
Source note. The textbook names De Moivre and Pascal as the two main contributors at the start. The founding pair were Pascal and Fermat; De Moivre's work came half a century later.
If a random experiment has \(n\) mutually exclusive, exhaustive and equally likely outcomes, of which \(m\) are favourable to event \(A\), then
Limitations: assumes all outcomes equally likely; fails when \(n\) is infinite or outcomes are not equally likely.
Probability of getting a king from a pack of 52 cards.
Favourable = 4 kings, total = 52. \(P(K) = 4/52 = 1/13\).
Probability of getting a sum of 7 on two dice.
Total outcomes = 36. Favourable: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6. \(P = 6/36 = 1/6\).
If a random experiment is repeated \(N\) times under identical conditions and event \(A\) occurs \(m\) times, then
A coin is tossed 10 000 times and 5 037 heads are observed. The empirical probability of head is \(5037/10000 = 0.5037\), close to the theoretical value 0.5.
From last 1 000 customers, 320 ordered tea. Empirical probability that a new customer orders tea = 0.32.
Let \(S\) be a sample space and \(\mathcal{F}\) a \(\sigma\)-field of events. A probability measure \(P\) is a function \(P : \mathcal{F} \to \mathbb{R}\) satisfying three axioms:
When \(S\) is finite (or countable), \(\mathcal{B}\) can be the class of all subsets of \(S\), and every subset is an event. That is the setting of every problem on this page.
Source note. The textbook describes \(\mathcal{B}\) as “the Borel \(\sigma\)-field, the set of all subsets of \(S\)”. For a finite sample space all subsets will do. For \(S = \mathbb{R}\) the Borel \(\sigma\)-field is the one generated by the intervals, and it is strictly smaller than the class of all subsets; that is why the \(\sigma\)-field is named in the definition. The textbook also states the axioms as \(P(E_i) > 0\) and with additivity for finitely many events. The first must be \(P(E) \ge 0\): the impossible event has probability 0 (Result 1 below). The third is Kolmogorov's countable additivity; additivity for \(n\) events follows from it by taking \(E_{n+1} = E_{n+2} = \cdots = \varnothing\).
Result 1: \(P(\varnothing) = 0\). \(S = S \cup \varnothing\), and \(S\) and \(\varnothing\) are disjoint, so by additivity and certainty
\[ 1 = P(S) = P(S) + P(\varnothing) = 1 + P(\varnothing) \quad\Rightarrow\quad P(\varnothing) = 0 . \]Result 2: \(P(\bar A) = 1 - P(A)\). \(A \cup \bar A = S\) with \(A\) and \(\bar A\) disjoint, so \(P(A) + P(\bar A) = P(S) = 1\).
Result 3: \(P(\bar A \cap B) = P(B) - P(A \cap B)\). Split \(B\) by whether \(A\) happens: \(B = (A \cap B) \cup (\bar A \cap B)\), a union of two disjoint events (Fig 1.1, in §5). By additivity \(P(B) = P(A \cap B) + P(\bar A \cap B)\); subtract. In the same way, \(P(A \cap \bar B) = P(A) - P(A \cap B)\).
Monotonicity. If \(A \subset B\) then \(A \cap B = A\), and Result 3 gives \(P(B) - P(A) = P(\bar A \cap B) \ge 0\). In particular \(A \subset S\) gives \(0 \le P(A) \le 1\).
Result 9: \(P(A \cap B) \le P(A) \le P(A \cup B) \le P(A) + P(B)\).
Source note. The textbook's proof of Result 9 says “since \(P(A \cap \bar B) > 0\)”. It is \(\ge 0\): when \(A \subset B\), \(A \cap \bar B\) is empty and the first inequality is an equality.
The conditional probability of event \(A\) given that event \(B\) has occurred (\(P(B) > 0\)) is
A die is rolled. Given that the outcome is even, what is the probability that it is 4?
\(B\) = even = {2,4,6}; \(A \cap B\) = {4}.
\(P(A|B) = (1/6)/(3/6) = 1/3\).
In a class of 60 students, 30 study Math, 24 study Statistics and 12 study both.
Find probability that a randomly selected student studies Statistics, given that he studies Math.
\(P(\text{Stat}|\text{Math}) = P(\text{Stat} \cap \text{Math})/P(\text{Math}) = (12/60)/(30/60) = 12/30 = 0.40\).
