Tasks 07-07 - Binomial Distribution

Section 07: Probability & Statistics

A: Skills

Problem 1: Binomial or Not? (x)

Give probabilities as fractions or as decimals with four decimal places, and keep all intermediate results unrounded. Give expected values and standard deviations to two decimal places. This rule holds for the whole sheet.

In each scenario, \(X\) counts something:

  1. A dry-cleaning shop cleans 15 shirts for a hotel. For each shirt, a stain cannot be removed with probability 4 %, independently of the other shirts. \(X\) is the number of shirts that still have a stain.
  2. A market researcher asks passers-by for an interview until 5 of them have agreed. \(X\) is the number of passers-by the researcher asks.
  3. A greengrocer (a shop that sells fruit and vegetables) has a crate with 22 melons, and 3 of them are not ripe yet. A customer picks 4 melons at random. \(X\) is the number of unripe melons among the 4.
  4. A tour operator counts how many of the 12 members of a travel group miss their flight. The whole group travels to the airport on the same bus. \(X\) is the number of members who miss the flight.

  1. Name the four requirements of a binomial experiment.

  2. Decide for each scenario whether \(X\) follows a binomial distribution. If yes, state \(n\), \(p\) and what counts as a success. If no, name a requirement that fails.

  3. For scenario 3, compute the probability that exactly one of the 4 melons is unripe and the probability that at least one of them is unripe.

  4. A classmate models scenario 3 with a binomial distribution with \(n = 4\) and \(p = \frac{3}{22}\). Compute \(P(X = 0)\) with this model and compare it with the exact value.

  1. A binomial experiment needs

    • a fixed number \(n\) of trials, known in advance,
    • two outcomes per trial: success (probability \(p\)) or failure (probability \(1 - p\)),
    • independent trials,
    • the same success probability \(p\) in every trial.
    • Scenario 1 is binomial: \(n = 15\) shirts, \(p = 0.04\), success = the shirt still has a stain. The shirts are independent, and the probability is the same for every shirt.
    • Scenario 2 is not binomial. The number of trials is not fixed: the researcher stops after the 5th agreement, so \(n\) is not known in advance.
    • Scenario 3 is not binomial. The customer picks the melons without replacement from a small crate, so the trials are not independent and \(p\) changes from melon to melon. This is a hypergeometric distribution with \(N = 22\), \(K = 3\) and \(n = 4\).
    • Scenario 4 is not binomial. The members are not independent: if the bus is stuck in a traffic jam, all 12 miss the flight together.

  3. Hypergeometric formula with \(N = 22\), \(K = 3\), \(n = 4\):

    \[P(X = 1) = \frac{\binom{3}{1}\binom{19}{3}}{\binom{22}{4}} = \frac{3 \cdot 969}{7315} = \frac{2907}{7315} = \frac{153}{385} \approx 0.3974\]

    “At least one” is the complement of “none”:

    \[P(X = 0) = \frac{\binom{3}{0}\binom{19}{4}}{\binom{22}{4}} = \frac{3876}{7315} = \frac{204}{385} \approx 0.5299, \qquad P(X \geq 1) = 1 - \frac{204}{385} = \frac{181}{385} \approx 0.4701\]

  4. Binomial model: \(P(X = 0) = \left(\frac{19}{22}\right)^4 \approx 0.5563\), compared with the exact \(0.5299\).

    The binomial model overestimates the chance of 4 ripe melons. Without replacement, every ripe melon the customer takes leaves a larger share of unripe melons in the crate. The sample of 4 is a large part of the 22 melons, so this effect is not small.

Typical mistake: in b), calling scenario 3 binomial because each melon is “ripe or unripe”. Two outcomes are only one of the four requirements. A draw without replacement from a small lot breaks independence.

Problem 2: Binomial Probabilities (x)

At an indoor swimming pool, 40 % of the visitors who come for a swim also buy something at the pool’s snack bar. On a quiet morning, 10 visitors come for a swim, independently of each other. \(X\) is the number of these visitors who buy something at the snack bar.

  1. State \(n\), \(p\) and the success, and write the binomial formula for \(P(X = k)\).

  2. Compute the probability that exactly 5 visitors buy something at the snack bar and the probability that nobody does.

  3. Compute the probability that at least one visitor buys something at the snack bar.

  4. Compute the probability that at most 2 visitors buy something at the snack bar.

  1. \(n = 10\) visitors, \(p = 0.4\), success = the visitor buys something at the snack bar, so \(X \sim B(10;\ 0.4)\):

    \[P(X = k) = \binom{10}{k} \cdot 0.4^k \cdot 0.6^{10 - k}\]

  2. \[P(X = 5) = \binom{10}{5} \cdot 0.4^5 \cdot 0.6^5 = 252 \cdot 0.01024 \cdot 0.07776 \approx 0.2007\]

    \[P(X = 0) = \binom{10}{0} \cdot 0.4^0 \cdot 0.6^{10} = 0.6^{10} \approx 0.0060\]

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

    \[P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6^{10} \approx 0.9940\]

  4. “At most 2” means 0, 1 or 2:

    \[P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\]

    \[= 0.6^{10} + 10 \cdot 0.4 \cdot 0.6^9 + 45 \cdot 0.4^2 \cdot 0.6^8 \approx 0.0060 + 0.0403 + 0.1209 = 0.1673\]

    (The sum uses the unrounded terms.)

Typical mistake: in b), forgetting the factor \(\binom{10}{5} = 252\). The product \(0.4^5 \cdot 0.6^5\) is the probability of one particular sequence, e.g. the first 5 visitors buy and the last 5 do not. There are 252 sequences with exactly 5 buyers.

Problem 3: Quality Control at a Tile Factory (xx)

A tile factory packs ceramic tiles in boxes of 24. Each tile is cracked with probability 6 %, independently of the other tiles. \(X\) is the number of cracked tiles in a box.

