Tasks 09-04 - Probability & Statistics Review

Section 09: Exam Preparation

A: Skills

Problem 1: Counting with the Calculator (xx)

Give probabilities as fractions or as decimals with four decimal places, and keep all intermediate results unrounded. Give amounts of money to the cent, and expected values and standard deviations to two decimal places. Round a minimum up and a maximum down. This rule holds for the whole sheet.

A hotel restaurant tests a new menu.

  1. The chef puts 9 new starters on the menu card, one below the other. Compute the number of possible orders.

  2. At a tasting evening, the guests choose the best, the second-best and the third-best of 13 new main courses. Compute the number of possible results.

  3. The restaurant forms a tasting panel of 5 people from its 10 cooks and 7 members of the service staff. Compute the number of possible panels.

  4. The panel is drawn by lot. Compute the probability that exactly 2 members of the service staff are on the panel and the probability that at least one member of the service staff is on it.

Hint: On the calculator, x! is SHIFT + x⁻¹, nPr is SHIFT + × and nCr is SHIFT + ÷. Type \(n\) first, then the key, then \(r\).

  1. All 9 starters are used and their order matters, so this is a permutation of all 9:

    \[9! = 9 \cdot 8 \cdot 7 \cdots 1 = 362{,}880\]

    Calculator: 9 SHIFT x⁻¹ = gives 362880.

  2. 3 of 13 courses, and the order matters (best, second-best, third-best are different results):

    \[P(13, 3) = \frac{13!}{10!} = 13 \cdot 12 \cdot 11 = 1{,}716\]

    Calculator: 13 SHIFT × 3 = gives 1716.

  3. 5 of 17 people, and the order does not matter (a panel has no ranks):

    \[\binom{17}{5} = \frac{17!}{5! \cdot 12!} = \frac{17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 6{,}188\]

    Calculator: 17 SHIFT ÷ 5 = gives 6188.

  4. Every panel is equally likely, so \(P = \frac{\text{favourable panels}}{\text{all panels}}\).

    Exactly 2 of the 7 service staff and 3 of the 10 cooks (counting principle):

    \[P(\text{exactly 2}) = \frac{\binom{7}{2} \cdot \binom{10}{3}}{\binom{17}{5}} = \frac{21 \cdot 120}{6188} = \frac{2520}{6188} = \frac{90}{221} \approx 0.4072\]

    “At least one” is the complement of “no member of the service staff”, i.e. 5 cooks:

    \[P(\text{at least one}) = 1 - \frac{\binom{10}{5}}{\binom{17}{5}} = 1 - \frac{252}{6188} = 1 - \frac{9}{221} = \frac{212}{221} \approx 0.9593\]

Typical mistake: in b), computing \(\binom{13}{3} = 286\). A ranking of the best three is an ordered result: the same three courses in another order are another result, so \(P(13, 3) = 3! \cdot \binom{13}{3}\).

Problem 2: Rules of Probability (xx)

An online shop for sports goods analyses its orders. 38 % of the orders contain shoes (\(S\)), 26 % contain clothing (\(C\)), and 11 % contain both.

  1. Compute the probability that an order contains shoes or clothing, and name the rule you use.

  2. Compute the probability that an order contains neither shoes nor clothing, and the probability that it contains exactly one of the two.

  3. Compute \(P(C \mid S)\) and \(P(S \mid C)\). Say in words what \(P(C \mid S)\) means for the shop.

  4. Decide whether \(S\) and \(C\) are mutually exclusive.

  5. In a second shop, 68 % of the orders contain shoes or clothing, 16 % contain both, and orders with shoes are three times as frequent as orders with clothing. Determine \(P(S)\) and \(P(C)\) for this shop.

Given: \(P(S) = 0.38\), \(P(C) = 0.26\), \(P(S \cap C) = 0.11\).

  1. Addition rule (the orders with both are counted once):

    \[P(S \cup C) = P(S) + P(C) - P(S \cap C) = 0.38 + 0.26 - 0.11 = 0.53\]

  2. “Neither … nor” is the complement of “shoes or clothing”:

    \[P(\bar{S} \cap \bar{C}) = 1 - P(S \cup C) = 1 - 0.53 = 0.47\]

    “Exactly one” is “at least one” without “both”:

    \[P(\text{exactly one}) = P(S \cup C) - P(S \cap C) = 0.53 - 0.11 = 0.42\]

    Check: shoes only \(0.38 - 0.11 = 0.27\), clothing only \(0.26 - 0.11 = 0.15\), and \(0.27 + 0.15 = 0.42\) ✓

  3. Conditional probability, the condition gives the base:

    \[P(C \mid S) = \frac{P(S \cap C)}{P(S)} = \frac{0.11}{0.38} = \frac{11}{38} \approx 0.2895, \qquad P(S \mid C) = \frac{0.11}{0.26} = \frac{11}{26} \approx 0.4231\]

    About 29 % of the orders with shoes also contain clothing.

  4. Mutually exclusive means \(P(S \cap C) = 0\). Here \(P(S \cap C) = 0.11 \neq 0\): 11 % of the orders contain both, so \(S\) and \(C\) are not mutually exclusive.

  5. Addition rule with \(P(S \cup C) = 0.68\) and \(P(S \cap C) = 0.16\), and the second condition \(P(S) = 3 \cdot P(C)\):

    \[\text{(I)}\;\; P(S) + P(C) - 0.16 = 0.68 \iff P(S) + P(C) = 0.84, \qquad \text{(II)}\;\; P(S) = 3 \cdot P(C)\]

    Insert (II) into (I): \(4 \cdot P(C) = 0.84\), so \(P(C) = 0.21\) and \(P(S) = 0.63\).

    Check: \(0.63 + 0.21 - 0.16 = 0.68\) ✓

Typical mistake: in a), adding \(0.38 + 0.26 = 0.64\). The orders with shoes and clothing are in both shares, so they would be counted twice.

Problem 3: Contingency Table, Independence and Raw Data (xx)

A kiosk at a railway station records its 600 sales of one day: whether the customer bought a hot drink (\(H\)) and the time of day of the sale.

morning (\(M\)) noon (\(N\)) afternoon (\(A\)) sum
hot drink (\(H\)) 100 45 95 240
no hot drink (\(\bar{H}\)) 150 105 105 360
sum 250 150 200 600

  1. Compute \(P(H)\), \(P(H \mid N)\) and \(P(M \mid H)\), and name the base of each probability.

  2. Decide with the test \(P(H \cap M) = P(H) \cdot P(M)\) whether buying a hot drink and a sale in the morning are independent.

