Solution. Let \(R\) be the event that the order is ready on time and \(D\) that it is delivered on time. We are given \(P(R)=0.80\) and \(P(R\cap D)=0.72\). By the definition of conditional probability, \[P(D\mid R) = \frac {P(R\cap D)}{P(R)} = \frac {0.72}{0.80} = 0.9.\]

Solution. The sets are picked in succession without replacement, so the second probability is conditional on the first. Let \(D_1\) and \(D_2\) be the events that the first and second set are defective. \[P(D_1\cap D_2) = P(D_1)\,P(D_2\mid D_1) = \frac {15}{240}\times \frac {14}{239} = \frac {7}{1912} \approx 0.0037.\] After one defective set is removed, 14 of the remaining 239 are defective – that is what makes the second factor \(\tfrac {14}{239}\) and not \(\tfrac {15}{240}\).

Solution. It is easiest to find \(P(A\cap B)\) first and answer the rest from it.

(b). From \(P(B\mid A)=\frac {P(A\cap B)}{P(A)}\), \[P(A\cap B) = P(B\mid A)\,P(A) = \frac {2}{3}\times \frac {1}{4} = \frac {1}{6}.\]

(a). \(A\) and \(B\) are independent only if \(P(A\mid B)=P(A)\). Here \(P(A\mid B)=\tfrac 12\) but \(P(A)=\tfrac 14\), so they are not independent. (Equivalently \(P(A)P(B)=\tfrac 14\times \tfrac 13=\tfrac 1{12}\neq \tfrac 16=P(A\cap B)\).)

(c). Mutually exclusive would mean \(P(A\cap B)=0\). Here \(P(A\cap B)=\tfrac 16\neq 0\), so they are not mutually exclusive.

(d). From \(P(A\mid B)=\frac {P(A\cap B)}{P(B)}\), \[P(B) = \frac {P(A\cap B)}{P(A\mid B)} = \frac {1/6}{1/2} = \frac {1}{3}.\]

(e). By the addition rule, \[P(A\cup B) = P(A)+P(B)-P(A\cap B) = \frac {1}{4}+\frac {1}{3}-\frac {1}{6} = \frac {5}{12}.\]

Solution. (a). Independence gives \(P(A\cap B)=P(A)P(B)\), and \(A\) is then also independent of \(B'\). \begin {align*} \text {(i)}\quad P(A\cap B) &= \frac {2}{5}\times \frac {1}{4} = \frac {1}{10}.\\ \text {(ii)}\quad P(A\cap B') &= P(A)\,P(B') = \frac {2}{5}\times \frac {3}{4} = \frac {3}{10}.\\ \text {(iii)}\quad P(A\cup B) &= \frac {2}{5}+\frac {1}{4}-\frac {1}{10} = \frac {11}{20}. \end {align*}

(b). Mutually exclusive means \(A\cap B=\emptyset \), so \(P(A\cap B)=0\). \begin {align*} \text {(i)}\quad P(A\cup B) &= P(A)+P(B) = \frac {2}{5}+\frac {1}{4} = \frac {13}{20}.\\ \text {(ii)}\quad P(A'\cap B) &= P(B)-P(A\cap B) = \frac {1}{4}-0 = \frac {1}{4}. \end {align*}

Note that the two parts describe different situations: events cannot in general be both independent and mutually exclusive.

Solution. (a). The events \(A\cap B\) and \(A\cap B'\) partition \(A\), so \[P(A\cap B) = P(A)-P(A\cap B') = \frac {4}{7}-\frac {1}{3} = \frac {12}{21}-\frac {7}{21} = \frac {5}{21}.\]

(b). From \(P(A\mid B)=\frac {P(A\cap B)}{P(B)}\), \[P(B) = \frac {P(A\cap B)}{P(A\mid B)} = \frac {5/21}{5/14} = \frac {5}{21}\times \frac {14}{5} = \frac {2}{3}.\]

(c). \[P(B\mid A) = \frac {P(A\cap B)}{P(A)} = \frac {5/21}{4/7} = \frac {5}{21}\times \frac {7}{4} = \frac {5}{12}.\]

Solution. There are \(4+6+2=12\) balls, and each pick removes one.

(a). \[P(B_1\cap W_2) = \frac {4}{12}\times \frac {6}{11} = \frac {24}{132} = \frac {2}{11}.\]

(b). The second ball can be white either after a white first ball or after a non-white one: \[P(W_2) = \frac {6}{12}\cdot \frac {5}{11} + \frac {6}{12}\cdot \frac {6}{11} = \frac {30+36}{132} = \frac {66}{132} = \frac {1}{2}.\] This is just \(\tfrac {6}{12}\), the same as \(P(W_1)\) – and that is no accident. Before any ball is seen, the second position is as likely to hold any particular ball as the first is, so drawing without replacement does not favour one position over another.