  1. Compute the probability that a box contains exactly 2 cracked tiles.

  2. Compute the probability that a box contains at most 1 cracked tile.

  3. Compute the probability that a box contains at least 2 cracked tiles. Explain why the complement is the shorter way.

  4. Compute the expected value and the standard deviation of \(X\) and interpret the expected value.

  5. A tiler needs at least 22 intact tiles from one box. Compute the probability that one box is enough.

\(X \sim B(24;\ 0.06)\) with \(P(X = k) = \binom{24}{k} \cdot 0.06^k \cdot 0.94^{24 - k}\).

  1. \[P(X = 2) = \binom{24}{2} \cdot 0.06^2 \cdot 0.94^{22} = 276 \cdot 0.0036 \cdot 0.94^{22} \approx 0.2547\]

  2. \[P(X \leq 1) = 0.94^{24} + 24 \cdot 0.06 \cdot 0.94^{23} \approx 0.2265 + 0.3470 = 0.5735\]

  3. “At least 2” means \(2, 3, \ldots, 24\), i.e. 23 terms. The complement “at most 1” has only 2 terms:

    \[P(X \geq 2) = 1 - P(X \leq 1) \approx 1 - 0.5735 = 0.4265\]

  4. \[\mu = np = 24 \cdot 0.06 = 1.44, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{24 \cdot 0.06 \cdot 0.94} = \sqrt{1.3536} \approx 1.16\]

    Over many boxes, a box contains 1.44 cracked tiles on average.

  5. At least 22 intact tiles out of 24 means at most 2 cracked tiles:

    \[P(X \leq 2) = P(X \leq 1) + P(X = 2) \approx 0.5735 + 0.2547 = 0.8282\]

    One box is enough with a probability of about 82.82 %.

Typical mistake: in e), computing \(P(X \geq 22)\) with \(X\) = number of cracked tiles. The event is about intact tiles, so translate it first: 22 or more intact tiles means 2 or fewer cracked tiles.

Problem 4: The Last Business-Class Seat (xx)

At the gate, an airline offers the last free business-class seat of a flight to economy passengers for an extra 120 EUR. A staff member asks the passengers one after another until one of them accepts. Each passenger accepts with probability 12.5 %, independently of the others. \(X\) is the number of the passenger who accepts first.

  1. Compute the probability that the 4th passenger is the first to accept.

  2. Compute the probability that none of the first 6 passengers accepts.

  3. Compute the expected number of passengers the staff member has to ask.

  4. Compute the probability that the seat is sold within the first 5 offers.

  5. The first 5 passengers have refused. The staff member says: “After five refusals, one of the next two passengers will almost surely accept.” Compute the probability that the next 2 passengers also refuse and compare it with the probability that the first 2 passengers refuse.

\(X\) follows a geometric distribution with \(p = 0.125\) and \(1 - p = 0.875\).

  1. The first 3 passengers refuse, the 4th accepts:

    \[P(X = 4) = 0.875^3 \cdot 0.125 \approx 0.0837\]

  2. \[P(X > 6) = 0.875^6 \approx 0.4488\]

  3. \[E[X] = \frac{1}{p} = \frac{1}{0.125} = 8 \text{ passengers}\]

  4. \[P(X \leq 5) = 1 - 0.875^5 \approx 1 - 0.5129 = 0.4871\]

  5. Memoryless property: after 5 refusals, the process starts again.

    \[P(X > 7 \mid X > 5) = P(X > 2) = 0.875^2 \approx 0.7656\]

    This is the same probability as for the first 2 passengers. The earlier refusals do not make an acceptance more likely, so the staff member is wrong.

Typical mistake: in a), computing \(0.125^3 \cdot 0.875\). The first 3 passengers refuse (probability 0.875 each), and only the 4th accepts (probability 0.125).

Problem 5: Balloon Rides and z-Scores (xx)

A balloon company offers hot-air balloon rides. The time in the air is normally distributed with mean \(\mu = 80\) minutes and standard deviation \(\sigma = 6\) minutes. Use the following values of the standard normal distribution:

\(z\) 1 1.5 1.645 2
\(P(Z \leq z)\) 0.8413 0.9332 0.95 0.9772

Hint: The standard normal curve is symmetric about 0, so \(P(Z \leq -z) = 1 - P(Z \leq z)\). For example, \(P(Z \leq -1.645) = 1 - 0.95 = 0.05\).

  1. Compute the z-scores of a ride of 71 minutes and of a ride of 92 minutes. Say in words what the z-score of 92 minutes means.

  2. Compute the probability that a ride lasts at most 71 minutes.

  3. Compute the probability that a ride lasts between 74 and 92 minutes.

  4. The company wants to promise “at least \(t\) minutes in the air” and keep the promise on 95 % of the rides. Determine \(t\).

  1. \[z_{71} = \frac{71 - 80}{6} = -1.5, \qquad z_{92} = \frac{92 - 80}{6} = 2\]

    A ride of 92 minutes lasts 2 standard deviations longer than the average ride.

  2. \[P(X \leq 71) = P(Z \leq -1.5) = 1 - 0.9332 = 0.0668\]

  3. \(z_1 = \frac{74 - 80}{6} = -1\) and \(z_2 = 2\):

    \[P(74 \leq X \leq 92) = P(Z \leq 2) - P(Z \leq -1) = 0.9772 - (1 - 0.8413) = 0.9772 - 0.1587 = 0.8185\]

  4. 95 % of the rides must last at least \(t\) minutes, so only 5 % may be shorter: \(P(X \leq t) = 0.05\). By symmetry, the z-score is \(-1.645\):

    \[t = \mu + z \cdot \sigma = 80 - 1.645 \cdot 6 = 70.13 \text{ minutes}\]

    The company can promise about 70 minutes in the air.