When the \(n(S)\) outcomes are equally likely, \(P(E) = n(E)/n(S)\) for every event, and the definition becomes a count:
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{n(A \cap B)/n(S)}{n(B)/n(S)} = \frac{n(A \cap B)}{n(B)} . \]The \(n(S)\) cancels: once \(B\) is known to have happened, \(B\) is the new sample space, and \(A\) is favoured by the outcomes of \(B\) that are also in \(A\). In Example 1 above, \(n(A \cap B) = 1\) and \(n(B) = 3\), so \(P(A \mid B) = 1/3\) at once. The counting form holds only for equally likely outcomes; the ratio of probabilities holds always.
For a fixed event \(C\) with \(P(C) > 0\), \(P(\,\cdot \mid C)\) obeys the same rules as \(P\).
Result 4: \(P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C)\). By the distributive law \((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\). Apply the addition theorem to the events \(A \cap C\) and \(B \cap C\), whose intersection is \((A \cap C) \cap (B \cap C) = A \cap B \cap C\):
\[ P[(A \cup B) \cap C] = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) . \]Divide every term by \(P(C)\); each ratio is a conditional probability.
Result 5: \(P(A \cap \bar B \mid C) + P(A \cap B \mid C) = P(A \mid C)\). \(A \cap C\) splits into the disjoint pieces \(A \cap \bar B \cap C\) and \(A \cap B \cap C\), so \(P(A \cap \bar B \cap C) + P(A \cap B \cap C) = P(A \cap C)\). Divide by \(P(C)\).
Source note. The textbook's proof of Result 4 writes the intersection as \(P[(A \cap B) \cap (B \cap C)]\); it is \((A \cap C) \cap (B \cap C)\), which gives the same \(A \cap B \cap C\). The condition \(P(C) > 0\), needed to divide, is not stated there.
Two events \(A\) and \(B\) are independent if the occurrence of one does not affect the probability of the other:
Equivalently, \(P(A|B) = P(A)\) and \(P(B|A) = P(B)\), provided both probabilities are positive.
For three events \(A,B,C\) — mutual independence requires:
Two coins are tossed. \(A\) = "first coin H", \(B\) = "second coin H". \(P(A)=P(B)=1/2\), \(P(A \cap B) = 1/4 = P(A)P(B)\). Independent ✓.
A die is rolled. \(A\) = "even", \(B\) = "≥ 4".
\(P(A) = 1/2,\; P(B) = 1/2,\; A \cap B = \{4,6\},\; P(A \cap B) = 2/6 = 1/3\).
\(P(A)P(B) = 1/4 \ne 1/3\). Hence not independent.
The two forms of the definition agree: if \(P(B \mid A) = P(B)\), the multiplication theorem (§6) gives \(P(A \cap B) = P(A)P(B \mid A) = P(A)P(B)\), and in the same way \(P(A \mid B) = P(A)\) gives the product. If \(A\) and \(B\) are independent, so are their complements, in every combination.
Result 6: \(A\) and \(\bar B\). By Result 3 and independence,
\[ P(A \cap \bar B) = P(A) - P(A \cap B) = P(A) - P(A)P(B) = P(A)\,[1 - P(B)] = P(A)\,P(\bar B) . \]Result 7: \(\bar A\) and \(B\). The same with the roles exchanged: \(P(\bar A \cap B) = P(B) - P(A)P(B) = P(B)\,P(\bar A)\).
Result 8: \(\bar A\) and \(\bar B\). By De Morgan's law \(\bar A \cap \bar B = \overline{A \cup B}\), so
\[ P(\bar A \cap \bar B) = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A)P(B) = [1 - P(A)]\,[1 - P(B)] = P(\bar A)\,P(\bar B) . \]The last step is the factorisation \(1 - a - b + ab = (1 - a)(1 - b)\).
Independent is not the same as mutually exclusive. If \(A\) and \(B\) are mutually exclusive with \(P(A), P(B) > 0\), then \(P(A \cap B) = 0 \ne P(A)P(B)\): knowing that \(A\) happened tells you that \(B\) did not, so they are dependent.
If \(A,B\) are mutually exclusive: \(P(A \cup B) = P(A) + P(B)\).
From a deck, find the probability of drawing a king or a heart.
\(P(K) = 4/52,\; P(H) = 13/52,\; P(K \cap H) = 1/52\).
\(P(K \cup H) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13\).
Probability that a student passes Math is 0.7, Physics 0.6 and both 0.5. Probability of passing at least one = 0.7 + 0.6 − 0.5 = 0.8.