  3. Decide in the same way for a hot drink and a sale at noon. Compute the number of sales with a hot drink at noon that you would expect if the two events were independent, and compare it with the table.

  4. Compute \(P(H \cup A)\) and \(P(\bar{H} \cap \bar{M})\).

  5. On 11 weekdays, the kiosk also counted the hot drinks sold in the morning:

\[96, \; 104, \; 88, \; 112, \; 100, \; 93, \; 121, \; 99, \; 107, \; 90, \; 101\]

Compute the mean, the median, the quartiles \(Q_1\) and \(Q_3\), the range and the interquartile range (IQR). Decide with the fences (1.5 times the IQR below \(Q_1\) and above \(Q_3\)) whether the day with 121 hot drinks is an outlier.

Hint: With an odd number of values, leave the median out of both halves.

  1. \[P(H) = \frac{240}{600} = 0.4 \quad \text{(base: all 600 sales)}\]

    \[P(H \mid N) = \frac{45}{150} = 0.3 \quad \text{(base: the 150 sales at noon)}\]

    \[P(M \mid H) = \frac{100}{240} = \frac{5}{12} \approx 0.4167 \quad \text{(base: the 240 sales with a hot drink)}\]

  2. \(P(M) = \frac{250}{600} = \frac{5}{12}\). The test:

    \[P(H \cap M) = \frac{100}{600} = \frac{1}{6}, \qquad P(H) \cdot P(M) = 0.4 \cdot \frac{5}{12} = \frac{1}{6}\]

    Both sides are equal, so \(H\) and \(M\) are independent. In words: in the morning, 40 % of the sales include a hot drink (\(\frac{100}{250} = 0.4\)), the same share as over the whole day.

  3. \(P(N) = \frac{150}{600} = 0.25\):

    \[P(H \cap N) = \frac{45}{600} = 0.075, \qquad P(H) \cdot P(N) = 0.4 \cdot 0.25 = 0.1\]

    \(0.075 \neq 0.1\), so \(H\) and \(N\) are not independent.

    Expected count if they were independent: \(600 \cdot P(H) \cdot P(N) = 600 \cdot 0.1 = 60\) sales. The table shows only 45, so at noon fewer customers buy a hot drink than the share of the whole day would suggest (30 % instead of 40 %).

  4. Addition rule: \(P(H \cup A) = \frac{240 + 200 - 95}{600} = \frac{345}{600} = 0.575\)

    No hot drink and not in the morning, i.e. noon or afternoon without a hot drink: \(P(\bar{H} \cap \bar{M}) = \frac{105 + 105}{600} = \frac{210}{600} = 0.35\)

  5. Sorted: \(88, 90, 93, 96, 99, 100, 101, 104, 107, 112, 121\) (\(n = 11\)).

    • Mean: \(\bar{x} = \frac{1111}{11} = 101\) hot drinks
    • Median: the 6th value, \(100\)
    • \(Q_1\): median of the lower half \(88, 90, 93, 96, 99\), so \(Q_1 = 93\)
    • \(Q_3\): median of the upper half \(101, 104, 107, 112, 121\), so \(Q_3 = 107\)
    • Range: \(121 - 88 = 33\), IQR: \(107 - 93 = 14\)

    Fences: \(93 - 1.5 \cdot 14 = 72\) and \(107 + 1.5 \cdot 14 = 128\). The value 121 lies inside \([72; 128]\), so it is not an outlier. It still pulls the mean (101) a little above the median (100).

Typical mistake: in c), calling \(H\) and \(N\) dependent because 45 sales have both. Events that happen together are not mutually exclusive, but that says nothing about independence. Independence is decided only by the test \(P(H \cap N) = P(H) \cdot P(N)\).

Problem 4: Binomial Probabilities and the Minimum Number (xx)

In a bakery, 14 % of the customers pay with their smartphone, independently of each other. \(X\) is the number of smartphone payments among the next 18 customers.

  1. State the distribution of \(X\) with \(n\) and \(p\). Compute the expected value and the standard deviation of \(X\).

  2. Compute the probability that exactly 3 of the 18 customers pay with their smartphone. Check your result with Binomial-Dichte (MENU → 7 → 4 → 2).

  3. Compute the probability that not more than 2 customers pay with their smartphone, first as a sum of binomial terms and then with the cumulative binomial distribution (MENU → 7 → ▼ → 1 → 2).

  4. Compute the probability that at least one of the 18 customers pays with the smartphone.

  5. Determine the minimum number of customers for which at least one of them pays with the smartphone with a probability of at least 99 %. Check the two whole numbers next to your result.

  1. 18 customers (fixed \(n\)), each pays with the smartphone or not, independently, always with \(p = 0.14\). So \(X \sim B(18;\ 0.14)\):

    \[\mu = np = 18 \cdot 0.14 = 2.52, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{18 \cdot 0.14 \cdot 0.86} = \sqrt{2.1672} \approx 1.47\]

  2. \[P(X = 3) = \binom{18}{3} \cdot 0.14^3 \cdot 0.86^{15} = 816 \cdot 0.14^3 \cdot 0.86^{15} \approx 0.2331\]

    Calculator: MENU → 7 → 4 → 2, then x = 3 =, N = 18 =, p = 0,14 =, then = gives about 0.2331.

