Tasks 07-04 - Conditional Probability

Section 07: Probability & Statistics

A: Skills

Problem 1: Conditional Probability (x)

For two events \(A\) and \(B\) of a random experiment, \(P(A) = 0.5\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.14\).

Give probabilities as fractions or decimals. Where a decimal has to be rounded, round to four decimal places and keep all intermediate results unrounded. This rule holds for the whole sheet.

  1. Compute \(P(A \mid B)\) and \(P(B \mid A)\). Explain why the two results differ.

  2. \(B'\) (also written \(\bar{B}\)) is the complement of \(B\). Compute \(P(A \cap \bar{B})\) and \(P(A \mid \bar{B})\).

  3. Decide with the test \(P(A \mid B) = P(A)\) whether \(A\) and \(B\) are independent.

  4. \(P(A)\) and \(P(B)\) stay the same. Determine the value of \(P(A \cap B)\) for which \(A\) and \(B\) would be independent, and give \(P(A \mid B)\) in this case.

  1. Definition of the conditional probability, the event behind the bar goes into the denominator:

    \[P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.14}{0.4} = 0.35, \qquad P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.14}{0.5} = 0.28\]

    Both use the same numerator, but the condition changes the base. \(P(A \mid B)\) only looks at the outcomes in \(B\), \(P(B \mid A)\) only at the outcomes in \(A\), and \(A\) is larger than \(B\).

  2. \(A \cap \bar{B} = A \setminus B\), so remove the overlap from \(A\):

    \[P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5 - 0.14 = 0.36\]

    \[P(A \mid \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})} = \frac{0.36}{1 - 0.4} = \frac{0.36}{0.6} = 0.6\]

  3. \(P(A \mid B) = 0.35 \neq 0.5 = P(A)\), so \(A\) and \(B\) are not independent. If \(B\) occurs, \(A\) becomes less likely.

    Check with \(P(A \cap B) = P(A) \cdot P(B)\): \(P(A) \cdot P(B) = 0.5 \cdot 0.4 = 0.2 \neq 0.14 = P(A \cap B)\) \(\checkmark\)

  4. Independence means \(P(A \cap B) = P(A) \cdot P(B)\):

    \[P(A \cap B) = 0.5 \cdot 0.4 = 0.2, \qquad P(A \mid B) = \frac{0.2}{0.4} = 0.5 = P(A)\]

    Knowing that \(B\) occurred would then not change the probability of \(A\).

Typical mistake: in a), computing \(P(A \mid B) = \frac{P(A \cap B)}{P(A)} = 0.28\). The event after the bar is the condition, and its probability is the denominator.

Problem 2: Drawing Without Replacement (x)

At the checkout of a bookshop, there is a bowl with 11 wrapped chocolates: 7 milk chocolates and 4 dark chocolates. From the outside, they all look the same. A customer takes two chocolates, one after the other, without looking.

  1. Draw a tree diagram for the two chocolates with all branch probabilities.

  2. Compute the probability that both chocolates are milk chocolates.

  3. Compute the probability that the customer gets one chocolate of each kind.

  4. Compute the probability that the second chocolate is dark. Compare it with the probability that the first chocolate is dark.

  5. On another day, the shop assistant puts a new chocolate of the same kind into the bowl after every chocolate taken, so the bowl always contains 7 milk and 4 dark chocolates. Compute the probability from b) for this case and explain why it differs from b).

Let \(M_1\), \(D_1\) be “the first chocolate is milk / dark” and \(M_2\), \(D_2\) the same for the second chocolate.

  1. After the first chocolate, only 10 chocolates are left, and one of them less of the kind taken first:

    • \(M_1\) (\(\frac{7}{11}\))
      • \(M_2\) (\(\frac{6}{10}\))
      • \(D_2\) (\(\frac{4}{10}\))
    • \(D_1\) (\(\frac{4}{11}\))
      • \(M_2\) (\(\frac{7}{10}\))
      • \(D_2\) (\(\frac{3}{10}\))

  2. Multiply along the path \(M_1 \to M_2\):

    \[P(M_1 \cap M_2) = P(M_1) \cdot P(M_2 \mid M_1) = \frac{7}{11} \cdot \frac{6}{10} = \frac{42}{110} = \frac{21}{55} \approx 0.3818\]

  3. Two paths, milk then dark or dark then milk. Add them:

    \[P(M_1 \cap D_2) + P(D_1 \cap M_2) = \frac{7}{11} \cdot \frac{4}{10} + \frac{4}{11} \cdot \frac{7}{10} = \frac{28}{110} + \frac{28}{110} = \frac{56}{110} = \frac{28}{55} \approx 0.5091\]

  4. Add the two paths that end in \(D_2\):

    \[P(D_2) = \frac{7}{11} \cdot \frac{4}{10} + \frac{4}{11} \cdot \frac{3}{10} = \frac{28}{110} + \frac{12}{110} = \frac{40}{110} = \frac{4}{11} \approx 0.3636\]

    This equals \(P(D_1) = \frac{4}{11}\). As long as nobody looks at the first chocolate, every chocolate in the bowl has the same chance to be the second one.