Write \(A \cup B\) as the union of two disjoint events, \(A\) and \(\bar A \cap B\) (Fig 1.1):
\[ A \cup B = A \cup (\bar A \cap B) \quad\Rightarrow\quad P(A \cup B) = P(A) + P(\bar A \cap B) \]by additivity. By Result 3 (§2), \(P(\bar A \cap B) = P(B) - P(A \cap B)\). Substituting,
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) . \;\square \]In words: adding \(P(A)\) and \(P(B)\) counts the overlap \(A \cap B\) twice, so it is taken away once.
Write \(S_k(A_1, \ldots, A_m)\) for the sum of \(P(A_{i_1} \cap \cdots \cap A_{i_k})\) over all choices \(i_1 < \cdots < i_k\) of \(k\) of the \(m\) events. The theorem is
\[ P\!\left(\bigcup_{i=1}^{n} A_i\right) = \sum_{k=1}^{n} (-1)^{k-1}\, S_k(A_1, \ldots, A_n) . \]\(n = 2\): \(S_1 = P(A_1) + P(A_2)\), \(S_2 = P(A_1 \cap A_2)\); this is the two-event theorem.
From \(r\) to \(r + 1\). Let \(U = A_1 \cup \cdots \cup A_r\). By the two-event theorem and the distributive law \(U \cap A_{r+1} = \bigcup_{i=1}^{r} (A_i \cap A_{r+1})\),
\[ P(U \cup A_{r+1}) = P(U) + P(A_{r+1}) - P\!\left(\bigcup_{i=1}^{r} (A_i \cap A_{r+1})\right). \]Both unions on the right have \(r\) members, so the hypothesis applies to each. Write \(T_k\) for \(S_k(A_1 \cap A_{r+1}, \ldots, A_r \cap A_{r+1})\): the sum over the \(k\)-fold intersections of \(A_1, \ldots, A_r\), each intersected with \(A_{r+1}\). Then, with \(T_0 = P(A_{r+1})\),
\[ P(U \cup A_{r+1}) = \sum_{k=1}^{r} (-1)^{k-1} S_k(A_1, \ldots, A_r) + T_0 + \sum_{k=1}^{r} (-1)^{k} T_k . \]The regrouping. A choice of \(k\) events from \(A_1, \ldots, A_{r+1}\) either leaves out \(A_{r+1}\) (those terms make up \(S_k(A_1, \ldots, A_r)\)) or includes it, together with \(k - 1\) of the others (those make up \(T_{k-1}\)). So \(S_k(A_1, \ldots, A_{r+1}) = S_k(A_1, \ldots, A_r) + T_{k-1}\). Shifting the index of the \(T\) sum by one, \((-1)^{k} T_k\) becomes \((-1)^{k-1} T_{k-1}\), and the right-hand side is
\[ \sum_{k=1}^{r+1} (-1)^{k-1}\left[S_k(A_1, \ldots, A_r) + T_{k-1}\right] = \sum_{k=1}^{r+1} (-1)^{k-1} S_k(A_1, \ldots, A_{r+1}), \]with \(S_{r+1}(A_1, \ldots, A_r) = 0\). That is the theorem for \(r + 1\) events; by induction it holds for every \(n\). \(\square\)
Source note. The textbook's proof passes over the regrouping in one line, and prints the last term as \(P(A_i \cap A_j \cap \cdots \cap A_{r+1})\); it is \(P(A_1 \cap A_2 \cap \cdots \cap A_{r+1})\).
If \(A,B\) are independent: \(P(A \cap B) = P(A) P(B)\).
If events are mutually independent: \(P\!\left(\bigcap A_i\right) = \prod P(A_i)\).
Two cards are drawn one after another without replacement. Probability both are aces.
\(P(A_1) = 4/52,\; P(A_2|A_1) = 3/51\). \(P(A_1 \cap A_2) = (4/52)(3/51) = 1/221\).
Probability of three independent machines functioning is 0.9, 0.8, 0.7. Probability all three function = 0.9 × 0.8 × 0.7 = 0.504.
For \(P(A) > 0\) the definition \(P(B \mid A) = P(A \cap B)/P(A)\) multiplied through by \(P(A)\) is \(P(A \cap B) = P(A)\,P(B \mid A)\); with the roles exchanged, \(P(A \cap B) = P(B)\,P(A \mid B)\) for \(P(B) > 0\).
With equally likely outcomes the same thing can be seen by counting. Multiply and divide by \(n(A)\):
\[ P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{n(A)}{n(S)} \cdot \frac{n(A \cap B)}{n(A)} = P(A)\,P(B \mid A) . \]Source note. The textbook prints the theorem as “\(P(A \cap B) = P(A), P(B \mid A)\)”; the comma is a product, \(P(A) \cdot P(B \mid A)\).