Typical mistake: in d), using \(z = +1.645\) and getting \(t = 89.87\) minutes. Only 5 % of the rides last longer than that, so the promise would fail on 95 % of the rides.

B: Set Up & Check

Problem 6: From Words to Binomial Probabilities (xx)

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

  1. A photo studio prints 12 wedding albums. Each album has a printing fault with probability 10 %, independently of the others. \(X\) is the number of albums with a fault. Write “not more than 3 albums have a fault” as a probability and compute it.

    Check: A classmate computes \(P(X < 3) \approx 0.8891\). Explain which number of faulty albums the classmate has left out.

  2. A tailor alters 11 suits per day. Each suit needs a second fitting with probability 20 %, independently of the others. \(X\) is the number of suits that need a second fitting. Write “fewer than 3 suits need a second fitting” as a probability and compute it.

    Check: Write “at least 3 suits need a second fitting” with the complement and compute it without a new sum.

  3. In a city hotel, 25 % of the guests book breakfast. On one evening, 16 guests check in, independently of each other. \(X\) is the number of these guests who book breakfast. Write “at least 2 but not more than 5 guests book breakfast” as a probability and as a sum of binomial terms, and compute it.

    Check: A classmate computes \(P(X \leq 5) - P(X \leq 2)\). Explain which term is missing.

  4. A restaurant gets deliveries from 8 suppliers every morning. Each delivery is incomplete with probability 3 %, independently of the others. \(X\) is the number of incomplete deliveries. Write “at least one delivery is incomplete” with the complement and compute it.

    Check: A classmate writes \(P(X \geq 1) = 1 - P(X \leq 1)\). Explain the mistake.

  5. A toy shop receives a carton of 15 puzzles, and 2 of them miss a piece. The shop checks 3 puzzles of the carton, chosen at random. \(X\) is the number of checked puzzles that miss a piece. Name the distribution of \(X\) and compute the probability that at least one checked puzzle misses a piece.

    Check: A classmate uses a binomial distribution with \(n = 3\) and \(p = \frac{2}{15}\) and gets about 0.3490. Explain why this model does not fit.

  1. “Not more than 3” means 0, 1, 2 or 3, with \(X \sim B(12;\ 0.1)\):

    \[P(X \leq 3) = 0.9^{12} + 12 \cdot 0.1 \cdot 0.9^{11} + 66 \cdot 0.1^2 \cdot 0.9^{10} + 220 \cdot 0.1^3 \cdot 0.9^9 \approx 0.2824 + 0.3766 + 0.2301 + 0.0852 \approx 0.9744\]

    (sum of the unrounded terms)

    Check: \(P(X < 3) = P(X \leq 2)\) leaves out the case of exactly 3 faulty albums, \(P(X = 3) \approx 0.0852\). “Not more than 3” includes 3.

    Typical mistake: reading “not more than 3” as “fewer than 3”. Only “fewer than” and “more than” leave out the endpoint.

  2. “Fewer than 3” means 0, 1 or 2, with \(X \sim B(11;\ 0.2)\):

    \[P(X < 3) = P(X \leq 2) = 0.8^{11} + 11 \cdot 0.2 \cdot 0.8^{10} + 55 \cdot 0.2^2 \cdot 0.8^9 \approx 0.0859 + 0.2362 + 0.2953 = 0.6174\]

    Check: every day has either fewer than 3 or at least 3 such suits, so the two events are complements:

    \[P(X \geq 3) = 1 - P(X \leq 2) \approx 1 - 0.6174 = 0.3826\]

    Typical mistake: writing “fewer than 3” as \(P(X \leq 3)\). This adds the days with exactly 3 second fittings, which are not “fewer than 3”.

  3. Both endpoints belong to the event, with \(X \sim B(16;\ 0.25)\):

    \[P(2 \leq X \leq 5) = \sum_{k=2}^{5} \binom{16}{k} \cdot 0.25^k \cdot 0.75^{16 - k} \approx 0.1336 + 0.2079 + 0.2252 + 0.1802 = 0.7469\]

    Check: \(P(X \leq 5) - P(X \leq 2) = P(3 \leq X \leq 5) \approx 0.6132\). Subtracting \(P(X \leq 2)\) also removes \(X = 2\), so the term \(P(X = 2) \approx 0.1336\) is missing. Correct: \(P(X \leq 5) - P(X \leq 1)\).

    Typical mistake: subtracting the CDF at the lower endpoint. For “at least \(a\)”, subtract \(P(X \leq a - 1)\).

  4. The complement of “at least one incomplete delivery” is “no incomplete delivery”, with \(X \sim B(8;\ 0.03)\):

    \[P(X \geq 1) = 1 - P(X = 0) = 1 - 0.97^8 \approx 0.2163\]

    Check: \(1 - P(X \leq 1)\) is \(P(X \geq 2)\), “at least two”. The complement of “at least 1” is “at most 0”, i.e. \(X = 0\).

    Typical mistake: computing \(8 \cdot 0.03 = 0.24\). Adding the 8 probabilities counts the mornings with two or more incomplete deliveries more than once.

  5. The shop checks 3 of 15 puzzles without replacement from a small lot, so \(X\) has a hypergeometric distribution with \(N = 15\), \(K = 2\), \(n = 3\). Complement:

    \[P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{\binom{2}{0}\binom{13}{3}}{\binom{15}{3}} = 1 - \frac{286}{455} = \frac{169}{455} = \frac{13}{35} \approx 0.3714\]

    Check: the binomial model needs independent draws with the same \(p\). After a complete puzzle is checked, the share of incomplete puzzles in the carton grows from \(\frac{2}{15}\) to \(\frac{2}{14}\), so \(p\) changes. The sample of 3 is a fifth of the carton, so the binomial value 0.3490 is clearly off.