    Check: all four paths add up to 1, \(\frac{42 + 28 + 28 + 12}{110} = 1\) \(\checkmark\)

  5. With replacement, the second draw does not depend on the first:

    \[P(M_1 \cap M_2) = \frac{7}{11} \cdot \frac{7}{11} = \frac{49}{121} \approx 0.4050\]

    This is larger than in b). Without replacement, a milk chocolate is missing after the first draw, so only 6 of 10 chocolates are milk chocolates, and \(\frac{6}{10} < \frac{7}{11}\).

Typical mistake: in b), computing \(\frac{7}{11} \cdot \frac{6}{11}\). After the first draw, one chocolate fewer is in the bowl, so both the numerator and the denominator go down by 1.

Problem 3: Law of Total Probability (xx)

A hotel gets its bookings through three channels: 25 % through its own website (\(W\)), 60 % through a booking platform (\(B\)) and 15 % through travel agencies (\(T\)). 6 % of the website bookings, 20 % of the platform bookings and 4 % of the travel-agency bookings are cancelled (\(C\)).

  1. Draw a tree diagram with all branch and path probabilities.

  2. Compute the probability that a booking is cancelled and name the rule you use.

  3. A booking was cancelled. Compute the probability that it came through the booking platform.

  4. A booking was not cancelled. Compute the probability that it came through the hotel’s own website.

  1. First stage: channel, second stage: cancelled or not. Path probability = product along the path:

    • \(W\) (\(0.25\))
      • \(C\) (\(0.06\)): \(0.25 \cdot 0.06 = 0.015\)
      • \(\bar{C}\) (\(0.94\)): \(0.25 \cdot 0.94 = 0.235\)
    • \(B\) (\(0.60\))
      • \(C\) (\(0.20\)): \(0.60 \cdot 0.20 = 0.12\)
      • \(\bar{C}\) (\(0.80\)): \(0.60 \cdot 0.80 = 0.48\)
    • \(T\) (\(0.15\))
      • \(C\) (\(0.04\)): \(0.15 \cdot 0.04 = 0.006\)
      • \(\bar{C}\) (\(0.96\)): \(0.15 \cdot 0.96 = 0.144\)

    Check: \(0.015 + 0.235 + 0.12 + 0.48 + 0.006 + 0.144 = 1\) \(\checkmark\)

  2. Law of total probability with three branches, add the paths that end in \(C\):

    \[P(C) = P(W) \cdot P(C \mid W) + P(B) \cdot P(C \mid B) + P(T) \cdot P(C \mid T) = 0.015 + 0.12 + 0.006 = 0.141\]

    So 14.1 % of all bookings are cancelled.

  3. The base is the cancelled bookings:

    \[P(B \mid C) = \frac{P(B \cap C)}{P(C)} = \frac{0.12}{0.141} = \frac{40}{47} \approx 0.8511\]

    About 85 % of the cancellations come from the platform, although the platform brings only 60 % of the bookings.

  4. \(P(\bar{C}) = 1 - 0.141 = 0.859\), and the base is now the bookings that were not cancelled:

    \[P(W \mid \bar{C}) = \frac{P(W \cap \bar{C})}{P(\bar{C})} = \frac{0.235}{0.859} = \frac{235}{859} \approx 0.2736\]

Typical mistake: in b), taking the mean of the three rates, \(\frac{6\,\% + 20\,\% + 4\,\%}{3} = 10\,\%\). The channels bring different shares of the bookings, so each rate must be weighted with its channel’s share.

Problem 4: Sequential Selection (xx)

An internal auditor checks 3 of the 16 travel expense reports that a department handed in last month. The auditor draws the reports at random, one after the other, and does not put a report back. 5 of the 16 reports are for trips abroad, the other 11 for domestic trips (trips within the country).

  1. Compute the probability that all 3 reports are for trips abroad and name the rule you use.

  2. Compute the probability that none of the 3 reports is for a trip abroad.

  3. Compute the probability that at least one of the 3 reports is for a trip abroad.

  4. Compute the probability that exactly 2 of the 3 reports are for trips abroad. Check your result with combinations (Session 07-03).

Hint: In the tree, “exactly 2” ends on three paths: abroad, abroad, domestic, then abroad, domestic, abroad, and then domestic, abroad, abroad.

Let \(A_i\) = “the \(i\)-th report is for a trip abroad”. After each draw, one report fewer is left.

  1. Extended multiplication rule:

    \[P(A_1 \cap A_2 \cap A_3) = P(A_1) \cdot P(A_2 \mid A_1) \cdot P(A_3 \mid A_1 \cap A_2) = \frac{5}{16} \cdot \frac{4}{15} \cdot \frac{3}{14} = \frac{60}{3360} = \frac{1}{56} \approx 0.0179\]

  2. All 3 reports come from the \(16 - 5 = 11\) domestic reports:

    \[P(\bar{A}_1 \cap \bar{A}_2 \cap \bar{A}_3) = \frac{11}{16} \cdot \frac{10}{15} \cdot \frac{9}{14} = \frac{990}{3360} = \frac{33}{112} \approx 0.2946\]

  3. “At least one abroad” is the complement of “none abroad”:

    \[P(\text{at least one}) = 1 - \frac{33}{112} = \frac{79}{112} \approx 0.7054\]