  3. “Not more than 2” means 0, 1 or 2:

    \[P(X \leq 2) = 0.86^{18} + 18 \cdot 0.14 \cdot 0.86^{17} + 153 \cdot 0.14^2 \cdot 0.86^{16} \approx 0.0662 + 0.1940 + 0.2685 = 0.5287\]

    Calculator: MENU → 7 → ▼ → 1 → 2, then x = 2 =, N = 18 =, p = 0,14 =, then = gives about 0.5287 ✓

  4. “At least one” is the complement of “nobody”:

    \[P(X \geq 1) = 1 - P(X = 0) = 1 - 0.86^{18} \approx 1 - 0.0662 = 0.9338\]

  5. With \(n\) customers: \(1 - 0.86^n \geq 0.99 \iff 0.86^n \leq 0.01\). Logarithm on both sides:

    \[n \cdot \ln(0.86) \leq \ln(0.01)\]

    \(\ln(0.86) < 0\), so dividing by it flips the inequality sign:

    \[n \geq \frac{\ln(0.01)}{\ln(0.86)} \approx \frac{-4.6052}{-0.1508} \approx 30.53\]

    Round up: at least 31 customers. Check: \(1 - 0.86^{30} \approx 0.9892 < 0.99\), but \(1 - 0.86^{31} \approx 0.9907 \geq 0.99\) ✓

Typical mistake: in e), keeping the inequality sign and getting \(n \leq 30.53\). This would say that fewer customers give a higher chance of a smartphone payment, which cannot be true.

Problem 5: Expected Value of a Game (xx)

At its summer party, a football club offers a game. A player pays 3 EUR and draws one ball from a box with 20 balls: 1 gold ball, 4 silver balls and 15 white balls. The gold ball wins 25 EUR, a silver ball wins 5 EUR, and a white ball wins nothing. \(X\) is the payout to the player in EUR.

  1. Give the probability distribution of \(X\) in a table and compute \(E[X]\).

  2. Compute the expected profit of a player per game and the expected profit of the club from 400 games. Determine the price at which the game would be fair (expected profit 0 for both sides).

  3. The club wants an expected profit of 1 EUR per game. The price of 3 EUR and the silver prizes stay the same. Determine the prize for the gold ball.

  4. The club also considers a second game: a player pays 3 EUR and rolls two dice. Two sixes win 60 EUR, everything else wins nothing. Compute the club’s expected profit per game and decide which game brings the club more on average. Name one reason why the club could still prefer the first game.

  1. Each ball is drawn with probability \(\frac{1}{20}\):

    \(x\) (EUR) 0 5 25
    \(P(X = x)\) \(\frac{15}{20} = 0.75\) \(\frac{4}{20} = 0.2\) \(\frac{1}{20} = 0.05\)

    Check: \(0.75 + 0.2 + 0.05 = 1\) ✓

    \[E[X] = 0 \cdot 0.75 + 5 \cdot 0.2 + 25 \cdot 0.05 = 0 + 1 + 1.25 = 2.25 \text{ EUR}\]

  2. The player’s profit is \(X - 3\). With \(E[aX + b] = aE[X] + b\):

    \[E[X - 3] = E[X] - 3 = 2.25 - 3 = -0.75 \text{ EUR}\]

    A player loses 0.75 EUR per game on average, so the club gains 0.75 EUR per game, and \(400 \cdot 0.75 = 300\) EUR from 400 games.

    Fair game: price \(= E[X] = 2.25\) EUR, because then \(E[X - 2.25] = 0\).

  3. Expected profit 1 EUR at a price of 3 EUR means an expected payout of 2 EUR. With the gold prize \(g\):

    \[5 \cdot 0.2 + g \cdot 0.05 = 2 \iff 0.05g = 1 \iff g = 20 \text{ EUR}\]

  4. Two sixes have the probability \(\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}\) (independent dice):

    \[E[Y] = 60 \cdot \frac{1}{36} = \frac{5}{3} \approx 1.67 \text{ EUR}, \qquad \text{club's expected profit } 3 - \frac{5}{3} = \frac{4}{3} \approx 1.33 \text{ EUR per game}\]

    On average, the second game brings the club more (1.33 EUR instead of 0.75 EUR per game).

    Reasons for the first game: players win more often (25 % of the games instead of about 3 %), which keeps them playing, and a single payout of 60 EUR is a larger risk for the club’s cash box on a short evening.

Typical mistake: in b), giving \(E[X] = 2.25\) EUR as the player’s profit. The payout is not the profit: the price of 3 EUR has to be subtracted.

B: Set Up & Check

Problem 6: From FSP Words to Probabilities (xx)

Translate each FSP statement into probability notation and compute what is asked. Then answer the check question: it tells you whether your translation works.

  1. A customs office scans containers in a port with an X-ray scanner. Let \(K\) = “the container holds smuggled goods” and \(T\) = “the scan raises an alarm (positive result)”. An FSP-style text says: “The prevalence of containers with smuggled goods is 0.06. The sensitivity of the scan is 0.82. The specificity of the scan is 0.92.” Here the prevalence is the share of containers with smuggled goods among all containers (the base rate), the sensitivity is the share of positive results among the containers with smuggled goods (the true positive rate), and the specificity is the share of negative results among the containers without smuggled goods (the true negative rate). Write the three values as probabilities and draw the tree diagram with all branch and path probabilities.

    Check: A classmate starts the tree with the scan result (\(T\) or \(\bar{T}\)) as the first stage. Explain why the three given values do not fit on this tree.

  2. A pharmacy offers a rapid test for a bacterial throat infection. Let \(K\) = “the customer has the infection” and \(T\) = “the test is positive”. Of 2,500 tested customers, 150 have the infection, and 126 of them test positive. Of the 2,350 customers without the infection, 141 test positive. Write the FSP phrases “a person with a positive test result is indeed ill” and “a person is ill but incorrectly identified as healthy” as probabilities and compute both.

    Check: A classmate answers 84 % for the first phrase and 16 % for the second. Find out which probabilities the classmate computed and name their bases.

  3. An online shop for craft supplies (wool, paper and paint) studies the people who visit its website. Let \(K\) = “has ordered from the shop in the last two years (is a customer)” and \(T\) = “has installed the shop’s app”. State the businesslike meaning of the quantities \(\bar{T}\) and \(\bar{K} \cap T\).

    Check: A classmate describes \(\bar{K} \cap T\) as “people who are neither customers nor app users”. Explain the mistake and write the classmate’s group in set notation.

  4. A courier delivers 90 parcels on one tour. For each parcel, the recipient is not at home with probability 7 %, independently of the other parcels. \(X\) is the number of parcels that cannot be delivered at the first attempt. Write the FSP events “not more than 5 parcels” and “at least 4 but not more than 10 parcels” with values \(P(X \leq k)\) of the cumulative binomial distribution and compute both.