\(n = 2\) is the theorem above. From \(r\) to \(r + 1\): treat \(A_1 \cap \cdots \cap A_r\) as one event and apply the two-event theorem, then the hypothesis:
\[ \begin{aligned} P(A_1 \cap \cdots \cap A_r \cap A_{r+1}) &= P(A_1 \cap \cdots \cap A_r)\; P(A_{r+1} \mid A_1 \cap \cdots \cap A_r) \\ &= P(A_1)\, P(A_2 \mid A_1) \cdots P(A_r \mid A_1 \cap \cdots \cap A_{r-1})\; P(A_{r+1} \mid A_1 \cap \cdots \cap A_r) . \end{aligned} \]So the theorem holds for \(r + 1\) events, and by induction for all \(n\). The condition \(P(A_1 \cap \cdots \cap A_{n-1}) > 0\) makes every conditional probability in the product defined. \(\square\)
For any events \(A_1, A_2, \ldots, A_n\) (not necessarily disjoint or independent):
Equivalently for the complement (Bonferroni's form):
\[ P\!\left(\bigcap_{i=1}^{n} A_i\right) \;\ge\; 1 - \sum_{i=1}^{n} P(\bar{A}_i) \]If \(P(A) = 0.4, P(B) = 0.5, P(C) = 0.3\), then \(P(A \cup B \cup C) \le 0.4 + 0.5 + 0.3 = 1.2\) (so at most 1).
If three events have probabilities 0.05, 0.04, 0.03 (rare failures), the probability that at least one occurs is at most 0.12. This is widely used in reliability engineering.
(i) The union form, \(P\!\left(\bigcup_{i=1}^{n} A_i\right) \le \sum_{i=1}^{n} P(A_i)\). For \(n = 2\), \(P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) \le P(A_1) + P(A_2)\), since \(P(A_1 \cap A_2) \ge 0\). If it holds for \(r\) events, then, using the case \(n = 2\) on \(\bigcup_{i=1}^{r} A_i\) and \(A_{r+1}\) and then the hypothesis,
\[ P\!\left(\bigcup_{i=1}^{r+1} A_i\right) \le P\!\left(\bigcup_{i=1}^{r} A_i\right) + P(A_{r+1}) \le \sum_{i=1}^{r} P(A_i) + P(A_{r+1}) = \sum_{i=1}^{r+1} P(A_i) . \](ii) The intersection form, \(P\!\left(\bigcap_{i=1}^{n} A_i\right) \ge \sum_{i=1}^{n} P(A_i) - (n - 1)\). For \(n = 2\), \(P(A_1 \cup A_2) \le 1\) gives \(P(A_1) + P(A_2) - P(A_1 \cap A_2) \le 1\), that is,
\[ P(A_1 \cap A_2) \ge P(A_1) + P(A_2) - 1 . \]If it holds for \(r\) events, apply the case \(n = 2\) to \(\bigcap_{i=1}^{r} A_i\) and \(A_{r+1}\), then the hypothesis:
\[ P\!\left(\bigcap_{i=1}^{r+1} A_i\right) \ge P\!\left(\bigcap_{i=1}^{r} A_i\right) + P(A_{r+1}) - 1 \ge \sum_{i=1}^{r} P(A_i) - (r - 1) + P(A_{r+1}) - 1 = \sum_{i=1}^{r+1} P(A_i) - r . \]This is the Bonferroni form stated above, written another way: since \(P(\bar A_i) = 1 - P(A_i)\), \(1 - \sum_{i=1}^{n} P(\bar A_i) = 1 - n + \sum_{i=1}^{n} P(A_i) = \sum_{i=1}^{n} P(A_i) - (n - 1)\). \(\square\)
Source note. In the induction step of (ii), the textbook writes “\(=\)” where the case \(n = 2\) gives only “\(\ge\)”.
Let \(B_1, B_2, \ldots, B_n\) be mutually exclusive and exhaustive events (a partition of \(S\)) with \(P(B_i) > 0\), and let \(A\) be any event with \(P(A) > 0\).
Prior \(P(B_k)\) updated by data \(A\) gives the posterior \(P(B_k|A)\).
Bayes' theorem is just conditional probability with its numerator and denominator each rewritten:
\[ P(B_k \mid A) = \frac{P(B_k \cap A)}{P(A)} = \frac{\overbrace{P(B_k)P(A\mid B_k)}^{\text{multiplication rule}}} {\underbrace{\sum_i P(B_i)P(A\mid B_i)}_{\text{total probability}}}. \]
The numerator is the multiplication theorem (Section 6); the denominator is the total-probability law above. Bayes therefore introduces nothing new — it reverses the conditioning from \(P(A\mid B_k)\) (which we can measure) to \(P(B_k\mid A)\) (which we want).