  4. Each of the three paths has the same product, only the order of the factors changes:

    \[\frac{5}{16} \cdot \frac{4}{15} \cdot \frac{11}{14} = \frac{5}{16} \cdot \frac{11}{15} \cdot \frac{4}{14} = \frac{11}{16} \cdot \frac{5}{15} \cdot \frac{4}{14} = \frac{220}{3360}\]

    \[P(\text{exactly } 2) = 3 \cdot \frac{220}{3360} = \frac{660}{3360} = \frac{11}{56} \approx 0.1964\]

    Check with combinations: 2 of the 5 reports abroad and 1 of the 11 domestic reports, out of all \(\binom{16}{3} = 560\) selections:

    \[\frac{\binom{5}{2} \cdot \binom{11}{1}}{\binom{16}{3}} = \frac{10 \cdot 11}{560} = \frac{110}{560} = \frac{11}{56} \;\checkmark\]

Typical mistake: in d), taking only one path, \(\frac{220}{3360} \approx 0.0655\). The report for the domestic trip can be the first, the second or the third one drawn.

Problem 5: Expected Value and Variance (xx)

A plumbing company looks at the working time of its standard repair jobs. \(X\) is the working time of a job in hours (a started hour counts as a full hour):

\(x\) 1 2 3 4
\(P(X = x)\) 0.1 0.3 0.4 0.2

Round expected values and standard deviations to two decimal places.

  1. Compute the expected value \(E[X]\) and interpret it.

  2. Compute the variance \(\mathrm{Var}(X)\) and the standard deviation \(\sigma_X\).

Hint: \(\mathrm{Var}(X) = E[(X - \mu)^2]\) is the expected value of the squared deviations, so multiply each \((x_i - \mu)^2\) by its probability \(p_i\) and add.

  1. For a job, the company charges a call-out fee (a fixed amount for coming to the customer) of 30 EUR and 55 EUR per hour. Give the price \(Y\) of a job in terms of \(X\). Compute \(E[Y]\), \(\mathrm{Var}(Y)\) and \(\sigma_Y\) with the rules for \(aX + b\).

  2. A plumber does two jobs on one morning. Their working times \(X_1\) and \(X_2\) are independent and both have the distribution from the table. Compute the expected total working time and its standard deviation.

  1. \[E[X] = 1 \cdot 0.1 + 2 \cdot 0.3 + 3 \cdot 0.4 + 4 \cdot 0.2 = 0.1 + 0.6 + 1.2 + 0.8 = 2.7\]

    Over many jobs, a job takes 2.7 hours on average. No single job takes 2.7 hours, because \(X\) only takes whole numbers.

  2. Deviations from \(\mu = 2.7\): \(-1.7\), \(-0.7\), \(0.3\) and \(1.3\).

    \[\mathrm{Var}(X) = (-1.7)^2 \cdot 0.1 + (-0.7)^2 \cdot 0.3 + 0.3^2 \cdot 0.4 + 1.3^2 \cdot 0.2 = 0.289 + 0.147 + 0.036 + 0.338 = 0.81\]

    \[\sigma_X = \sqrt{0.81} = 0.9 \text{ hours}\]

  3. \(Y = 55X + 30\), so \(a = 55\) and \(b = 30\):

    \[E[Y] = 55 \cdot E[X] + 30 = 55 \cdot 2.7 + 30 = 178.50 \text{ EUR}\]

    \[\mathrm{Var}(Y) = 55^2 \cdot \mathrm{Var}(X) = 3025 \cdot 0.81 = 2450.25, \qquad \sigma_Y = \sqrt{2450.25} = 49.50 \text{ EUR}\]

    Check: \(\sigma_Y = 55 \cdot \sigma_X = 55 \cdot 0.9 = 49.5\) \(\checkmark\) The call-out fee shifts the mean but does not change the spread.

  4. For independent variables, expected values and variances add:

    \[E[X_1 + X_2] = 2.7 + 2.7 = 5.4 \text{ hours}, \qquad \mathrm{Var}(X_1 + X_2) = 0.81 + 0.81 = 1.62\]

    \[\sigma_{X_1 + X_2} = \sqrt{1.62} \approx 1.27 \text{ hours}\]

    This is less than \(0.9 + 0.9 = 1.8\): variances add, standard deviations do not.

Typical mistake: in c), computing \(\mathrm{Var}(Y) = 55 \cdot 0.81 + 30\). The factor \(a\) is squared, and the constant \(b\) drops out, because adding the same amount to every price does not change the spread.

B: Set Up & Check

Problem 6: From Words to Conditional Probabilities (xx)

Translate each statement into probability notation. Then answer the check question: it tells you whether your translation works.

  1. At a garden centre, 40 % of the customers buy plants (\(G\)). Of the customers who buy plants, 15 % also buy fertiliser (\(F\)). Write both statements in probability notation and compute the probability that a customer buys plants and fertiliser.

    Check: A classmate writes the second statement as \(P(G \mid F) = 0.15\). Explain which event belongs behind the bar and why.

  2. At a motorway service station, 45 % of the customers fill up their car (\(T\)). 18 % of all customers fill up and buy a coffee (\(K\)). Write both statements in probability notation and compute the probability that a customer who fills up also buys a coffee.

    Check: The station manager says: “18 % of the customers who fill up also buy a coffee.” Decide whether this is right and explain the difference between the two statements.

  3. At a car dealership, 25 % of the visitors take a test drive (\(T\)), 10 % buy a car (\(B\)), and 8 % do both. Compute the probability that a visitor who takes a test drive buys a car, and the probability that a buyer took a test drive.