    Typical mistake: using the binomial distribution whenever each item has two outcomes. The words “from a carton of 15” signal a draw without replacement from a small lot.

Problem 7: One Paragraph, All Events (xx)

A language school prepares adults for a German certificate exam. From experience, 70 % of its participants pass the exam at the first attempt, independently of each other. In the next course, 9 participants take the exam. The school wants to know the probability that exactly 7 participants pass, the probability that not more than 2 participants fail and the probability that at least 5 but not more than 8 participants pass. It also wants to know how many participants it can expect to pass and, for its advertising, the probability that all 9 pass. Another school sends its participants to the exam one by one and wants to know the probability that the fourth participant is the first one who fails.

Name the random variable \(X\) with its distribution, \(n\) and \(p\). Then write every question of the schools as a mathematical statement: the probability or expected value, and which values of \(X\) it contains. Do not compute anything.

Check: An assistant writes “not more than 2 participants fail” as \(P(X \leq 2)\). Explain the mistake.

  1. \(X\) = number of the 9 participants who pass, \(X \sim B(9;\ 0.7)\): fixed \(n = 9\), pass or fail, independent participants, the same \(p = 0.7\) for everybody
  2. Exactly 7 pass: \(P(X = 7) = \binom{9}{7} \cdot 0.7^7 \cdot 0.3^2\)
  3. Not more than 2 fail: at least 7 pass, \(P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)\)
  4. At least 5 but not more than 8 pass: \(P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)\), both endpoints included
  5. Expected number of participants who pass: \(E[X] = \mu = n \cdot p = 9 \cdot 0.7\)
  6. All 9 pass: \(P(X = 9) = 0.7^9\)
  7. Second school: the number \(T\) of the participant who fails first is geometric with \(p = 0.3\) (a failure counts as the “success” here): \(P(T = 4) = 0.7^3 \cdot 0.3\)

Check: \(X\) counts the participants who pass, but the question is about the participants who fail. Not more than 2 fail means at least 7 pass, so the correct statement is \(P(X \geq 7)\). Alternatively, count the failures: \(Y = 9 - X \sim B(9;\ 0.3)\) and \(P(Y \leq 2)\).

Typical mistake: mixing the two counts in one problem. Fix one random variable at the start, say what it counts, and translate every question into that variable.

C: Applied

Problem 8: Planning Appointments at a Kitchen Studio (xx)

A kitchen studio (a shop that plans and sells fitted kitchens) offers free planning appointments, in which a designer plans a new kitchen with the customer. From experience, 35 % of the appointments end with an order, independently of each other. Next month, the studio has 14 planning appointments. \(X\) is the number of these appointments that end with an order.

  1. Explain why \(X\) follows a binomial distribution and state \(n\) and \(p\). Compute the expected value and the standard deviation of \(X\).

  2. Compute the probability that exactly 5 of the 14 appointments end with an order.

  3. Compute the probability that fewer than 3 appointments end with an order.

(for verification: \(P(X < 3) \approx 0.0839\))

  1. The owner keeps the free appointments only if at least 4 of the 14 appointments of the month end with an order. Compute the probability that the studio keeps them.

  2. Compute the probability that at least 4 but not more than 7 appointments end with an order.

  3. Determine the most likely number of orders and compare it with the expected value.

Hint: The probabilities \(P(X = k)\) first grow and then fall, as the bar charts of the lecture show. Compare \(P(X = k)\) for the whole numbers next to the expected value.

a) The 14 appointments are a fixed number of trials, each appointment ends with an order or not (two outcomes), the customers decide independently of each other, and every appointment ends with an order with the same probability 0.35. So \(X \sim B(14;\ 0.35)\) with \(n = 14\) and \(p = 0.35\):

\[\mu = np = 14 \cdot 0.35 = 4.9, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{14 \cdot 0.35 \cdot 0.65} = \sqrt{3.185} \approx 1.78\]

b) \[P(X = 5) = \binom{14}{5} \cdot 0.35^5 \cdot 0.65^9 = 2002 \cdot 0.35^5 \cdot 0.65^9 \approx 0.2178\]

c) “Fewer than 3” means 0, 1 or 2:

\[P(X \leq 2) = 0.65^{14} + 14 \cdot 0.35 \cdot 0.65^{13} + 91 \cdot 0.35^2 \cdot 0.65^{12} \approx 0.0024 + 0.0181 + 0.0634 = 0.0839\]

d) “At least 4” has 11 values (\(4, 5, \ldots, 14\)), its complement “at most 3” only 4. With c):

\[P(X = 3) = 364 \cdot 0.35^3 \cdot 0.65^{11} \approx 0.1366\]

\[P(X \geq 4) = 1 - P(X \leq 3) = 1 - (P(X \leq 2) + P(X = 3)) \approx 1 - (0.0839 + 0.1366) = 1 - 0.2205 = 0.7795\]

The studio keeps the free appointments with a probability of about 77.95 %.

e) Both endpoints belong to the event:

\[P(4 \leq X \leq 7) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)\]

\[\approx 0.2022 + 0.2178 + 0.1759 + 0.1082 \approx 0.7042\]

(sum of the unrounded terms, with \(\binom{14}{4} = 1001\), \(\binom{14}{6} = 3003\), \(\binom{14}{7} = 3432\))

f) The expected value is 4.9, so compare the values next to it:

\[P(X = 4) \approx 0.2022, \qquad P(X = 5) \approx 0.2178, \qquad P(X = 6) \approx 0.1759\]

\(P(X = 5)\) is larger than both neighbours, so 5 orders is the most likely number. It is the whole number closest to the expected value 4.9. The expected value itself cannot occur, because \(X\) only takes whole numbers.

Typical mistake: in d), computing \(1 - P(X \leq 4)\). This is “at least 5”. The complement of “at least 4” is “at most 3”.