Three machines \(M_1,M_2,M_3\) produce 25 %, 35 %, 40 % of items with defective rates 5 %, 4 %, 2 % respectively. A random item is found defective. Find probability it came from \(M_1\).
\(P(D) = 0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345\).
\(P(M_1 | D) = 0.0125 / 0.0345 = 0.362\) (≈ 36.2 %).
1 % of population has a disease. Test sensitivity = 0.99 (true positive), false-positive rate = 0.05. Given a positive test, what is the probability the person has the disease?
\(P(D|+) = \dfrac{(0.01)(0.99)}{(0.01)(0.99) + (0.99)(0.05)} = \dfrac{0.0099}{0.0594} \approx 0.167\).
Despite the test being 99 % sensitive, only ~17 % of positives are truly diseased — the rare base rate matters!
The proof needs a little less than a partition: \(E_1, \ldots, E_n\) mutually exclusive with \(P(E_i) > 0\), and an event \(A\) with \(P(A) > 0\) that lies inside \(E_1 \cup \cdots \cup E_n\). (A partition of \(S\) satisfies this for every \(A\).)
Step 1: split \(A\). Since \(A \subset \bigcup_i E_i\),
\[ A = A \cap \bigcup_{i=1}^{n} E_i = (A \cap E_1) \cup (A \cap E_2) \cup \cdots \cup (A \cap E_n), \]by the distributive law. The \(E_i\) are disjoint, so the pieces \(A \cap E_i\) are disjoint too.
Step 2: total probability. By additivity, then the multiplication theorem \(P(A \cap E_i) = P(E_i)\,P(A \mid E_i)\),
\[ P(A) = \sum_{i=1}^{n} P(A \cap E_i) = \sum_{i=1}^{n} P(E_i)\,P(A \mid E_i) . \]Step 3: reverse the conditioning.
\[ P(E_i \mid A) = \frac{P(A \cap E_i)}{P(A)} = \frac{P(E_i)\,P(A \mid E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A \mid E_j)}, \qquad i = 1, \ldots, n . \;\square \]The summation index in the denominator is a dummy, \(j\), so that it does not clash with the fixed \(i\) in the numerator.
Twenty-two problems in the order the textbook sets them: nine on the classical definition, four on the addition theorem, six on the multiplication theorem and independence, and three on Bayes' theorem, followed by the exercises with their answers checked. Each one names its sample space, counts (or multiplies) the favourable cases, and states the answer as a fraction in lowest terms. The counting tools are the ones from algebra: \(^{n}C_r = \binom{n}{r}\) ways to choose \(r\) things from \(n\), and \(n!\) ways to arrange \(n\) different things.
Source note. Every answer below was recomputed exactly, as a fraction, and wherever the sample space is small enough to list (coins, dice, cards, letters, urns) by listing it and counting. Twenty-one of the twenty-two worked answers agree exactly. The twenty-second (Worked Problem 22) differs in the third decimal place, a rounding artefact of the book's four-place intermediates, and is left standing with a note. Two of the exercise answers are wrong and are corrected.
Two coins are tossed. Find the probability of at least one head.
\(S = \{HH, HT, TH, TT\}\), \(n = 4\) equally likely outcomes. Let \(E\) = at least one head. Its complement is easier: \(\bar E\) = no head \(= \{TT\}\), \(m = 1\).
\[ P(\bar E) = \frac{1}{4}, \qquad P(E) = 1 - P(\bar E) = 1 - \frac14 = \frac34 . \]Two dice are thrown. Find the probability that the sum is 10.
Each die has 6 faces, so \(n = 6 \times 6 = 36\) ordered pairs. Sum 10: \(E = \{(4,6), (5,5), (6,4)\}\), \(m = 3\).
\[ P(E) = \frac{3}{36} = \frac{1}{12} . \]Note that \((4,6)\) and \((6,4)\) are different outcomes: the dice are told apart (say, red and blue), which is what makes the 36 outcomes equally likely.
A bag contains 3 red, 6 white and 7 blue balls. Two balls are drawn. Find the probability that they are one white and one red.
16 balls; two can be chosen in \(n = \binom{16}{2} = \dfrac{16 \times 15}{2} = 120\) ways. One white from 6 and one red from 3: \(m = \binom61\binom31 = 6 \times 3 = 18\).
\[ P(E) = \frac{18}{120} = \frac{3}{20} . \]Two cards are drawn from a well-shuffled pack of 52. Find the probability that they are (i) two aces, (ii) a king and a queen.