    Check: The sales manager writes in a report: “80 % of the visitors who take a test drive buy a car.” Find the mix-up.

  4. A pharmacy receives 70 % of its prescriptions as e-prescriptions (\(E\)), the rest on paper. For 2 % of the e-prescriptions and 6 % of the paper prescriptions, the pharmacy has to call the doctor’s practice with a question (\(Q\)). Write the given values in probability notation and compute the probability that a prescription leads to a call.

    Check: A classmate takes the mean \(\frac{2\,\% + 6\,\%}{2} = 4\,\%\). Explain why this is wrong and in which case it would be right.

  5. A marketing team has a list of 25 customers, 10 of them new customers. It draws 2 customers at random for a phone interview. Write the probability that both are new customers, first if a customer can be drawn only once, and then if the team draws twice from the full list, so that the same customer could be drawn twice. Compute both.

    Check: A classmate uses \(\left(\frac{10}{25}\right)^2\) for the first case. Explain which case this formula belongs to.

  1. \(P(G) = 0.4\) and \(P(F \mid G) = 0.15\). Multiplication rule:

    \[P(G \cap F) = P(G) \cdot P(F \mid G) = 0.4 \cdot 0.15 = 0.06\]

    So 6 % of all customers buy plants and fertiliser.

    Check: “of the customers who buy plants” names the group that the 15 % refer to, i.e. the condition. The condition goes behind the bar, so it is \(P(F \mid G)\). \(P(G \mid F)\) would be the share of plant buyers among the fertiliser buyers.

    Typical mistake: writing \(P(G \cap F) = 0.15\). The 15 % refer only to the plant buyers, not to all customers.

  2. \(P(T) = 0.45\) and, because “of all customers” is the base, \(P(T \cap K) = 0.18\) (a joint probability). Definition of the conditional probability:

    \[P(K \mid T) = \frac{P(T \cap K)}{P(T)} = \frac{0.18}{0.45} = 0.4\]

    Check: the manager is wrong. The 18 % refer to all customers of the station, the manager’s sentence refers only to the customers who fill up. Of these, 40 % buy a coffee.

    Typical mistake: multiplying, \(0.18 \cdot 0.45 = 0.081\). The joint probability is already the product \(P(T) \cdot P(K \mid T)\), so \(P(K \mid T)\) is found by dividing.

  3. \(P(T) = 0.25\), \(P(B) = 0.1\) and \(P(T \cap B) = 0.08\):

    \[P(B \mid T) = \frac{0.08}{0.25} = 0.32, \qquad P(T \mid B) = \frac{0.08}{0.1} = 0.8\]

    Check: 80 % is \(P(T \mid B)\): of the buyers, 80 % took a test drive. Of the visitors who take a test drive, only 32 % buy a car. The manager swapped the condition.

    Typical mistake: assuming \(P(B \mid T) = P(T \mid B)\). Both have the numerator \(0.08\), but different bases (denominators).

  4. \(P(E) = 0.7\), \(P(\bar{E}) = 0.3\), \(P(Q \mid E) = 0.02\) and \(P(Q \mid \bar{E}) = 0.06\). Law of total probability:

    \[P(Q) = P(E) \cdot P(Q \mid E) + P(\bar{E}) \cdot P(Q \mid \bar{E}) = 0.7 \cdot 0.02 + 0.3 \cdot 0.06 = 0.014 + 0.018 = 0.032\]

    Check: the plain mean gives both kinds of prescriptions the same weight, but 70 % are e-prescriptions with the lower rate. The mean would only be right if both kinds made up 50 %: \(0.5 \cdot 0.02 + 0.5 \cdot 0.06 = 0.04\).

    Typical mistake: adding the rates, \(0.02 + 0.06 = 0.08\). The two rates refer to different groups, so each must be weighted with the share of its group.

  5. Drawn only once (without replacement): after a new customer, 9 new customers of 24 are left.

    \[P(\text{both new}) = \frac{10}{25} \cdot \frac{9}{24} = \frac{90}{600} = 0.15\]

    Drawn twice from the full list (with replacement): the two draws are independent.

    \[P(\text{both new}) = \frac{10}{25} \cdot \frac{10}{25} = \frac{100}{625} = 0.16\]

    Check: \(\left(\frac{10}{25}\right)^2\) still has 10 new customers among 25 at the second draw. That is only true if the first customer goes back onto the list, i.e. in the second case.

    Typical mistake: \(\frac{10}{25} \cdot \frac{9}{25}\). If one new customer is gone, the list has only 24 customers left, so the denominator goes down as well.

Problem 7: One Paragraph, One Tree (xx)

A used-car dealer buys cars at auctions and from private sellers. 45 % of the cars come from auctions (\(A\)), the rest from private sellers. Before a car is sold, the workshop checks it. 28 % of the auction cars and 12 % of the cars from private sellers need a repair (\(R\)). The dealer wants to know the probability that a car needs a repair, the probability that a car that needs a repair came from an auction, and the probability that a car comes from a private seller and needs no repair. The dealer also asks whether the need for a repair is independent of where the car was bought.

Draw the skeleton of a tree diagram for this paragraph: name the events at each stage and label every branch with its probability. Then write every question of the dealer as a mathematical statement and name the rule or test you would use. Do not compute anything.