    Check: Compute also “not more than 3 parcels” and “more than 10 parcels”. Check that these two events and the second FSP event together have the probability 1.

  5. A charity calls former donors and asks them for a new donation. Each former donor donates again with probability 4.5 %, independently of the others. Determine the number of former donors the charity has to call to get at least one new donation “with a likelihood (probability) of at least 95 %”.

    Check: A classmate gets \(n \approx 65.06\) and answers 65 donors. Compute the probability of at least one new donation for 65 and for 66 donors and decide whether the classmate is right.

  1. “Of the containers with smuggled goods” and “of the containers without smuggled goods” name the conditions \(K\) and \(\bar{K}\):

    \[P(K) = 0.06 \;\text{(prevalence)}, \qquad P(T \mid K) = 0.82 \;\text{(sensitivity)}, \qquad P(\bar{T} \mid \bar{K}) = 0.92 \;\text{(specificity)}\]

    Tree diagram (first stage: smuggled goods or not, second stage: scan result):

    • \(K\) (\(0.06\))
      • \(T\) (\(0.82\)): \(0.06 \cdot 0.82 = 0.0492\)
      • \(\bar{T}\) (\(0.18\)): \(0.06 \cdot 0.18 = 0.0108\)
    • \(\bar{K}\) (\(0.94\))
      • \(T\) (\(0.08\)): \(0.94 \cdot 0.08 = 0.0752\)
      • \(\bar{T}\) (\(0.92\)): \(0.94 \cdot 0.92 = 0.8648\)

    The path probabilities add up to 1: \(0.0492 + 0.0108 + 0.0752 + 0.8648 = 1\) ✓

    Check: the sensitivity and the specificity are shares of the containers with or without smuggled goods, so they are conditional on \(K\) or \(\bar{K}\). On a tree, the branches of the second stage carry probabilities under the condition of the first stage. With \(T\) as the first stage, the second stage would need \(P(K \mid T)\) and \(P(K \mid \bar{T})\), which are not given.

    Typical mistake: writing the specificity 0.92 on the branch \(\bar{K} \to T\). This branch carries the false positive rate \(1 - 0.92 = 0.08\), a false alarm for a clean container.

  2. “A person with a positive test result is indeed ill”: the positive result is known, so the base is the \(126 + 141 = 267\) positive results:

    \[P(K \mid T) = \frac{126}{267} = \frac{42}{89} \approx 0.4719\]

    “A person is ill but incorrectly identified as healthy”: ill and a negative result, one path of the tree, with all 2,500 customers as the base:

    \[P(K \cap \bar{T}) = \frac{150 - 126}{2500} = \frac{24}{2500} = 0.0096\]

    Check: \(\frac{126}{150} = 0.84 = P(T \mid K)\) is the sensitivity, and \(\frac{24}{150} = 0.16 = P(\bar{T} \mid K)\) is the false negative rate. Both have the 150 ill customers as the base. The first phrase starts with the positive result (base 267), the second speaks about all tested customers (base 2,500).

    Typical mistake: dividing by the ill customers whenever the word “ill” appears. The base is the group that is known: the positive results in the first phrase and all customers in the second.

  3. \(\bar{T}\): the person has not installed the shop’s app, so the shop cannot reach this person with offers in the app.

    \(\bar{K} \cap T\): the person has installed the app, but has not ordered in the last two years. These are app users who are not (or no longer) customers, a group that the shop could try to win back with offers in the app.

    Check: the bar stands only over \(K\), so \(T\) (“has the app”) still holds. “Neither customer nor app user” is \(\bar{K} \cap \bar{T}\), the complement of \(K \cup T\).

    Typical mistake: answering with letters only (“not \(K\) and \(T\)”) or with a probability. The FSP asks who these people are for the business, in words.

  4. \(X \sim B(90;\ 0.07)\). “Not more than 5” is already a cumulative value:

    \[P(X \leq 5) \approx 0.3916\]

    “At least 4 but not more than 10” includes both endpoints, so subtract the cumulative value at \(4 - 1 = 3\):

    \[P(4 \leq X \leq 10) = P(X \leq 10) - P(X \leq 3) \approx 0.9503 - 0.1174 = 0.8329\]

    (cumulative binomial distribution with N = 90, p = 0,07 and x = 5, 10, 3)

    Check: \(P(X \leq 3) \approx 0.1174\) and \(P(X > 10) = 1 - P(X \leq 10) \approx 0.0497\). The three events cover every value of \(X\) exactly once: \(0.1174 + 0.8329 + 0.0497 = 1\) ✓

    Typical mistake: computing \(P(X \leq 10) - P(X \leq 4)\). This leaves out the tours with exactly 4 parcels that cannot be delivered, which belong to the event.

  5. Complement: \(1 - 0.955^n \geq 0.95 \iff 0.955^n \leq 0.05\). Logarithm on both sides:

    \[n \cdot \ln(0.955) \leq \ln(0.05)\]

    \(\ln(0.955) < 0\), so dividing by it flips the sign:

    \[n \geq \frac{\ln(0.05)}{\ln(0.955)} \approx \frac{-2.9957}{-0.0460} \approx 65.06\]

    Round up: the charity has to call at least 66 former donors.

    Check: \(1 - 0.955^{65} \approx 0.9499 < 0.95\) and \(1 - 0.955^{66} \approx 0.9521 \geq 0.95\). The classmate is wrong: with 65 donors, the likelihood stays just below 95 %.

    Typical mistake: rounding to the nearest whole number. A minimum is always rounded up, because the rounded number must still meet the demand.

Problem 7: One Paragraph, All Questions (xx)

A regional energy supplier studies the households it supplies. 13 % of the households have solar panels on the roof (\(K\)). Of the households with solar panels, 45 % use the supplier’s dynamic tariff (\(T\)), in which the electricity price changes every hour. Of the households without solar panels, 8 % use the dynamic tariff. The supplier wants to know the share of the dynamic-tariff households that have solar panels, the share of all households that have solar panels but do not use the dynamic tariff, and the businesslike meaning of the quantity \(\bar{K} \cap \bar{T}\). For a random sample of 25 households, it asks for the likelihood that exactly 5 households have solar panels, that not more than 3 households have solar panels, and that at least 4 but not more than 8 households have solar panels. Finally, it wants the number of households it has to ask to find at least one household with the dynamic tariff with a likelihood of at least 90 %.