(c). \[P(B_1\cap W_2\cap R_3) = \frac {4}{12}\times \frac {6}{11}\times \frac {2}{10} = \frac {48}{1320} = \frac {2}{55}.\]

(d). Three different colours means one black, one white and one red in some order. Each particular order has the same probability \(\tfrac {48}{1320}\) as in (c), and there are \(3!=6\) orders, so \[P = 6\times \frac {48}{1320} = \frac {288}{1320} = \frac {12}{55}.\]

Solution. The three agencies partition the sample space, so this is the law of total probability. Let \(A_1, A_2, A_3\) be the agencies and \(C\) the event that the car needs an oil change. \begin {align*} P(C) &= P(A_1)P(C\mid A_1)+P(A_2)P(C\mid A_2)+P(A_3)P(C\mid A_3)\\ &= (0.60)(0.09)+(0.30)(0.20)+(0.10)(0.06)\\ &= 0.054+0.060+0.006 = 0.12. \end {align*}

Solution. Note that \(0.01\) percent means \(P(D)=0.0001\), so \(P(D')=0.9999\). We are given \(P(p\mid D)=0.98\) and \(P(p\mid D')=0.03\). By the law of total probability, \begin {align*} P(p) &= P(D)P(p\mid D)+P(D')P(p\mid D')\\ &= (0.0001)(0.98)+(0.9999)(0.03)\\ &= 0.000098+0.029997 = 0.030095. \end {align*}

Then by Bayes’ theorem, \[P(D'\mid p) = \frac {P(D')P(p\mid D')}{P(p)} = \frac {0.029997}{0.030095} \approx 0.9967.\] Almost every positive result is a false alarm. This is not a fault in the test – it is what happens when a disease is rare: the 3% of a very large healthy group vastly outnumbers the 98% of a very small sick one.

Solution. (a). There are \(45+50+17 = 112\) females out of 200: \[P(F) = \frac {112}{200} = \frac {14}{25} = 0.56.\]

(b). Conditioning on secondary education restricts attention to that row, which has \(28+50=78\) people, of whom 28 are male: \[P(M\mid S) = \frac {28}{78} = \frac {14}{39} \approx 0.359.\]

(c). Conditioning on female restricts attention to that column, which has 112 people, of whom \(45+50=95\) are not college graduates: \[P(C'\mid F) = \frac {95}{112} \approx 0.848.\]

Solution. Let \(R\), \(S\) and \(L\) be the three motels and \(F\) the event of faulty plumbing.

(a). The three motels partition the clients, so by the law of total probability \begin {align*} P(F) &= P(R)P(F\mid R)+P(S)P(F\mid S)+P(L)P(F\mid L)\\ &= (0.20)(0.05)+(0.50)(0.04)+(0.30)(0.08)\\ &= 0.010+0.020+0.024 = 0.054. \end {align*}

(b). This reverses the conditioning, so it is Bayes’ theorem: \[P(L\mid F) = \frac {P(L)P(F\mid L)}{P(F)} = \frac {0.024}{0.054} = \frac {4}{9} \approx 0.444.\] The Lakeview Lodge takes only 30% of the clients but accounts for 44% of the complaints, because its rooms are the worst.

Solution. Let \(T\), \(C\) and \(H\) be the events that each man is elected, and \(I\) the event of an increase in fees. Exactly one of the three is elected, so they partition the sample space – note \(0.3+0.4+0.3 = 1\).

(a). By the law of total probability, \begin {align*} P(I) &= P(T)P(I\mid T)+P(C)P(I\mid C)+P(H)P(I\mid H)\\ &= (0.3)(0.6)+(0.4)(0.2)+(0.3)(0.4)\\ &= 0.18+0.08+0.12 = 0.38. \end {align*}

(b). The fee increase is known and the election is what we are asking about, so the conditioning is reversed and this is Bayes’ theorem: \[P(C\mid I) = \frac {P(C)P(I\mid C)}{P(I)} = \frac {0.08}{0.38} = \frac {4}{19} \approx 0.211.\] Mr Chela was the most likely to be elected, at \(0.4\), yet learning that the fees went up makes him the least likely of the three to have won – he was the candidate least inclined to raise them.

Solution. The four cells are the joint probabilities and they sum to \(1\).

(a). Summing the first-born column and the aggressive row, \begin {align*} \text {(i)}\quad P(B) &= 0.25+0.15 = 0.40,\\ \text {(ii)}\quad P(A) &= 0.25+0.20 = 0.45,\\ \text {(iii)}\quad P(A\mid B) &= \frac {P(A\cap B)}{P(B)} = \frac {0.25}{0.40} = 0.625. \end {align*}

(b). Independence would require \(P(A\cap B) = P(A)P(B)\). Here \[P(A)P(B) = (0.45)(0.40) = 0.18, \qquad \text {but}\qquad P(A\cap B) = 0.25.\] These differ, so \(A\) and \(B\) are not independent. Equivalently, \(P(A\mid B)=0.625\) is not equal to \(P(A)=0.45\): knowing a student is first born changes the probability that they are aggressive.