Check: An assistant labels the branch from \(A\) to \(R\) with “\(P(A \cap R) = 0.28\)”. Explain what the 28 % are and where \(P(A \cap R)\) appears in the tree.

Tree skeleton (first stage: where the car was bought, second stage: repair or not):

  • \(A\): \(P(A) = 0.45\)
    • \(R\): \(P(R \mid A) = 0.28\)
    • \(\bar{R}\): \(P(\bar{R} \mid A) = 1 - 0.28\)
  • \(\bar{A}\): \(P(\bar{A}) = 1 - 0.45\)
    • \(R\): \(P(R \mid \bar{A}) = 0.12\)
    • \(\bar{R}\): \(P(\bar{R} \mid \bar{A}) = 1 - 0.12\)

Questions as mathematical statements:

  1. Random experiment: one randomly chosen car of the dealer, with the events \(A\) = “the car comes from an auction” and \(R\) = “the car needs a repair”
  2. A car needs a repair: \(P(R) = P(A) \cdot P(R \mid A) + P(\bar{A}) \cdot P(R \mid \bar{A})\), law of total probability (add the two paths that end in \(R\))
  3. A car that needs a repair came from an auction: \(P(A \mid R) = \frac{P(A \cap R)}{P(R)}\), definition of the conditional probability (the path \(A \to R\) divided by all paths that end in \(R\))
  4. A car comes from a private seller and needs no repair: \(P(\bar{A} \cap \bar{R}) = P(\bar{A}) \cdot P(\bar{R} \mid \bar{A})\), multiplication rule (one path)
  5. Independence: test whether \(P(R \mid A) = P(R)\)

Check: the 28 % refer only to the auction cars, so they are the conditional probability \(P(R \mid A)\) and belong on the branch from \(A\) to \(R\). \(P(A \cap R)\) is the path probability at the end of the path \(A \to R\), i.e. the product \(P(A) \cdot P(R \mid A)\).

Typical mistake: writing “a car that needs a repair came from an auction” as \(P(R \mid A)\). Here the repair is known, so \(R\) is the condition and goes behind the bar: \(P(A \mid R)\).

C: Applied

Problem 8: Drinking Glasses from a Glassworks (xx)

A glassworks makes drinking glasses in two ways: 25 % of the glasses are mouth-blown by glassblowers (\(H\)), the other 75 % are made on a pressing line. Every glass is inspected. A glass with a small flaw, e.g. a tiny air bubble, is sold as a second (\(F\)), i.e. a product with a small flaw that is sold at a lower price. 8 % of the mouth-blown glasses and 4 % of the pressed glasses are seconds.

  1. Create a tree diagram with all branch and path probabilities.

  2. Compute the probability that a glass is a second and name the rule you use.

(for verification: 5 % of the glasses are seconds)

  1. A customer buys a second in the factory outlet (the glassworks’ own shop for seconds). Compute the probability that the glass was mouth-blown. Compare the result with the share of mouth-blown glasses in the production and explain the difference.

  2. The glassworks makes 24,000 glasses per month. Estimate the number of seconds per month and how many of them come from each way of production.

  3. The pressing line gets a new mould. Afterwards, 4.4 % of all glasses are seconds. The shares of the two ways of production and the rate of seconds of the glassblowers stay the same. Show that only 3.2 % of the pressed glasses are now seconds.

Hint: Call the new rate of seconds of the pressing line \(x\). Write the law of total probability for the new share of seconds with \(x\), then solve the linear equation.

  1. Compute the probability from c) for the situation with the new mould. Explain why it has increased, although the glassblowers work as before.

a) First stage: way of production, second stage: second or not. The branches at each fork add up to 1, and each path probability is the product along the path:

  • \(H\) (\(0.25\))
    • \(F\) (\(0.08\)): \(0.25 \cdot 0.08 = 0.02\)
    • \(\bar{F}\) (\(0.92\)): \(0.25 \cdot 0.92 = 0.23\)
  • \(\bar{H}\) (\(0.75\))
    • \(F\) (\(0.04\)): \(0.75 \cdot 0.04 = 0.03\)
    • \(\bar{F}\) (\(0.96\)): \(0.75 \cdot 0.96 = 0.72\)

Check: \(0.02 + 0.23 + 0.03 + 0.72 = 1\) \(\checkmark\)

b) Law of total probability: add the two paths that end in \(F\):

\[P(F) = P(H) \cdot P(F \mid H) + P(\bar{H}) \cdot P(F \mid \bar{H}) = 0.02 + 0.03 = 0.05\]

c) The glass is known to be a second, so the base is \(P(F)\):

\[P(H \mid F) = \frac{P(H \cap F)}{P(F)} = \frac{0.02}{0.05} = 0.4\]

Mouth-blown glasses make up 25 % of the production, but 40 % of the seconds. A mouth-blown glass is a second twice as often as a pressed glass (8 % instead of 4 %), so the mouth-blown glasses are overrepresented among the seconds.

d) Relative frequency ≈ probability:

\[24{,}000 \cdot 0.05 = 1{,}200 \text{ seconds per month}\]

Mouth-blown: \(24{,}000 \cdot 0.02 = 480\), pressed: \(24{,}000 \cdot 0.03 = 720\).