\(n = \binom{52}{2} = \dfrac{52 \times 51}{2} = 1326\).
(i) Two of the 4 aces: \(m = \binom42 = 6\), so \(P = \dfrac{6}{1326} = \dfrac{1}{221}\).
(ii) One of 4 kings and one of 4 queens: \(m = \binom41\binom41 = 16\), so \(P = \dfrac{16}{1326} = \dfrac{8}{663}\).
Check of (i) by the multiplication theorem (§6): \(\dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221}\), as in §6's Example 1.
Find the probability that a leap year has 53 Sundays.
366 days \(= 52\) weeks \(+ 2\) days. The 52 weeks give 52 Sundays whatever happens; a 53rd needs one of the two extra, consecutive days to be a Sunday. The pair can be
(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun),
\(n = 7\), taken as equally likely. Two contain a Sunday: (Sun, Mon) and (Sat, Sun), so \(m = 2\) and
\[ P = \frac{2}{7} . \]The model takes the year to start on any day of the week with equal chance. (Over the Gregorian calendar's 400-year cycle the actual share is \(28/97 \approx 0.289\), close to \(2/7 \approx 0.286\).)
Five-digit numbers are formed from the digits 0, 1, 2, 3, 4, each used once. Find the probability that 2 is in the tens place and 0 in the units place.
The digits can be arranged in \(5! = 120\) ways, but the \(4! = 24\) arrangements starting with 0 are not five-digit numbers. So \(n = 5! - 4! = 96\) (equivalently, 4 choices for the first digit, then \(4!\) for the rest: \(4 \times 4! = 96\)).
Fix 2 in the tens place and 0 in the units place; 1, 3, 4 fill the first three places in \(m = 3! = 6\) ways, none of them starting with 0.
\[ P = \frac{6}{96} = \frac{1}{16} . \]Four letters are placed at random in four addressed envelopes. Find the probability that at least one letter is in the wrong envelope.
\(n = 4! = 24\) ways to place the letters. The complement, “every letter is in its own envelope”, happens in exactly \(m = 1\) way.
\[ P(\text{at least one wrong}) = 1 - \frac{1}{24} = \frac{23}{24} . \]The letters of MISSISSIPPI are arranged at random. Find the probability that the four S's come together.
11 letters: M once, I four times, S four times, P twice. Arrangements of letters with repeats divide out the orders of the identical letters:
\[ n = \frac{11!}{4!\,4!\,2!} = \frac{39916800}{24 \times 24 \times 2} = 34650 . \]Glue the four S's into one block. Then there are 8 units (the block, M, four I's, two P's):
\[ m = \frac{8!}{4!\,2!} = \frac{40320}{48} = 840, \qquad P = \frac{840}{34650} = \frac{4}{165} . \]The letters of UNIVERSITY are arranged at random. Find the probability that the two I's do not come together.
10 letters with I twice: \(n = 10!/2!\). For the complement, glue the two I's together: 9 units, all different, \(9!\) arrangements.
\[ P(\text{together}) = \frac{9!}{10!/2!} = \frac{2}{10} = \frac15, \qquad P(\text{apart}) = 1 - \frac15 = \frac45 . \]A quicker view: only the two positions of the I's matter. Of the \(\binom{10}{2} = 45\) pairs of positions, 9 are next to each other, and \(9/45 = 1/5\).
Two dice are thrown. Find the probability that the sum is either 10 or 11.
\(n(S) = 36\). \(A\) = sum 10 \(= \{(4,6), (5,5), (6,4)\}\), \(n(A) = 3\); \(B\) = sum 11 \(= \{(5,6), (6,5)\}\), \(n(B) = 2\). No pair has both sums, so \(A \cap B = \varnothing\).
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{36} + \frac{2}{36} - 0 = \frac{5}{36} . \]A card is drawn from a pack of 52. Find the probability that it is red or an ace.
\(A\) = red: 26 cards; \(B\) = ace: 4 cards; \(A \cap B\) = red ace: 2 cards (hearts and diamonds).
\[ P(A \cup B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13} . \]Without the subtraction the two red aces would be counted twice.
A ticket is drawn at random from 20 tickets numbered 1 to 20. Find the probability that its number is divisible by 3 or 4.
\(A = \{3, 6, 9, 12, 15, 18\}\), \(B = \{4, 8, 12, 16, 20\}\), \(A \cap B = \{12\}\) (divisible by both, i.e. by 12).
\[ P(A \cup B) = \frac{6}{20} + \frac{5}{20} - \frac{1}{20} = \frac{10}{20} = \frac12 . \]A box contains 6 white and 4 green balls. Two are drawn at random. Find the probability that they are both white or both green.