List every given value in probability notation. Then write every question of the supplier as a mathematical statement and name the rule or the distribution you would use. Do not compute anything.

Check: An assistant writes the first question as \(P(T \mid K)\). Explain the mistake.

Given values:

  1. \(P(K) = 0.13\), so \(P(\bar{K}) = 1 - 0.13\)
  2. \(P(T \mid K) = 0.45\)
  3. \(P(T \mid \bar{K}) = 0.08\)

Questions as mathematical statements:

  1. Share of the dynamic-tariff households with solar panels: \(P(K \mid T) = \frac{P(T \mid K) \cdot P(K)}{P(T \mid K) \cdot P(K) + P(T \mid \bar{K}) \cdot P(\bar{K})}\), Bayes’ theorem with the law of total probability in the denominator
  2. Solar panels but no dynamic tariff: \(P(K \cap \bar{T}) = P(K) \cdot (1 - P(T \mid K))\), multiplication rule along one path (all households as the base)
  3. \(\bar{K} \cap \bar{T}\): households without solar panels that do not use the dynamic tariff, given in words
  4. \(X\) = number of households with solar panels among the 25, \(X \sim B(25;\ 0.13)\): \(P(X = 5)\), \(P(X \leq 3)\) and \(P(4 \leq X \leq 8) = P(X \leq 8) - P(X \leq 3)\)
  5. Minimum number \(n\): \(1 - (1 - P(T))^n \geq 0.9\) with \(P(T) = P(T \mid K) \cdot P(K) + P(T \mid \bar{K}) \cdot P(\bar{K})\) (law of total probability), solved with logarithms and rounded up

Check: \(P(T \mid K) = 0.45\) is a given value: the share of the households with solar panels that use the tariff. The question starts from the households with the dynamic tariff, so \(T\) is the condition and goes behind the bar: \(P(K \mid T)\).

Typical mistake: in 8, using \(P(T \mid K) = 0.45\) as \(p\). The supplier asks random households, not only households with solar panels, so \(p\) is the share \(P(T)\) of all households.

C: Applied

Problem 8: Listeners of a Regional Radio Station (xxx)

A regional radio station wants to know how many adults in its region it reaches and how many of them follow the station on social media. Consider the following probabilities:

  • The probability that a random adult in the region listens to the station at least once a week (a regular listener) is 24 %.
  • A regular listener follows the station on social media with a probability of 70 %.
  • An adult who is not a regular listener follows the station on social media with a probability of 6 %.

Use the following notation: \(K\): “is a regular listener of the station”, \(T\): “follows the station on social media”.

  1. The region has 80,000 adults. Display the fractions in a contingency table. Fill in the blank fields.
\(K\) \(\bar{K}\) sum
\(T\)
\(\bar{T}\)
sum 80,000

(for verification: 17,088 adults follow the station on social media)

  1. State the businesslike meaning of the quantities \(\bar{K}\) and \(K \cap \bar{T}\).

  2. Compute the probabilities \(P(K \cap \bar{T})\) and \(P(K \mid \bar{T})\) with the table. Explain why the two values differ.

  3. A person follows the station on social media. Compute the probability that this person is a regular listener of the station.

  4. The station asks a random sample of 13 adults from the region. \(X\) is the number of adults in the sample who follow the station on social media. Determine the likelihood of the following events:

(1) Exactly 4 adults follow the station on social media.

(2) Not more than 1 adult follows the station on social media.

(3) At least 3 but not more than 5 adults follow the station on social media.

a) Given: \(P(K) = 0.24\), \(P(T \mid K) = 0.7\), \(P(T \mid \bar{K}) = 0.06\). Out of 80,000 adults:

  • \(K\): \(80{,}000 \cdot 0.24 = 19{,}200\), so \(\bar{K}\): \(80{,}000 - 19{,}200 = 60{,}800\)
  • \(T \cap K\): \(19{,}200 \cdot 0.7 = 13{,}440\), \(\bar{T} \cap K\): \(19{,}200 \cdot 0.3 = 5{,}760\)
  • \(T \cap \bar{K}\): \(60{,}800 \cdot 0.06 = 3{,}648\), \(\bar{T} \cap \bar{K}\): \(60{,}800 \cdot 0.94 = 57{,}152\)
\(K\) \(\bar{K}\) sum
\(T\) 13,440 3,648 17,088
\(\bar{T}\) 5,760 57,152 62,912
sum 19,200 60,800 80,000

Check: \(17{,}088 + 62{,}912 = 80{,}000\) and \(19{,}200 + 60{,}800 = 80{,}000\) ✓

b) \(\bar{K}\): the adult does not listen to the station at least once a week. For the station, these are the adults it does not reach regularly, i.e. possible new listeners.

\(K \cap \bar{T}\): the adult listens to the station at least once a week, but does not follow it on social media. The station reaches these listeners on the radio, but not with its posts.

c) Both values have the same numerator 5,760, but different bases:

\[P(K \cap \bar{T}) = \frac{5760}{80000} = 0.072, \qquad P(K \mid \bar{T}) = \frac{5760}{62912} = \frac{90}{983} \approx 0.0916\]

\(P(K \cap \bar{T})\) is a joint probability with all 80,000 adults as the base: 7.2 % of all adults are regular listeners who do not follow the station. \(P(K \mid \bar{T})\) is a conditional probability with the 62,912 adults who do not follow the station as the base: about 9.16 % of them are regular listeners. The base of the conditional probability is smaller, so its value is larger.

d) The condition is \(T\), so the base is the 17,088 followers (Bayes’ theorem in the table):

\[P(K \mid T) = \frac{P(K \cap T)}{P(T)} = \frac{13440}{17088} = \frac{70}{89} \approx 0.7865\]

With the formula: \(\frac{0.24 \cdot 0.7}{0.24 \cdot 0.7 + 0.76 \cdot 0.06} = \frac{0.168}{0.2136}\), the same value ✓. About 79 % of the followers are regular listeners.

e) An adult follows the station with \(p = P(T) = \frac{17088}{80000} = 0.2136\). The 13 adults are a random sample from a large population, so they count as independent: \(X \sim B(13;\ 0.2136)\) with

\[P(X = k) = \binom{13}{k} \cdot 0.2136^k \cdot 0.7864^{13 - k}\]