Check: \(480 + 720 = 1{,}200\) \(\checkmark\)

e) Let \(x = P(F \mid \bar{H})\) be the new rate of the pressing line. Law of total probability with the new total \(P(F) = 0.044\):

\[0.044 = 0.25 \cdot 0.08 + 0.75 \cdot x = 0.02 + 0.75x\]

\[0.75x = 0.024 \iff x = \frac{0.024}{0.75} = 0.032\]

So 3.2 % of the pressed glasses are now seconds \(\checkmark\)

f) The path \(H \to F\) still has the probability \(0.02\):

\[P(H \mid F) = \frac{0.02}{0.044} = \frac{5}{11} \approx 0.4545\]

The number of mouth-blown seconds has not changed, but there are fewer pressed seconds. So the base, all seconds, is smaller, and the same numerator divided by a smaller denominator gives a larger share. About 45 % of the seconds are now mouth-blown.

Typical mistake: in c), answering \(P(F \mid H) = 0.08\) or \(P(H \cap F) = 0.02\). The question says that the glass is a second, so the base is the seconds: divide the path \(H \to F\) by all paths that end in \(F\).

Problem 9: Skate Rental at an Ice Rink (xx)

An ice rink sells single tickets (\(S\)), ten-visit cards (\(T\)) and season passes (\(M\)). Visitors can rent skates (\(R\)) or bring their own. On one weekend, the ice rink counted 800 visits:

single ticket (\(S\)) ten-visit card (\(T\)) season pass (\(M\)) sum
rents skates (\(R\)) 120 140 20 280
own skates (\(\bar{R}\)) 80 260 180 520
sum 200 400 200 800

One of the 800 visits is chosen at random. Use the relative frequencies as probabilities.

  1. Compute \(P(R)\), \(P(R \mid S)\), \(P(R \mid T)\) and \(P(R \mid M)\). Write \(P(R \mid S)\) as a sentence about the visitors.

(for verification: \(P(R) = 0.35\))

  1. Decide with the test \(P(A \mid B) = P(A)\) whether renting skates is independent of having a single ticket, and whether it is independent of having a ten-visit card. Interpret both results for the ice rink.

  2. Check your result for the ten-visit card with \(P(R \cap T) = P(R) \cdot P(T)\). Decide whether \(R\) and \(T\) are mutually exclusive.

  3. A visitor at the rental counter is chosen at random. Compute the probability that the visitor has a single ticket. Explain why this is not the value \(P(R \mid S)\) from a).

  4. For a Christmas skating event, the ice rink expects 1,200 visitors on the ice at the same time, with the same shares as on the weekend. It has 400 pairs of skates for rent. Estimate the number of visitors who want to rent skates and decide whether the skates are enough. Estimate how many of the renters have a single ticket.

a) Every visit is equally likely, so \(P(R) = \frac{280}{800} = 0.35\). For a conditional probability, the condition shrinks the base to its column:

\[P(R \mid S) = \frac{P(R \cap S)}{P(S)} = \frac{120/800}{200/800} = \frac{120}{200} = 0.6\]

\[P(R \mid T) = \frac{140}{400} = 0.35, \qquad P(R \mid M) = \frac{20}{200} = 0.1\]

\(P(R \mid S) = 0.6\): 60 % of the visitors with a single ticket rent skates.

b) Single ticket: \(P(R \mid S) = 0.6 \neq 0.35 = P(R)\), so \(R\) and \(S\) are not independent. Visitors with a single ticket rent skates much more often than the average visitor, probably because many of them skate only now and then and have no skates of their own.

Ten-visit card: \(P(R \mid T) = 0.35 = P(R)\), so \(R\) and \(T\) are independent. Knowing that a visitor has a ten-visit card does not change the probability that the visitor rents skates.

c) \(P(T) = \frac{400}{800} = 0.5\) and

\[P(R) \cdot P(T) = 0.35 \cdot 0.5 = 0.175 = \frac{140}{800} = P(R \cap T) \;\checkmark\]

\(R\) and \(T\) are not mutually exclusive: \(P(R \cap T) = 0.175 \neq 0\), 140 visits had both.

d) Now the renting is known, so the base is the 280 renters:

\[P(S \mid R) = \frac{P(R \cap S)}{P(R)} = \frac{120}{280} = \frac{3}{7} \approx 0.4286\]

\(P(R \mid S) = 0.6\) answers “of the visitors with a single ticket, how many rent skates?” (base 200). \(P(S \mid R)\) answers “of the renters, how many have a single ticket?” (base 280). Same numerator, different bases.

e) Renters: \(1200 \cdot 0.35 = 420 > 400\). The skates are not enough, about 20 pairs are missing.

Renters with a single ticket: \(P(R \cap S) = \frac{120}{800} = 0.15\), so about \(1200 \cdot 0.15 = 180\) visitors.

Check: \(420 \cdot \frac{3}{7} = 180\) \(\checkmark\)

Typical mistake: in a), computing \(P(R \mid S) = \frac{120}{800} = 0.15\). This is the joint probability \(P(R \cap S)\). The condition “single ticket” restricts the base to the 200 visits with a single ticket.

Problem 10: Coffee Machines for Offices (xxx)

A company rents out coffee machines to offices. A contract runs for one year. At the end of the year, the office renews the contract (\(R\)) or not. The company’s data from the last years: 80 % of the offices renewed their contract. Of the offices that renewed, 15 % had needed a service call (\(S\)) during the year, i.e. a technician had to come because the machine did not work. Of the offices that did not renew, 60 % had needed a service call.