(c). In these data the psychologists’ belief is supported. A first born is aggressive with probability \(0.625\), against \(0.45\) for a student chosen without regard to birth order, so the two characteristics are associated.

Association is not causation, though, and it is worth being careful here: these figures show only that aggressiveness and birth order occur together more often than chance would produce. They say nothing about which, if either, brings about the other.

Solution. The design is deliberate: because nobody knows which question a slip refers to, no individual answer reveals anything – yet the aggregate is still informative, because the two questions are mixed in known proportions.

Of the hundred numbers \(00\) to \(99\), seventy fall in \(00\)–\(69\), so a student answers the cannabis question with probability \(0.7\) and the computer-number question with probability \(0.3\). A computer number is as likely to end in an even digit as an odd one, so the non-sensitive question yields “yes” with probability \(0.5\).

Let \(p\) be the proportion who would answer “yes” to the cannabis question. By the law of total probability, \[P(\text {yes}) = 0.7p + 0.3(0.5) = 0.7p + 0.15.\] Setting this equal to the observed \(0.44\), \[0.7p = 0.44-0.15 = 0.29 \implies p = \frac {0.29}{0.7} = \frac {29}{70} \approx 0.414.\] So an estimated \(41.4\%\) of students have smoked cannabis.

Note what the randomisation costs. The \(0.3\) of responses answering the harmless question carry no information about cannabis at all, so the estimate rests on fewer usable answers than the hundred collected – the price paid for honest ones.

Solution. In both parts start from the addition rule \(P(A\cup B) = P(A)+P(B)-P(A\cap B)\). The two assumptions differ only in what they say about \(P(A\cap B)\). Also \(P(A\setminus B) = P(A\cap B') = P(A)-P(A\cap B)\).

(a). Independence gives \(P(A\cap B) = P(A)P(B) = 2p^{2}\), so \[p+2p-2p^{2} = 0.75 \implies 2p^{2}-3p+0.75 = 0 \implies p = \frac {3\pm \sqrt {3}}{4}.\] The root \(\frac {3+\sqrt 3}{4}\approx 1.18\) would make \(P(B) = 2p\) exceed 1, so \(p = \frac {3-\sqrt 3}{4}\approx 0.3170\). Then \[P(A\setminus B) = p-2p^{2} = p(1-2p) = \frac {2\sqrt 3-3}{4}\approx 0.1160.\]

(b). Mutually exclusive events have \(P(A\cap B) = 0\), so \(3p = 0.75\), giving \(p = 0.25\), and \(P(A\setminus B) = P(A) = 0.25\): when \(A\) and \(B\) cannot occur together, removing \(B\) from \(A\) removes nothing.

The two answers differ, which is the point: independent and mutually exclusive are different conditions, and two events with positive probability cannot be both. If they are mutually exclusive, the occurrence of one tells you the other did not occur, which is as far from independent as events can be.

Solution. The data are \(P(B) = 0.0001\), \(P(B') = 0.9999\) and \(P(A\mid B) = 0.99\). Specificity is about the healthy, \(P(A'\mid B') = 0.99\), so the false-positive rate is \(P(A\mid B') = 0.01\).

(a). \(B\) and \(B'\) partition the population, so by total probability \begin {align*} P(A) &= P(B)P(A\mid B)+P(B')P(A\mid B')\\ &= (0.0001)(0.99)+(0.9999)(0.01) = 0.000099+0.009999 = 0.010098. \end {align*}

(b). By Bayes’ theorem, \[P(B\mid A) = \frac {P(B)P(A\mid B)}{P(A)} = \frac {0.000099}{0.010098}\approx 0.0098.\] A test that is right 99% of the time, applied to a positive patient, leaves fewer than a 1% chance of disease. The two terms in (a) explain it: in a population of a million, about 99 sick people test positive, but about \(9\,999\) healthy people do too. When a condition is rare the false positives swamp the true ones, which is why a positive screening result is followed by a second, independent test rather than a diagnosis.

Solution. Let \(M_j\) be the event that the bolt came from machine \(j\) and \(D\) the event that it is defective. Every bolt comes from exactly one machine, so \(M_1, M_2, M_3\) partition the output.

(a). \begin {align*} P(D) &= \sum _{j=1}^{3}P(M_j)P(D\mid M_j) = (0.2)(0.05)+(0.3)(0.03)+(0.5)(0.02)\\ &= 0.010+0.009+0.010 = 0.029. \end {align*}

(b). \[P(M_1\mid D) = \frac {P(M_1)P(D\mid M_1)}{P(D)} = \frac {0.010}{0.029} = \frac {10}{29}\approx 0.345.\] Machine 1 makes only a fifth of the bolts but about a third of the defective ones, because its defect rate is the highest. The same reasoning gives \(P(M_2\mid D) = \frac {9}{29}\) and \(P(M_3\mid D) = \frac {10}{29}\). As a check, the three posteriors sum to 1.