Problem 9: Complaints at a Drinks Delivery Service (xx)

A drinks delivery service brings crates of water, juice and soft drinks to offices. Every day, it delivers 125 orders. In 4 % of the orders, a crate is missing, independently of the other orders, and the customer complains. For every complaint, the service gives the customer a credit note (money off the next order) of 12 EUR. \(X\) is the number of complaints about the orders of one day.

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

  2. Compute the probability of exactly 5 complaints on one day. Explain why this probability is below 20 %, although 5 is the expected value.

  3. Compute the probability of not more than 7 complaints on one day.

(for verification: \(P(X \leq 7) \approx 0.8709\))

  1. Compute the probability of more than 7 complaints and the probability of at least 4 but not more than 10 complaints on one day.

  2. \(K = 12X\) is the daily cost of the credit notes in EUR. Compute the expected value and the standard deviation of \(K\).

  3. The service brings every missing crate to the customer on the same day with a spare van, which can make at most 9 of these trips per day. Compute the probability that the spare van is enough on one day. The service wants to take more orders per day, but the probability that the spare van is not enough must stay below 5 %. Determine the largest number of orders per day for which this holds.

Hint: Keep \(x = 9\) and \(p = 0.04\) in the cumulative binomial distribution and try different numbers of orders N, starting with N = 135.

a) The 125 orders of a day are a fixed number of trials, each order leads to a complaint or not, the orders are independent, and each order has the same probability 0.04. So \(X \sim B(125;\ 0.04)\):

\[\mu = 125 \cdot 0.04 = 5, \qquad \sigma = \sqrt{125 \cdot 0.04 \cdot 0.96} = \sqrt{4.8} \approx 2.19\]

Over many days, the company gets 5 complaints per day on average.

b) \[P(X = 5) = \binom{125}{5} \cdot 0.04^5 \cdot 0.96^{120} = 234{,}531{,}275 \cdot 0.04^5 \cdot 0.96^{120} \approx 0.1791\]

\(X\) can take many values near 5 (e.g. 3, 4, 6, 7) with similar probabilities, e.g. \(P(X = 4) \approx 0.1776\). The total probability of 1 is spread over all these values, so no single value gets much of it. On more than 80 % of the days, the number of complaints is not exactly 5.

c) “Not more than 7” means \(0, 1, \ldots, 7\), i.e. 8 terms. The cumulative binomial distribution with x = 7, N = 125 and p = 0.04 gives

\[P(X \leq 7) \approx 0.8709\]

d) “More than 7” is the complement of c):

\[P(X > 7) = 1 - P(X \leq 7) \approx 1 - 0.8709 = 0.1291\]

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

\[P(4 \leq X \leq 10) = P(X \leq 10) - P(X \leq 3) \approx 0.9881 - 0.2593 = 0.7288\]

e) With \(E[aX] = a \cdot E[X]\) and \(\mathrm{Var}(aX) = a^2 \cdot \mathrm{Var}(X)\):

\[E[K] = 12 \cdot 5 = 60 \text{ EUR}, \qquad \sigma_K = 12 \cdot \sigma_X = 12 \cdot \sqrt{4.8} \approx 26.29 \text{ EUR}\]

f) The spare van is enough if there are not more than 9 complaints. Cumulative binomial distribution with x = 9, N = 125, p = 0.04:

\[P(X \leq 9) \approx 0.9711\]

The spare van is enough on about 97.11 % of the days.

It is not enough if \(X > 9\), so the condition is \(P(X > 9) = 1 - P(X \leq 9) < 0.05\). With more orders, more complaints are likely, so \(P(X > 9)\) grows with N. Keep x = 9 and p = 0.04 and try N:

Orders per day N 135 136 137 138
\(P(X \leq 9)\) 0.9548 0.9529 0.9509 0.9489
\(P(X > 9)\) 0.0452 0.0471 0.0491 0.0511

With 137 orders, the spare van is not enough on 4.91 % of the days, which is below 5 %. With 138 orders, it is 5.11 %. The service can take at most 137 orders per day, i.e. 12 more than now.

Typical mistake: in e), computing \(\sigma_K = 12^2 \cdot \sigma_X\). The factor \(12^2\) belongs to the variance: \(\mathrm{Var}(K) = 144 \cdot 4.8 = 691.2\), and its square root is \(12 \cdot \sigma_X\).

Problem 10: Big Successes of a Venture Capital Fund (xxx)

A venture capital fund (an investor that buys shares in young companies, called start-ups) knows from experience that 7 % of the start-ups it invests in become a big success, i.e. the fund later sells its shares for at least ten times the money it paid. Assume that the start-ups succeed independently of each other.

  1. The fund invests in 20 start-ups. Compute the probability that exactly 2 of them become a big success and the probability that at least one of them does.

(for verification: \(P(\text{at least one}) \approx 0.7658\))

  1. A smaller fund invests in one start-up after another. \(Y\) is the number of the investment that brings the first big success. Compute the probability that the 6th investment is the first big success and the expected number of investments until the first big success.

  2. The smaller fund has made 12 investments without a big success. A partner says: “After so many failures, our next start-up is due to be a big success.” Assess the statement.

  3. The fund wants at least one big success with a probability of at least 90 %. Determine the minimum number of start-ups it has to invest in. Explain why you have to round up.