\(n = \binom{10}{2} = 45\). \(A\) = both white: \(\binom62 = 15\); \(B\) = both green: \(\binom42 = 6\). Two balls cannot be both white and both green, so \(A \cap B = \varnothing\).
\[ P(A \cup B) = \frac{15}{45} + \frac{6}{45} = \frac{21}{45} = \frac{7}{15} . \]A bag contains 4 red and 5 black balls. Two are drawn in succession. Find the probability that both are red if the first ball is (i) replaced, (ii) not replaced.
Let \(A\) = red at the first draw, \(B\) = red at the second.
(i) Replaced. The bag is the same for both draws, so \(A\) and \(B\) are independent: \(P(A) = P(B) = 4/9\) and
\[ P(A \cap B) = P(A)\,P(B) = \frac49 \cdot \frac49 = \frac{16}{81} . \](ii) Not replaced. After a red ball is taken, 8 balls remain, 3 of them red: \(P(B \mid A) = 3/8\). By the multiplication theorem,
\[ P(A \cap B) = P(A)\,P(B \mid A) = \frac49 \cdot \frac38 = \frac{12}{72} = \frac16 . \]Printing note. The textbook writes (i) as \(P(A \cup B) = P(A) \cdot P(B)\); it is \(P(A \cap B)\).
Three students are selected one after another from a class of 12 boys and 8 girls. Find the probability that all three are girls.
\(A\), \(B\), \(C\) = a girl at the first, second, third selection. Each selection leaves one girl and one student fewer:
\[ P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B) = \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} = \frac{336}{6840} = \frac{14}{285} . \]Check by counting: \(\binom83 \big/ \binom{20}{3} = 56/1140 = 14/285\).
Two aeroplanes attack a target in turn, hitting it with probabilities 0.3 and 0.2. The second attacks only if the first misses. Find the probability that (i) the target is hit, (ii) both fail.
\(A\) = first hits, \(B\) = second hits, the two attempts independent.
(i) The target is hit if the first hits, or the first misses and the second hits. These are mutually exclusive, so
\[ P = P(A) + P(\bar A \cap B) = P(A) + P(\bar A)\,P(B) = 0.3 + (0.7)(0.2) = 0.44 . \](ii) \(P(\bar A \cap \bar B) = P(\bar A)\,P(\bar B) = (0.7)(0.8) = 0.56\) (Result 8). As a check, (i) and (ii) are complements: \(0.44 + 0.56 = 1\).
Three students A, B and C, working independently, solve a problem with probabilities \(\tfrac12\), \(\tfrac34\) and \(\tfrac14\). Find the probability that it is solved.
It is solved if at least one solves it: \(P(A \cup B \cup C)\). By the addition theorem for three events, with independence turning each intersection into a product,
\[ P(A \cup B \cup C) = \frac12 + \frac34 + \frac14 - \frac12\cdot\frac34 - \frac34\cdot\frac14 - \frac12\cdot\frac14 + \frac12\cdot\frac34\cdot\frac14 . \]Over the common denominator 32,
\[ P(A \cup B \cup C) = \frac{16 + 24 + 8 - 12 - 6 - 4 + 3}{32} = \frac{29}{32} . \]Another way. It is unsolved only if all three fail, and (Result 8) the failures are independent too:
\[ P(\bar A \cap \bar B \cap \bar C) = \frac12 \cdot \frac14 \cdot \frac34 = \frac{3}{32}, \qquad P(A \cup B \cup C) = 1 - \frac{3}{32} = \frac{29}{32} . \]A box contains 6 red, 4 white and 5 black balls. Four are drawn at random. Find the probability that there is at least one of each colour.
\(n = \binom{15}{4} = 1365\). Four balls with all three colours means one colour twice and the others once — three disjoint cases:
\(P(A) = \tfrac34\) and \(P(B) = \tfrac58\). Show that (i) \(P(A \cup B) \ge \tfrac34\), (ii) \(\tfrac38 \le P(A \cap B) \le \tfrac58\).
(i) \(A \subset A \cup B\), so by monotonicity (§2) \(P(A \cup B) \ge P(A) = \tfrac34\).
(ii) \(A \cap B \subset B\), so \(P(A \cap B) \le P(B) = \tfrac58\). For the lower bound, \(P(A \cup B) \le 1\) and the addition theorem give
\[ P(A \cap B) \ge P(A) + P(B) - 1 = \frac34 + \frac58 - 1 = \frac38 , \]which is Boole's inequality in its intersection form (§7) with \(n = 2\). The two probabilities add to more than 1, so the events must overlap by at least the excess, \(\tfrac38\).