(1) \[P(X = 4) = \binom{13}{4} \cdot 0.2136^4 \cdot 0.7864^9 = 715 \cdot 0.2136^4 \cdot 0.7864^9 \approx 0.1712\]

(2) “Not more than 1” means 0 or 1:

\[P(X \leq 1) = 0.7864^{13} + 13 \cdot 0.2136 \cdot 0.7864^{12} \approx 0.0440 + 0.1553 = 0.1993\]

(3) Both endpoints belong to the event:

\[P(3 \leq X \leq 5) = P(X = 3) + P(X = 4) + P(X = 5) \approx 0.2521 + 0.1712 + 0.0837 = 0.5070\]

(with \(\binom{13}{3} = 286\) and \(\binom{13}{5} = 1287\), and the sum of the unrounded terms). Check with the cumulative binomial distribution: \(P(X \leq 5) - P(X \leq 2) \approx 0.9595 - 0.4525 = 0.5070\) ✓

Typical mistake: in e), using \(p = 0.24\), the share of regular listeners. \(X\) counts the followers, and an adult of the sample follows the station with \(P(T) = 0.2136\), whether or not the adult is a regular listener.

Problem 9: Faulty Charging Cables (xxx)

A company makes USB charging cables and delivers them to a chain of mobile phone shops in cartons of 250 cables. Each cable is faulty with probability 1.8 %, independently of the other cables. \(X\) is the number of faulty cables in a carton. The contract with the chain says: the company replaces every faulty cable, which costs it 9 EUR per cable. If a carton contains more than 8 faulty cables, the chain also charges a penalty of 600 EUR for the extra work in its shops.

  1. Give reasons why \(X\) follows a binomial distribution. Compute the expected value and the standard deviation of \(X\).

  2. Compute the probability for the following events:

(1) Exactly 5 cables in a carton are faulty.

(2) Not more than 3 cables in a carton are faulty.

(3) At least 3 but not more than 8 cables in a carton are faulty.

(for verification: \(P(X \leq 8) \approx 0.9612\))

  1. Compute the probability that the chain charges the penalty for a carton. Compute the expected costs per carton for replacements and penalties.

Hint: The expected replacement costs are 9 EUR times the expected number of faulty cables. The penalty is a cost of 600 EUR that occurs with a certain probability.

  1. A final test of every cable before packing costs 0.30 EUR per cable. After the test, only 0.3 % of the cables in a carton are still faulty, because the test misses a few faults. Compute the expected costs per carton with the test (test, replacements and penalties) and decide whether the test pays off.

  2. Another supplier offers the same test at a lower price. Determine the highest price per cable at which the test pays off.

a) A carton has a fixed number of \(n = 250\) cables, each cable is faulty or not (two outcomes), the cables are faulty independently of each other, and every cable has the same probability \(p = 0.018\). So \(X \sim B(250;\ 0.018)\):

\[\mu = np = 250 \cdot 0.018 = 4.5, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{250 \cdot 0.018 \cdot 0.982} = \sqrt{4.419} \approx 2.10\]

b) Events (2) and (3) have many terms, so use the cumulative binomial distribution \(P(X \leq x)\) (MENU → 7 → ▼ → 1 → 2, N = 250, p = 0,018).

(1) Binomial-Dichte (MENU → 7 → 4 → 2) with x = 5, N = 250, p = 0,018:

\[P(X = 5) = \binom{250}{5} \cdot 0.018^5 \cdot 0.982^{245} \approx 0.1725\]

(2) “Not more than 3” means \(0, 1, 2, 3\): \(P(X \leq 3) \approx 0.3400\)

(3) Both endpoints belong to the event, so subtract the cumulative value at \(3 - 1 = 2\):

\[P(3 \leq X \leq 8) = P(X \leq 8) - P(X \leq 2) \approx 0.9612 - 0.1710 = 0.7902\]

c) The penalty is charged for “more than 8”, i.e. \(X \geq 9\), the complement of \(X \leq 8\):

\[P(X > 8) = 1 - P(X \leq 8) \approx 1 - 0.9612 = 0.0388\]

Expected costs per carton: the replacements cost \(9 \cdot E[X]\), and the penalty of 600 EUR occurs with the probability \(P(X > 8)\):

\[E[\text{costs}] = 9 \cdot 4.5 + 600 \cdot P(X > 8) \approx 40.50 + 600 \cdot 0.038792 \approx 40.50 + 23.28 = 63.78 \text{ EUR}\]

d) With the test, \(Y\) = number of faulty cables in a carton, \(Y \sim B(250;\ 0.003)\) with \(E[Y] = 250 \cdot 0.003 = 0.75\). The cumulative binomial distribution with x = 8, N = 250, p = 0,003 gives \(P(Y \leq 8) \approx 1.0000\), so \(P(Y > 8) \approx 0.0000\) and the penalty practically never occurs.

\[E[\text{costs with test}] = 250 \cdot 0.30 + 9 \cdot 0.75 + 600 \cdot P(Y > 8) \approx 75 + 6.75 + 0 = 81.75 \text{ EUR}\]

\(81.75 > 63.78\): the test does not pay off. It costs about 17.97 EUR more per carton than it saves.

e) With a price of \(c\) EUR per cable, the test pays off if the expected costs with the test are not higher than without it:

\[250c + 6.75 \leq 63.7754 \iff 250c \leq 57.0254 \iff c \leq 0.2281\]

(with the unrounded value \(600 \cdot 0.038792 \approx 23.2754\)). The term \(600 \cdot P(Y > 8)\) is below 0.0001 EUR and does not change the result. A highest price is rounded down: the test pays off up to 0.22 EUR per cable. Check: at 0.22 EUR, \(55 + 6.75 = 61.75 < 63.78\) ✓, at 0.23 EUR, \(57.50 + 6.75 = 64.25 > 63.78\).

Typical mistake: in c), adding the full penalty of 600 EUR to the costs of every carton. The chain charges it only for about 3.88 % of the cartons, so the penalty adds \(600 \cdot 0.0388 \approx 23.28\) EUR per carton on average.