  4. The fund’s budget allows 25 investments. Compute the probability of at least one big success with this budget and decide whether it reaches the goal from d).

a) \(X\) = number of big successes, \(X \sim B(20;\ 0.07)\):

\[P(X = 2) = \binom{20}{2} \cdot 0.07^2 \cdot 0.93^{18} = 190 \cdot 0.0049 \cdot 0.93^{18} \approx 0.2521\]

“At least one” is the complement of “none”:

\[P(X \geq 1) = 1 - 0.93^{20} \approx 1 - 0.2342 = 0.7658\]

b) \(Y\) is geometric with \(p = 0.07\). The first 5 start-ups fail, the 6th succeeds:

\[P(Y = 6) = 0.93^5 \cdot 0.07 \approx 0.0487\]

\[E[Y] = \frac{1}{p} = \frac{1}{0.07} \approx 14.29 \text{ investments}\]

(The probability from a) is the same as \(P(Y \leq 20) = 1 - 0.93^{20}\): at least one success in 20 investments means that the first success comes within the first 20.)

c) Wrong. The start-ups are independent, so the geometric distribution is memoryless: after 12 failures, the next start-up is a big success with probability 0.07, as at the start. In formulas, \(P(Y > 12 + n \mid Y > 12) = P(Y > n)\). Nothing is “due”.

d) \[1 - 0.93^n \geq 0.9 \iff 0.93^n \leq 0.1\]

Take the logarithm on both sides and use \(\ln(a^n) = n \cdot \ln(a)\):

\[n \cdot \ln(0.93) \leq \ln(0.1)\]

\(\ln(0.93) < 0\), so dividing by it flips the inequality sign:

\[n \geq \frac{\ln(0.1)}{\ln(0.93)} \approx \frac{-2.3026}{-0.0726} \approx 31.73\]

The fund must invest in at least 32 start-ups.

Why round up: \(n\) is a whole number, and \(1 - 0.93^n\) grows with \(n\). With 31 start-ups, \(1 - 0.93^{31} \approx 0.8946 < 0.9\), which misses the goal. With 32, \(1 - 0.93^{32} \approx 0.9019 \geq 0.9\) \(\checkmark\). Rounding down to 31 would give a probability below 90 %.

e) \[P(\text{at least one in 25}) = 1 - 0.93^{25} \approx 0.8370 < 0.9\]

The budget does not reach the goal. By d), the fund needs 32 investments, i.e. 7 more than the budget allows.

Typical mistake: in d), not flipping the inequality sign and getting \(n \leq 31.73\). This would say that fewer investments give a higher chance of success, which cannot be true.

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: Recall of a Compact Car [22 pts. total]

A car manufacturer has found out that the fuel hose of its compact car model can crack after a few years. A cracked hose cannot be seen from outside, so a workshop needs a pressure test to find it. The manufacturer calls the cars of this model back to its workshops (a recall) to test their fuel hoses. Round probabilities to four decimal places where rounding is necessary.

Part a)

Before the recall starts, engineers test 32 cars of the model, and exactly 2 of them have a cracked hose. For a fault rate \(p\) with \(0 \leq p \leq 1\), let \(f(p)\) be the probability that exactly 2 of 32 cars have a cracked hose. Give a term for \(f(p)\). Determine the fault rate \(p\) for which the engineers’ result is most likely and prove that \(f\) has its maximum there.

Hint: Use the product rule and the chain rule, then factor out \(p \cdot (1 - p)^{29}\).

(for verification: \(p = 0.0625\)) [4 pts.]

Term: 32 independent cars, each with a cracked hose with probability \(p\), so the binomial formula with \(n = 32\) and \(k = 2\) gives

\[f(p) = \binom{32}{2} \cdot p^2 \cdot (1 - p)^{30} = 496 \, p^2 (1 - p)^{30}\]

Derivative (product rule, chain rule with the inner derivative \(-1\)):

\[f'(p) = 496 \left[ 2p \, (1 - p)^{30} + p^2 \cdot 30 \, (1 - p)^{29} \cdot (-1) \right]\]

Factor out \(p \, (1 - p)^{29}\):

\[f'(p) = 496 \, p \, (1 - p)^{29} \left[ 2(1 - p) - 30p \right] = 496 \, p \, (1 - p)^{29} \, (2 - 32p)\]

Critical point: for \(0 < p < 1\), the factor \(496 \, p \, (1 - p)^{29}\) is positive, so \(f'(p) = 0\) only if

\[2 - 32p = 0 \iff p = \frac{2}{32} = \frac{1}{16} = 0.0625\]

Proof of the maximum: the sign of \(f'\) is the sign of \(2 - 32p\). For \(0 < p < 0.0625\), \(f'(p) > 0\) and \(f\) increases. For \(0.0625 < p < 1\), \(f'(p) < 0\) and \(f\) decreases. So \(f'\) changes its sign from \(+\) to \(-\), and \(f\) has a maximum at \(p = 0.0625\). At the ends of the interval, \(f(0) = f(1) = 0 < f(0.0625) \approx 0.2795\), so it is the global maximum on \([0; 1]\).

The most likely fault rate is 6.25 %, the share \(\frac{2}{32}\) of cracked hoses in the engineers’ test.

Grading: 1 Pt. \(f(p)\) with \(\binom{32}{2} = 496\), 1 Pt. derivative with the product and the chain rule, 1 Pt. factoring and \(p = 0.0625\), 1 Pt. sign change of \(f'\) (or comparison with the ends of the interval) as proof.

Part b)

From now on, the fault rate is \(p = 0.0625\). A workshop tests the cars of the model one after another. Compute the probability that the 5th car is the first one with a cracked hose and the probability that the workshop finds a cracked hose among the first 10 cars. Compute the expected number of cars the workshop tests until it finds the first cracked hose. [3 pts.]

\(Y\) = number of the first car with a cracked hose. \(Y\) is geometric with \(p = 0.0625\) and \(1 - p = 0.9375\).

First cracked hose at the 5th car: the first 4 cars are fine, the 5th is not:

\[P(Y = 5) = 0.9375^4 \cdot 0.0625 \approx 0.0483\]

Within the first 10 cars:

\[P(Y \leq 10) = 1 - 0.9375^{10} \approx 1 - 0.5245 = 0.4755\]

Expected value:

\[E[Y] = \frac{1}{p} = \frac{1}{0.0625} = 16 \text{ cars}\]

Grading: 1 Pt. per value.