Urn I holds 1 white, 2 black and 3 red balls; urn II 2 white, 1 black and 1 red; urn III 4 white, 5 black and 3 red. An urn is chosen at random and two balls drawn: one white, one red. Find the probability that they came from each urn.
\(E_1, E_2, E_3\) = urn I, II, III chosen, each with prior \(\tfrac13\). \(A\) = one white and one red.
\[ P(A \mid E_1) = \frac{\binom11\binom31}{\binom62} = \frac{3}{15} = \frac15, \quad P(A \mid E_2) = \frac{\binom21\binom11}{\binom42} = \frac26 = \frac13, \quad P(A \mid E_3) = \frac{\binom41\binom31}{\binom{12}{2}} = \frac{12}{66} = \frac{2}{11} . \]Total probability:
\[ P(A) = \frac13\left(\frac15 + \frac13 + \frac{2}{11}\right) = \frac13 \cdot \frac{33 + 55 + 30}{165} = \frac{118}{495} . \]Bayes' theorem divides each path by this total; the common \(\tfrac13\) and \(\tfrac{1}{165}\) cancel, leaving the numerators 33, 55, 30:
\[ P(E_1 \mid A) = \frac{33}{118}, \qquad P(E_2 \mid A) = \frac{55}{118}, \qquad P(E_3 \mid A) = \frac{30}{118} = \frac{15}{59} . \]They add to 1, as they must. Urn II, the most likely source, is the one where a white-red pair is easiest to draw.
Printing note. The textbook's statement reads “urns I, II and II”; the third is urn III.
Machines A, B, C make 20%, 30% and 50% of a factory's bolts; 6%, 3% and 2% of their output is defective. A bolt drawn at random is defective. Find the probability that it was made by each machine.
Priors \(P(E_1) = 0.2\), \(P(E_2) = 0.3\), \(P(E_3) = 0.5\); likelihoods \(P(A \mid E_i) = 0.06, 0.03, 0.02\).
\[ P(A) = (0.2)(0.06) + (0.3)(0.03) + (0.5)(0.02) = 0.012 + 0.009 + 0.010 = 0.031 , \] \[ \begin{aligned} P(E_1 \mid A) &= \frac{0.012}{0.031} = \frac{12}{31} = 0.3871, \\ P(E_2 \mid A) &= \frac{0.009}{0.031} = \frac{9}{31} = 0.2903, \\ P(E_3 \mid A) &= \frac{0.010}{0.031} = \frac{10}{31} = 0.3226 . \end{aligned} \]Machine A makes the fewest bolts but is the most likely culprit: its defect rate is three times C's.
Printing note. The textbook writes \(P(E) = 0.5\) for \(P(E_3) = 0.5\).
Urn I holds 46 red, 28 white and 26 black balls (100); urn II 28 red, 46 white and 15 black (89); urn III 42 red, 56 white and 64 black (162). An urn is chosen at random and 8 balls drawn: 5 red and 3 white. Find the probability that they came from urn I, and from urn III.
Priors \(\tfrac13\) each. The likelihoods choose 5 of the reds and 3 of the whites out of all \(\binom{N}{8}\) samples:
\[ P(A \mid E_1) = \frac{\binom{46}{5}\binom{28}{3}}{\binom{100}{8}} = 0.0241, \quad P(A \mid E_2) = \frac{\binom{28}{5}\binom{46}{3}}{\binom{89}{8}} = 0.0211, \quad P(A \mid E_3) = \frac{\binom{42}{5}\binom{56}{3}}{\binom{162}{8}} = 0.0024 , \] \[ P(A) = \tfrac13(0.0241 + 0.0211 + 0.0024) = 0.0159 . \]Exactly, without rounding the intermediates,
\[ P(E_1 \mid A) = 0.5065, \qquad P(E_2 \mid A) = 0.4434, \qquad P(E_3 \mid A) = 0.0501, \]and the probability that the balls came from urn I or urn III is \(0.5065 + 0.0501 = 0.5566\).
Rounding note. The textbook gives \(P(E_1 \mid A) = 0.5052\) and \(P(E_3 \mid A) = 0.0503\). Those follow exactly from its four-place intermediates (\(\tfrac13(0.0241)/0.0159\) and \(\tfrac13(0.0024)/0.0159\)). Rounding the likelihoods and \(P(A) = 0.015882\) to four places is enough to move the third decimal of the answers.