Problem 10: A Money-Back Guarantee for Dolphin Tours (xxx)

A boat company on the Atlantic coast offers dolphin-watching tours. A ticket costs 52 EUR. From past seasons, the sea is calm (\(C\)) on 72 % of the tour days and rough on the other days. On a calm day, the guests see dolphins (\(D\)) on 90 % of the tours, on a rough day only on 55 % of the tours. The company plans a guarantee: every guest who sees no dolphins gets the full ticket price back.

  1. Create a tree diagram with all branch and path probabilities and compute the probability that the guests of a tour see no dolphins.

(for verification: 19.8 %)

  1. The guests of a tour have seen no dolphins. Compute the probability that the sea was rough on that day. Compare the result with the share of rough days and explain the difference.

  2. \(R\) is the company’s revenue from one ticket after the guarantee, in EUR. Give the probability distribution of \(R\) and compute \(E[R]\). The company wants to keep an expected revenue of 52 EUR per ticket with the guarantee. Determine the new ticket price.

  3. Instead of the money back, the company could give every guest who sees no dolphins a voucher for a free second tour. A guest with a voucher costs the company 14 EUR on a tour that runs anyway (fuel, crew and insurance per guest), and 75 % of the guests use their voucher. Compute the expected costs of the voucher per ticket. Determine the ticket price that keeps an expected revenue of 52 EUR per ticket after these costs, and compare it with the price from c).

a) First stage: state of the sea, second stage: dolphins seen or not.

  • \(C\) (\(0.72\))
    • \(D\) (\(0.9\)): \(0.72 \cdot 0.9 = 0.648\)
    • \(\bar{D}\) (\(0.1\)): \(0.72 \cdot 0.1 = 0.072\)
  • \(\bar{C}\) (\(0.28\))
    • \(D\) (\(0.55\)): \(0.28 \cdot 0.55 = 0.154\)
    • \(\bar{D}\) (\(0.45\)): \(0.28 \cdot 0.45 = 0.126\)

Check: \(0.648 + 0.072 + 0.154 + 0.126 = 1\) ✓

Law of total probability:

\[P(\bar{D}) = 0.072 + 0.126 = 0.198\]

On 19.8 % of the tours, the guests see no dolphins.

b) Bayes’ theorem, the base is all tours without dolphins:

\[P(\bar{C} \mid \bar{D}) = \frac{P(\bar{D} \mid \bar{C}) \cdot P(\bar{C})}{P(\bar{D})} = \frac{0.126}{0.198} = \frac{7}{11} \approx 0.6364\]

Only 28 % of the tour days are rough, but about 64 % of the tours without dolphins happen on rough days. On a rough day, a tour misses the dolphins far more often (45 % instead of 10 %), so the rough days are overrepresented among the tours without dolphins.

c) The company keeps the ticket price if the guests see dolphins and pays everything back otherwise:

\(r\) (EUR) 52 0
\(P(R = r)\) 0.802 0.198

\[E[R] = 52 \cdot 0.802 + 0 \cdot 0.198 = 41.704 \approx 41.70 \text{ EUR}\]

The guarantee costs the company about 10.30 EUR per ticket on average. With a new price \(p\), the expected revenue is \(p \cdot 0.802\):

\[p \cdot 0.802 \geq 52 \iff p \geq \frac{52}{0.802} \approx 64.8379\]

The price is a minimum, so round up: 64.84 EUR. Check: \(64.84 \cdot 0.802 \approx 52.0017 \geq 52\) ✓, but \(64.83 \cdot 0.802 \approx 51.9937 < 52\).

d) A guest causes voucher costs if the guest sees no dolphins and uses the voucher:

\[P(\text{voucher used}) = 0.198 \cdot 0.75 = 0.1485, \qquad E[\text{voucher costs}] = 14 \cdot 0.1485 = 2.079 \approx 2.08 \text{ EUR per ticket}\]

With a new price \(p\), the expected revenue after these costs is \(p - 2.079\):

\[p - 2.079 \geq 52 \iff p \geq 54.079\]

Round up: 54.08 EUR. Check: \(54.08 - 2.079 = 52.001 \geq 52\) ✓, \(54.07 - 2.079 = 51.991 < 52\).

The voucher needs a price increase of only 2.08 EUR, the money-back guarantee one of 12.84 EUR. The voucher is much cheaper for the company because a guest on a tour that runs anyway costs only 14 EUR, not the full ticket price, and not every guest uses the voucher.

Typical mistake: in c), adding the expected refund to the price: \(52 + 52 \cdot 0.198 = 62.30\) EUR. The refund grows with the new price, because the company pays back the new price, so the equation is \(p \cdot 0.802 = 52\).

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: Glaucoma Screening at a Health Insurance Fund [26 pts. total]

Glaucoma is an eye disease that damages the optic nerve (the nerve between the eye and the brain). Without early treatment, it can lead to blindness. A health insurance fund offers its insured persons aged 40 and over a free screening with an eye-pressure test. Persons with a positive test result (\(+\)) are referred to an eye doctor for a full examination. In one year, 75,000 insured persons take the test. Let \(G\) be the event “the person has glaucoma”.

  • The prevalence (the share of persons with glaucoma among all tested persons, the base rate) is 1.2 %.
  • In 86 % of all persons with glaucoma, the test is positive. (The sensitivity, i.e. the true positive rate, of the test is 0.86.)
  • In 93 % of all persons without glaucoma, the test is negative (\(-\)). (The specificity, i.e. the true negative rate, of the test is 0.93.)

Round probabilities to four decimal places where rounding is necessary.

Part a)

Create a tree diagram for the screening and complete the following table.

glaucoma (\(G\)) no glaucoma (\(\bar{G}\)) sum
test positive (\(+\))
test negative (\(-\))
sum 75,000

[7 pts.]

Given: \(P(G) = 0.012\), \(P(+ \mid G) = 0.86\), \(P(- \mid \bar{G}) = 0.93\). Complements: \(P(\bar{G}) = 0.988\), \(P(- \mid G) = 0.14\) (false negative rate), \(P(+ \mid \bar{G}) = 0.07\) (false positive rate).