  1. Create a tree diagram with all branch and path probabilities. Explain why the first stage of the tree is the renewal, although the service calls happen earlier in the year.

  2. Compute the probability that an office needed a service call.

(for verification: 24 %)

  1. The sales manager says: “An office with a service call renews its contract with a probability of only 15 %.” Check the claim: compute the probability that an office with a service call renews and the probability that an office without a service call renews. Explain the manager’s mistake.

  2. Decide whether renewing the contract and needing a service call are independent. Interpret the result for the company.

  3. This year, 1,200 contracts end. The sales team wants to call every office that needed a service call and offer a discount for the renewal. Estimate the number of offices the team calls and how many of them would have renewed without the discount.

a) The 15 % and the 60 % are given “of the offices that renewed” and “of the offices that did not renew”. So they are conditional probabilities \(P(S \mid R)\) and \(P(S \mid \bar{R})\) with the renewal as the condition. On a tree, the second-stage branches are conditional on the first stage, so the tree must start with \(R\) and \(\bar{R}\), whatever happened first in time.

  • \(R\) (\(0.8\))
    • \(S\) (\(0.15\)): \(0.8 \cdot 0.15 = 0.12\)
    • \(\bar{S}\) (\(0.85\)): \(0.8 \cdot 0.85 = 0.68\)
  • \(\bar{R}\) (\(0.2\))
    • \(S\) (\(0.6\)): \(0.2 \cdot 0.6 = 0.12\)
    • \(\bar{S}\) (\(0.4\)): \(0.2 \cdot 0.4 = 0.08\)

Check: \(0.12 + 0.68 + 0.12 + 0.08 = 1\) \(\checkmark\)

b) Law of total probability, add the paths that end in \(S\):

\[P(S) = 0.12 + 0.12 = 0.24\]

c) The service call is known, so the base is \(P(S)\), or \(P(\bar{S}) = 1 - 0.24 = 0.76\) for the offices without a service call:

\[P(R \mid S) = \frac{P(R \cap S)}{P(S)} = \frac{0.12}{0.24} = 0.5, \qquad P(R \mid \bar{S}) = \frac{P(R \cap \bar{S})}{P(\bar{S})} = \frac{0.68}{0.76} = \frac{17}{19} \approx 0.8947\]

The claim is wrong. The 15 % are \(P(S \mid R)\), the share of offices with a service call among the offices that renewed. The manager read them as \(P(R \mid S)\). In fact, half of the offices with a service call (50 %) still renew.

d) \(P(R \mid S) = 0.5 \neq 0.8 = P(R)\), so renewal and service calls are not independent.

For the company: offices that needed a service call renew much less often than the others (50 % instead of about 89 %). Reliable machines and a fast service are important to keep customers. The data show a connection, but not that the service call is the reason: an office could, for example, also be unhappy with the coffee.

e) Offices with a service call: \(1200 \cdot P(S) = 1200 \cdot 0.24 = 288\) calls.

Of these, \(P(R \mid S) = 0.5\) would have renewed anyway: \(288 \cdot 0.5 = 144\) offices (check: \(1200 \cdot P(R \cap S) = 1200 \cdot 0.12 = 144\) \(\checkmark\)).

So 144 of the 288 discounts go to offices that would have renewed anyway. Only the other \(288 - 144 = 144\) offices can be won back.

Typical mistake: in a), starting the tree with \(S\) and writing 15 % on the branch from \(S\) to \(R\). The 15 % describe the renewing offices, so they belong on a branch that starts at \(R\).

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Used Handbags from an Online Shop [20 pts. total]

An online shop resells used handbags. Private sellers send their bags to the shop, and 20 % of these bags are from a luxury brand (\(L\)). The shop checks every bag. It accepts a bag for sale (\(A\)) if the bag is complete and in good condition, and it sends the other bags back to their sellers. 90 % of the luxury bags and 50 % of the other bags are accepted. Give probabilities as decimals with four decimal places.

Part a)

Create a tree diagram for the check of a bag with all branch and path probabilities. [4 pts.]

Tree diagram (first stage: brand, second stage: accepted or not. The branches at each fork add up to 1, path probability = product along the path.):

  • \(L\) (\(0.2\))
    • \(A\) (\(0.9\)): \(0.2 \cdot 0.9 = 0.18\)
    • \(\bar{A}\) (\(0.1\)): \(0.2 \cdot 0.1 = 0.02\)
  • \(\bar{L}\) (\(0.8\))
    • \(A\) (\(0.5\)): \(0.8 \cdot 0.5 = 0.4\)
    • \(\bar{A}\) (\(0.5\)): \(0.8 \cdot 0.5 = 0.4\)

Check: \(0.18 + 0.02 + 0.4 + 0.4 = 1\) \(\checkmark\)

Grading: 1 Pt. first stage with \(P(\bar{L}) = 0.8\), 1 Pt. second stage with both complements, 2 Pts. the four path probabilities with the check.

Part b)

Compute the probability that a bag is accepted. [2 pts.]

Law of total probability: add the two paths that end in \(A\):

\[P(A) = P(L) \cdot P(A \mid L) + P(\bar{L}) \cdot P(A \mid \bar{L}) = 0.18 + 0.4 = 0.58\]

So 58 % of the bags sent in are accepted.