Part c)

The manufacturer’s lab needs at least one cracked hose to find the cause of the cracks. Determine the number of cars which need to be tested to find at least one car with a cracked fuel hose with a likelihood (probability) of 98 %. Explain why the result has to be rounded up. [4 pts.]

Complement: “at least one cracked hose among \(n\) cars” \(= 1 -\) “no cracked hose”:

\[1 - 0.9375^n \geq 0.98 \iff 0.9375^n \leq 0.02\]

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

\[n \cdot \ln(0.9375) \leq \ln(0.02)\]

\(\ln(0.9375) < 0\), so dividing by it flips the inequality sign:

\[n \geq \frac{\ln(0.02)}{\ln(0.9375)} \approx \frac{-3.9120}{-0.0645} \approx 60.62\]

At least 61 cars have to be tested.

Why round up: the number of cars is a whole number, and \(1 - 0.9375^n\) grows with \(n\). With 60 cars, \(1 - 0.9375^{60} \approx 0.9792 < 0.98\), which is not enough. With 61 cars, \(1 - 0.9375^{61} \approx 0.9805 \geq 0.98\) \(\checkmark\). Rounding down would miss the required likelihood.

Grading: 1 Pt. complement inequality, 1 Pt. logarithm with the sign flip, 1 Pt. \(n \geq 60.62\) and \(n = 61\), 1 Pt. reason for rounding up with the check of 60 and 61.

Part d)

In one month, a large workshop tests 144 cars of the model. \(X\) is the number of these cars with a cracked hose. Show that the four requirements of a binomial distribution hold for \(X\). Compute the expected value and the standard deviation of \(X\) and state what the expected value means for the workshop. [3 pts.]

Requirements:

  • fixed number of trials: \(n = 144\) cars, known in advance,
  • two outcomes per car: cracked hose (success) or intact hose,
  • the cars are independent: one car’s hose does not affect another car’s hose,
  • the same probability \(p = 0.0625\) for every car.

So \(X \sim B(144;\ 0.0625)\).

\[\mu = np = 144 \cdot 0.0625 = 9, \qquad \sigma = \sqrt{np(1 - p)} = \sqrt{144 \cdot 0.0625 \cdot 0.9375} = \sqrt{8.4375} \approx 2.90\]

Meaning: over many months, the workshop finds 9 cracked hoses per month on average, so it should plan spare hoses and repair time for about 9 cars per month.

Grading: 1 Pt. four requirements applied to the cars, 1 Pt. \(\mu = 9\) and \(\sigma \approx 2.90\), 1 Pt. meaning of \(\mu\).

Part e)

Compute the probability for the following events:

(1) exactly 7 of the 144 cars have a cracked hose

(2) not more than 10 of the 144 cars have a cracked hose

(3) at least 5 but not more than 12 of the 144 cars have a cracked hose

[6 pts.]

\(X \sim B(144;\ 0.0625)\) from part d).

(1) Exactly 7:

\[P(X = 7) = \binom{144}{7} \cdot 0.0625^7 \cdot 0.9375^{137} \approx 0.1183\]

(Binomial-Dichte with x = 7, N = 144, p = 0.0625)

(2) “Not more than 10” means \(0, 1, \ldots, 10\), i.e. 11 terms. The cumulative binomial distribution with x = 10, N = 144, p = 0.0625 gives

\[P(X \leq 10) \approx 0.7099\]

(3) “At least 5 but not more than 12” includes both endpoints, so subtract the cumulative probability at \(5 - 1 = 4\):

\[P(5 \leq X \leq 12) = P(X \leq 12) - P(X \leq 4) \approx 0.8829 - 0.0497 = 0.8332\]

Grading: 1 Pt. (1). 2 Pts. (2): 1 Pt. \(P(X \leq 10)\) as a cumulative probability, 1 Pt. value. 3 Pts. (3): 1 Pt. \(P(X \leq 12) - P(X \leq 4)\) with the lower endpoint \(5 - 1 = 4\), 1 Pt. both cumulative values, 1 Pt. result.

Part f)

The workshop can repair at most 12 cracked hoses per month. Further cars with a cracked hose have to wait until the next month. Compute the probability that in this month, at least one car with a cracked hose has to wait. [2 pts.]

A car has to wait if more than 12 cars have a cracked hose, i.e. \(X \geq 13\). \(P(X \leq 12) \approx 0.8829\) is already known from e) (3):

\[P(X \geq 13) = 1 - P(X \leq 12) \approx 1 - 0.8829 = 0.1171\]

In about 11.71 % of the months, at least one car has to wait.

Grading: 1 Pt. \(X \geq 13\) as the complement of \(X \leq 12\) with \(P(X \leq 12)\) from (3), 1 Pt. 0.1171.

Calculator Practice

The FX-991DE X computes binomial probabilities in the menu Verteilungsfunkt. (distribution functions):

  • \(P(X = k)\): MENU → 7 → 4 (Binomial-Dichte) → 2 (Variable), then enter x (your \(k\)), N (your \(n\)) and p, each with =, and press = again.
  • \(P(X \leq k)\): MENU → 7 → ▼ → 1 (cumulative binomial distribution) → 2 (Variable), then x, N and p as above. 1 (Liste) instead of 2 computes several x at once.
  • Every other question becomes \(P(X \leq k)\) first: \(P(X \geq k) = 1 - P(X \leq k - 1)\) and \(P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)\).

Note: The German model has a decimal comma key, so you type \(p = 0.4\) as 0,4.

Problem 11: Calculator: Single Probabilities (x)

  1. Check your result from Problem 2 b), \(P(X = 5) \approx 0.2007\) for \(X \sim B(10;\ 0.4)\), with Binomial-Dichte.

  2. \(X \sim B(30;\ 0.15)\). Compute \(P(X = 4)\) and \(P(X = 5)\).