Tree diagram (first stage: glaucoma or not, second stage: test result, path probability = product along the path):

  • \(G\) (\(0.012\))
    • \(+\) (\(0.86\)): \(0.012 \cdot 0.86 = 0.01032\)
    • \(-\) (\(0.14\)): \(0.012 \cdot 0.14 = 0.00168\)
  • \(\bar{G}\) (\(0.988\))
    • \(+\) (\(0.07\)): \(0.988 \cdot 0.07 = 0.06916\)
    • \(-\) (\(0.93\)): \(0.988 \cdot 0.93 = 0.91884\)

Table (path probability \(\cdot\) 75,000):

\(G\) \(\bar{G}\) sum
\(+\) 774 5,187 5,961
\(-\) 126 68,913 69,039
sum 900 74,100 75,000

Check: \(5{,}961 + 69{,}039 = 75{,}000\) and \(900 + 74{,}100 = 75{,}000\) ✓

Grading: 1 Pt. complements, 2 Pts. tree diagram with branch and path probabilities, 3 Pts. inner fields, 1 Pt. sums with the check.

Part b)

(for verification: 5,961 of the 75,000 persons test positive)

Compute the probability that a person with a positive test result does indeed suffer from glaucoma. Interpret the result for the persons who are referred to an eye doctor.

[3 pts.]

Bayes’ theorem, the base is all positive results (row sum 5,961):

\[P(G \mid +) = \frac{P(+ \mid G) \cdot P(G)}{P(+ \mid G) \cdot P(G) + P(+ \mid \bar{G}) \cdot P(\bar{G})} = \frac{0.01032}{0.01032 + 0.06916} = \frac{774}{5961} = \frac{258}{1987} \approx 0.1298\]

Interpretation: only about 13 % of the referred persons have glaucoma, so about 87 of 100 referrals are false alarms. Glaucoma is rare (1.2 %), and 7 % of the many persons without glaucoma still test positive. A positive eye-pressure test is a reason for the full examination, not a diagnosis.

Grading: 1 Pt. Bayes’ theorem with the correct base, 1 Pt. \(P(G \mid +) \approx 0.1298\), 1 Pt. interpretation.

Part c)

Compute the probability that a person does not suffer from glaucoma but is incorrectly identified as ill. The full examination at the eye doctor costs the fund 85 EUR per referred person. Determine the number of referred persons without glaucoma in one year and the costs of their examinations.

[4 pts.]

“No glaucoma and a positive result” is one path of the tree, a joint probability with all tested persons as the base:

\[P(\bar{G} \cap +) = P(\bar{G}) \cdot P(+ \mid \bar{G}) = 0.988 \cdot 0.07 = 0.06916\]

So 6.916 % of all tested persons are referred although they do not have glaucoma: \(75{,}000 \cdot 0.06916 = 5{,}187\) persons (the field \(\bar{G} \cap +\) of the table).

Costs of their examinations: \(5{,}187 \cdot 85 = 440{,}895\) EUR per year.

Grading: 1 Pt. joint probability along the path \(\bar{G} \to +\), 1 Pt. \(0.06916\), 1 Pt. 5,187 persons, 1 Pt. 440,895 EUR.

Part d)

In one town, 750 insured persons take the test. Compute the probability for the following events:

(1) exactly 11 of the 750 persons have glaucoma

(2) not more than 6 of the 750 persons have glaucoma

(3) at least 8 but not more than 14 of the 750 persons have glaucoma

[6 pts.]

\(X\) = number of persons with glaucoma among the 750. The 750 persons are a fixed number, each has glaucoma or not, independently of the others, with the same probability \(p = 0.012\) (the prevalence). So \(X \sim B(750;\ 0.012)\) with

\[P(X = k) = \binom{750}{k} \cdot 0.012^k \cdot 0.988^{750 - k}\]

(1) Exactly 11, Binomial-Dichte with x = 11, N = 750, p = 0,012:

\[P(X = 11) \approx 0.0975\]

Events (2) and (3) have many terms, so use the cumulative binomial distribution \(P(X \leq x)\) (MENU → 7 → ▼ → 1 → 2, N = 750, p = 0,012).

(2) “Not more than 6” means \(0, 1, \ldots, 6\):

\[P(X \leq 6) \approx 0.2051\]

(3) “At least 8 but not more than 14” includes both endpoints, so subtract the cumulative value at \(8 - 1 = 7\):

\[P(8 \leq X \leq 14) = P(X \leq 14) - P(X \leq 7) \approx 0.9595 - 0.3225 = 0.6370\]

Grading: 1 Pt. model \(B(750;\ 0.012)\) with the reason, 1 Pt. (1), 1 Pt. (2). 3 Pts. (3): 1 Pt. \(P(X \leq 14) - P(X \leq 7)\) with the lower endpoint \(8 - 1 = 7\), 1 Pt. both cumulative values, 1 Pt. result.

Part e)

Determine the number of persons which need to be tested to find at least one person testing positive with a likelihood (probability) of 99 %. Check your result with the two whole numbers next to it.

[6 pts.]

Probability of a positive result (law of total probability, or the row sum from part a)):

\[P(+) = 0.01032 + 0.06916 = 0.07948 = \frac{5961}{75000}\]

Complement: “at least one positive result among \(n\) persons” \(= 1 -\) “no positive result”:

\[1 - (1 - 0.07948)^n \geq 0.99 \iff 0.92052^n \leq 0.01\]

Logarithm on both sides with \(\ln(a^n) = n \cdot \ln(a)\):

\[n \cdot \ln(0.92052) \leq \ln(0.01)\]

\(\ln(0.92052) < 0\), so dividing by it flips the inequality sign:

\[n \geq \frac{\ln(0.01)}{\ln(0.92052)} \approx \frac{-4.6052}{-0.0828} \approx 55.61\]

At least 56 persons have to be tested. The number of persons is a whole number and must meet the demand, so round up.

Check: with 55 persons, \(1 - 0.92052^{55} \approx 0.9895 < 0.99\), which is not enough. With 56 persons, \(1 - 0.92052^{56} \approx 0.9903 \geq 0.99\) ✓

Grading: 1 Pt. \(p = P(+) = 0.07948\), 1 Pt. complement inequality, 1 Pt. logarithm, 1 Pt. sign flip with \(n \geq 55.61\), 1 Pt. \(n = 56\) rounded up, 1 Pt. check of 55 and 56.