Grading: 1 Pt. law of total probability with both paths, 1 Pt. \(P(A) = 0.58\).

Part c)

A new employee writes on the shop’s website: “20 % of the bags in our shop are from a luxury brand.” Compute the probability that an accepted bag is from a luxury brand and explain the employee’s mistake. [3 pts.]

The shop only sells accepted bags, so the base is \(P(A)\) from part b):

\[P(L \mid A) = \frac{P(L \cap A)}{P(A)} = \frac{0.18}{0.58} = \frac{9}{29} \approx 0.3103\]

About 31.03 % of the bags in the shop are from a luxury brand.

Mistake: the 20 % are \(P(L)\), the share of luxury bags among all bags sent in. The bags in the shop are only the accepted ones, and luxury bags are accepted more often (90 % instead of 50 %). So their share in the shop is larger than 20 %.

Grading: 1 Pt. \(P(L \mid A)\) with the base \(P(A)\), 1 Pt. \(0.3103\), 1 Pt. explanation of the wrong base.

Part d)

Decide whether the events \(L\) and \(A\) are independent and interpret your result. [3 pts.]

Test: \(L\) and \(A\) are independent exactly if \(P(A \mid L) = P(A)\):

\[P(A \mid L) = 0.9 \neq 0.58 = P(A)\]

So \(L\) and \(A\) are not independent.

Check with \(P(L \cap A) = P(L) \cdot P(A)\): \(P(L) \cdot P(A) = 0.2 \cdot 0.58 = 0.116 \neq 0.18 = P(L \cap A)\) \(\checkmark\)

Interpretation: whether a bag is accepted depends on its brand. A luxury bag passes the check with 90 %, clearly more often than the average bag (58 %) and than the other bags (50 %). Owners of expensive bags probably treat them more carefully.

Grading: 1 Pt. test with \(P(A \mid L)\) and \(P(A)\) (or \(P(L \cap A)\) with \(P(L) \cdot P(A)\)), 1 Pt. “not independent”, 1 Pt. interpretation.

Part e)

\(X\) is the sale price of a bag that is sent in: an accepted luxury bag sells for 400 EUR, an accepted other bag for 120 EUR, and a bag that goes back to its seller brings 0 EUR. The shop keeps 40 % of the sale price and pays the rest to the seller. For every bag sent in, the shop has costs of 36 EUR for the check, the photos and the shipping. Give the probability distribution of \(X\) and compute \(E[X]\). Show that the shop’s profit per bag is \(Y = 0.4X - 36\) and compute the expected profit per bag. [4 pts.]

Distribution of \(X\) from the paths of the tree (a bag that is sent back is \(\bar{A}\), with \(P(\bar{A}) = 1 - 0.58 = 0.42\)):

\(x\) (EUR) 0 120 400
\(P(X = x)\) 0.42 0.40 0.18

Check: \(0.42 + 0.40 + 0.18 = 1\) \(\checkmark\)

Expected sale price:

\[E[X] = 0 \cdot 0.42 + 120 \cdot 0.40 + 400 \cdot 0.18 = 0 + 48 + 72 = 120 \text{ EUR}\]

Profit per bag: the shop earns 40 % of the sale price, \(0.4X\), and pays the costs of 36 EUR for every bag, sold or not:

\[Y = 0.4X - 36\]

Expected profit with \(E[aX + b] = a \cdot E[X] + b\):

\[E[Y] = 0.4 \cdot 120 - 36 = 48 - 36 = 12 \text{ EUR per bag}\]

Grading: 1 Pt. distribution of \(X\), 1 Pt. \(E[X] = 120\), 1 Pt. profit term with reason, 1 Pt. \(E[Y] = 12\) with the rule for \(aX + b\).

Part f)

In summer, fewer luxury bags are sent in. Let \(p\) be the share of luxury bags among the bags sent in, and all other values stay the same. Show that \(E[X] = 60 + 300p\). Determine the smallest share \(p\) for which the shop does not make a loss per bag on average. [4 pts.]

Expected sale price with share \(p\): the paths now have the probabilities \(P(L \cap A) = 0.9p\) and \(P(\bar{L} \cap A) = 0.5 \cdot (1 - p)\):

\[E[X] = 400 \cdot 0.9p + 120 \cdot 0.5 \cdot (1 - p) = 360p + 60 - 60p = 60 + 300p \;\checkmark\]

Expected profit with the term from part e):

\[E[Y] = 0.4 \cdot (60 + 300p) - 36 = 24 + 120p - 36 = 120p - 12\]

No loss on average means \(E[Y] \geq 0\):

\[120p - 12 \geq 0 \iff 120p \geq 12 \iff p \geq 0.1\]

(Dividing by the positive number 120 keeps the inequality sign.)

At least 10 % of the bags sent in must be luxury bags.

Check: \(p = 0.1\) gives \(E[Y] = 12 - 12 = 0\), and \(p = 0.2\) gives \(E[Y] = 24 - 12 = 12\) EUR as in part e) \(\checkmark\)

Grading: 1 Pt. \(E[X] = 60 + 300p\) from the path probabilities, 1 Pt. \(E[Y] = 120p - 12\), 1 Pt. inequality \(E[Y] \geq 0\) solved, 1 Pt. \(p \geq 0.1\) with the meaning and the check.