  1. \(P(X = 5)\) for \(B(10;\ 0.4)\)
    • Enter: MENU → 7 → 4 → 2, then x = 5 =, N = 10 =, p = 0,4 =, then =
    • Result: 0.2006581248, so \(P(X = 5) \approx 0.2007\) ✓
  2. \(P(X = 4)\) and \(P(X = 5)\) for \(B(30;\ 0.15)\)
    • Enter: x = 4 =, N = 30 =, p = 0,15 =, then =
    • Result: 0.2028088541, so \(P(X = 4) \approx 0.2028\)
    • Enter: x = 5 =, N = 30 =, p = 0,15 =, then =
    • Result: 0.1861069485, so \(P(X = 5) \approx 0.1861\)

Problem 12: Calculator: Cumulative Probabilities (x)

  1. Check your hand-computed sum from Problem 6, item 1, \(P(X \leq 3) \approx 0.9744\) for \(X \sim B(12;\ 0.1)\), with the cumulative binomial distribution.

  2. \(X \sim B(60;\ 0.08)\). Compute \(P(X \leq 4)\) and \(P(X \leq 7)\). State how many terms \(P(X \leq 7)\) would have by hand.

  1. \(P(X \leq 3)\) for \(B(12;\ 0.1)\)
    • Enter: MENU → 7 → ▼ → 1 → 2, then x = 3 =, N = 12 =, p = 0,1 =, then =
    • Result: 0.9743625298, so \(P(X \leq 3) \approx 0.9744\), the same as the sum of the four terms ✓
  2. \(P(X \leq 4)\) and \(P(X \leq 7)\) for \(B(60;\ 0.08)\)
    • Enter: x = 4 =, N = 60 =, p = 0,08 =, then =
    • Result: 0.4701713601, so \(P(X \leq 4) \approx 0.4702\)
    • Enter: x = 7 =, N = 60 =, p = 0,08 =, then =
    • Result: 0.8955593223, so \(P(X \leq 7) \approx 0.8956\)
    • By hand, \(P(X \leq 7)\) is the sum of the 8 terms \(P(X = 0)\) to \(P(X = 7)\).

Problem 13: Calculator: At Least, More Than and Between (xx)

A bike-sharing company has 200 bikes in a city. On a given morning, each bike has a flat tyre with probability 3.5 %, independently of the other bikes. \(X\) is the number of bikes with a flat tyre. Write each event with \(P(X \leq k)\) and compute it with the cumulative binomial distribution.

  1. At least 5 bikes have a flat tyre.

  2. More than 10 bikes have a flat tyre.

  3. At least 4 but not more than 9 bikes have a flat tyre.

  4. Fewer than 6 bikes have a flat tyre.

\(X \sim B(200;\ 0.035)\). Every value comes from MENU → 7 → ▼ → 1 → 2 with N = 200 and p = 0,035.

  1. At least 5: \(P(X \geq 5) = 1 - P(X \leq 4)\)
    • Enter: x = 4 =, N = 200 =, p = 0,035 =, then =
    • Result: 0.1681408968
    • \(P(X \geq 5) \approx 1 - 0.1681 = 0.8319\)
  2. More than 10: \(P(X > 10) = P(X \geq 11) = 1 - P(X \leq 10)\)
    • Enter: x = 10 =, N = 200 =, p = 0,035 =, then =
    • Result: 0.9052778218
    • \(P(X > 10) \approx 1 - 0.9053 = 0.0947\)
  3. At least 4 but not more than 9: \(P(4 \leq X \leq 9) = P(X \leq 9) - P(X \leq 3)\)
    • Enter: x = 9 =, N and p as above, then =
    • Result: 0.83413915
    • Enter: x = 3 =, N and p as above, then =
    • Result: 0.07810227479
    • \(P(4 \leq X \leq 9) \approx 0.8341 - 0.0781 = 0.7560\)
  4. Fewer than 6: \(P(X < 6) = P(X \leq 5)\)
    • Enter: x = 5 =, N and p as above, then =
    • Result: 0.2961543572
    • \(P(X < 6) \approx 0.2962\)

Typical mistake: in c), computing \(P(X \leq 9) - P(X \leq 4)\). This removes the case \(X = 4\), which belongs to the event. Subtract \(P(X \leq a - 1)\) for “at least \(a\)”.

Problem 14: Calculator: A Table with Liste (xx)

\(X \sim B(28;\ 0.3)\).

  1. Use the Liste option to make a table of \(P(X \leq x)\) for \(x = 5, 6, \ldots, 12\).

  2. Compute \(P(6 \leq X \leq 11)\) and \(P(X > 8)\) with values from your table.

  3. Determine the smallest \(k\) with \(P(X \leq k) \geq 0.9\).

  1. Table with Liste

    • Enter: MENU → 7 → ▼ → 1 → 1 (Liste)
    • Type the x values into the x column: 5 = 6 = 7 = … 12 =
    • Enter N = 28 =, p = 0,3 =, then =
    • Result (rounded to four decimal places):
    \(x\) 5 6 7 8 9 10 11 12
    \(P(X \leq x)\) 0.1128 0.2202 0.3648 0.5275 0.6825 0.8087 0.8972 0.9509

  2. \[P(6 \leq X \leq 11) = P(X \leq 11) - P(X \leq 5) \approx 0.8972 - 0.1128 = 0.7844\]

    \[P(X > 8) = 1 - P(X \leq 8) \approx 1 - 0.5275 = 0.4725\]

  3. Read down the table: \(P(X \leq 11) \approx 0.8972 < 0.9\) and \(P(X \leq 12) \approx 0.9509 \geq 0.9\). The smallest \(k\) is 12.