Comprehensive Study Guide on Binomial Distribution

Theoretical Foundations of the Binomial Distribution

• A Bernoulli experiment is a random experiment that has exactly two mutually exclusive outcomes: Success (Erfolg) and Failure (Misserfolg).
• When a Bernoulli experiment is repeated $n$ times independently under identical conditions with a constant probability of success $p$, the discrete random variable $X$ represents the total number of successes achieved across the $n$ trials.
• The success probability is bounded such that $0 \le p \le 1$, and the variable $k$ denoting the specific number of successful outcomes must satisfy $k \in \{0, 1, 2, \dots, n\}$ with $k \le n$.
• A discrete random variable $X$ adhering to these conditions is defined as binomially distributed (binomialverteilt).
• The underlying probability distribution function $f$ is designated as the Binomial Distribution $B(n, p)$ with parameters $n$ and $p$:

$f(k) = P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n-k}$

• Parameter breakdown:

  • $n$: Total number of trials (Anzahl der Versuche).
  • $p$: Success probability for a single individual trial (Erfolgswahrscheinlichkeit im Einzelversuch).
  • $1 - p$: Failure probability for a single individual trial (Wahrscheinlichkeit für Misserfolg).
  • $k$: Target frequency of successful outcomes (Anzahl der Erfolge).
  • $n - k$: Target frequency of failed outcomes (Anzahl der Misserfolge).
  • $\binom{n}{k}$: Binomial coefficient (Binomialkoeffizient), which calculates the total number of distinct orderings/ways that $k$ successes can occur in $n$ trials.

Standard Model Example and Probability Distribution

• Consider a game of 10 played sets ($n = 10$) between Sarah and Robin, where Sarah has a individual set win probability of $p = 0.7$:

  • Target outcome: Sarah wins exactly 7 sets ($k = 7$).
  • Number of set losses: $n - k = 10 - 7 = 3$.
  • Individual set loss probability: $1 - p = 1 - 0.7 = 0.3$
  • Total combinations to win 7 out of 10 sets: $\binom{10}{7} = 120$
  • Probability calculation:

$P(X = 7) = \binom{10}{7} \cdot 0.7^7 \cdot 0.3^3 = 120 \cdot 0.0823543 \cdot 0.027 = 0.2668 = 26.7\%$

• Contextual Interpretation of Expressions:

  • The expression $\binom{9}{2} \cdot 0.7^2 \cdot 0.3^7$ represents the exact probability that Sarah wins exactly 2 sets out of 9 total played sets against Robin.

• Tabular and Graphical Probability Distribution for $n = 10$ and $p = 0.7$:

  • $P(X = 0) = 0.0000059$
  • $P(X = 1) = 0.00014$
  • $P(X = 2) = 0.0014$
  • $P(X = 3) = 0.009$
  • $P(X = 4) = 0.037$
  • $P(X = 5) = 0.103$
  • $P(X = 6) = 0.200$
  • $P(X = 7) = 0.267$
  • $P(X = 8) = 0.233$
  • $P(X = 9) = 0.121$
  • $P(X = 10) = 0.028$

• Critical Rule: The success probability $p$ ALWAYS specifies the probability of success in a SINGLE isolated trial (Einzelversuch).

Determining Minimum Trial Sample Size (n)

• Problem Model: A basketball player converts a half-court shot with a success probability of $24\%$ ($p = 0.24$). How many times must the player throw ($n$) to hit at least once with a cumulative probability of at least $90\%$ ($P(X \ge 1) \ge 0.90$)?

• Solution Procedure:

  • The complementary event (Gegenereignis) of "at least one hit" ($X \ge 1$) is "zero hits" ($X = 0$).
  • Probability of missing a single shot: $1 - p = 1 - 0.24 = 0.76$
  • Probability of missing all $n$ consecutive shots: $0.76^n$
  • Probability of hitting at least once expressed via the complementary event: $P(X \ge 1) = 1 - 0.76^n$
  • Set up the inequality:

$1 - 0.76^n \ge 0.90$

• Rearrange algebraically:

$0.10 \ge 0.76^n$

• Apply natural logarithm to both sides:

$\ln(0.10) \ge n \cdot \ln(0.76)$

• Note that $\ln(0.76) \approx -0.2744$. Dividing an inequality by a negative number reverses the inequality direction:

$\frac{\ln(0.10)}{\ln(0.76)} \le n$

$16.78 \le n$

• Conclusion: The player must throw at least 17 times ($n \ge 17$) to ensure a $\ge 90\%$ probability of scoring at least one basket.

Expected Value, Variance, and Standard Deviation

• For a binomially distributed discrete random variable $X \sim B(n, p)$, the key parameters are defined as follows:

  • Expected Value (Erwartungswert): $E(X) = \mu = n \cdot p$
  • Variance (Varianz): $V(X) = \sigma^2 = n \cdot p \cdot (1 - p)$
  • Standard Deviation (Standardabweichung): $\sigma = \sqrt{V(X)} = \sqrt{n \cdot p \cdot (1 - p)}$

• Comprehensive Sample Problem (Handball Penalty Throws):

  • Context: A handball player converts 7-meter penalty throws with a success probability of $77\%$ ($p = 0.77$). The player takes $n = 30$ penalty throws during a training session. $X$ counts the total number of goals scored.

  • Expected Value Calculation:

$E(X) = n \cdot p = 30 \cdot 0.77 = 23.1$

• Expected Value Interpretation: If the player repeats this exact 30-throw training session very many times, they will score an average of approximately 23.1 goals per session.

• Variance Calculation:

$V(X) = n \cdot p \cdot (1 - p) = 30 \cdot 0.77 \cdot 0.23 = 5.313$

• Variance Interpretation: Upon repeating this training session very often, the empirical variance of the recorded scores converges toward $V(X) = 5.313$

• Standard Deviation Calculation:

$\sigma = \sqrt{5.313} \approx 2.305 \approx 2.3$

• 1-Sigma Interval Calculation:

$[\mu - \sigma, \mu + \sigma] = [23.1 - 2.3, 23.1 + 2.3] = [20.8, 25.4]$

• 1-Sigma Interval Interpretation: Repeating the training session very frequently will often result in scoring between 21 and 25 goals.

• 2-Sigma Interval Calculation:

$[\mu - 2\sigma, \mu + 2\sigma] = [23.1 - 2(2.3), 23.1 + 2(2.3)] = [18.5, 27.7]$

• 2-Sigma Interval Interpretation: Upon frequent repetitions, scoring 18 or fewer goals, or 28 or more goals, occurs rarely. In the vast majority of sessions, the player scores between 19 and 27 goals.

Applied Practice Exercises

• Exercise 7: Drawing Urn Balls

  • Setup: An urn contains 40 balls total. 30 balls say "du gewinnst leider nichts" (loss), and 10 balls say "du gewinnst 100 €" (win $100\,\text{€}$). A person draws 7 times with replacement ($n = 7$). $X$ is the number of winning balls drawn.
  • a. Is $X$ binomially distributed? Yes, because there are exactly two outcomes (winning/losing) and drawing with replacement keeps the probability $p$ constant across trials.
  • b. Parameters: $n = 7$, $p = \frac{10}{40} = 0.25$. Distribution: $B(7, 0.25)$.
  • c. Probability of winning $200\,\text{€}$: Requires drawing exactly 2 winning balls ($k = 2$), evaluated via $P(X = 2)$.
  • d. Probability of winning at least $600\,\text{€}$: Requires drawing at least 6 winning balls ($k \ge 6$), evaluated via $P(X \ge 6)$.
  • e. Probability of drawing at most 3 winning balls: Evaluated via $P(X \le 3)$.
  • f. Probability distribution represented as table and graph.

• Exercise 8: Tennis Match (Markus vs. Paul)

  • Setup: Markus wins a set against Paul with $80\%$ probability ($p_{\text{Markus}} = 0.8$). They play 4 sets in the morning and 3 sets in the afternoon ($n = 7$). $X$ counts the sets won by Paul ($p_{\text{Paul}} = 0.2$).
  • a. Is $X$ binomially distributed? Yes, fixed number of independent sets with two outcomes per set.
  • b. Probability distribution table and diagram.
  • c. Probability Paul wins at most 3 sets: $P(X \le 3)$.
  • d. Probability Paul wins at least 5 sets: $P(X \ge 5)$.
  • e. Interpretation of expression for a 5-set match ($n = 5$):

$\binom{5}{3} \cdot 0.8^3 \cdot 0.2^2 + \binom{5}{4} \cdot 0.8^4 \cdot 0.2^1 + 0.8^5$

    This expression yields the probability that Markus wins at least 3 sets out of 5 played sets (or equivalently, that Paul wins at most 2 sets).

• Exercise 9: Game Console Production Quality

  • Setup: Defect rate $p = 0.01$. $n = 200$ consoles tested. $X$ counts defective consoles.
  • a. Is $X$ binomially distributed? Yes.
  • b. Probability that at most 3 devices are defective: $P(X \le 3)$.
  • c. Probability that zero devices are defective: $P(X = 0) = 0.99^{200}$.

• Exercise 10: Biathlon Shooting Performance

  • Setup: Prone shooting hit probability $p_{\text{prone}} = 0.95$; Standing shooting hit probability $p_{\text{standing}} = 0.80$.
  • a. Sprint event (5 prone shots, $n = 5$): Probability of exactly 4 hits ($P(X = 4)$).
  • b. Individual 20 km event (10 standing shots, $n = 10$): Probability of (i) exactly 7 hits ($P(X = 7)$), (ii) at most 3 hits ($P(X \le 3)$), (iii) at least 8 hits ($P(X \ge 8)$).
  • c. Training session (40 prone shots, $n = 40$): Probability of achieving at least 38 hits ($P(X \ge 38)$).
  • d. Interpretation of expression for a 20-shot standing series ($n = 20$):

$\binom{20}{18} \cdot 0.8^{18} \cdot 0.2^2 + \binom{20}{19} \cdot 0.8^{19} \cdot 0.2^1 + \binom{20}{20} \cdot 0.8^{20} \cdot 0.2^0$

    This expression calculates the probability that the biathlete hits at least 18 out of 20 targets in standing position.

• Exercise 11: Term Equivalence

  • Target Expression: $\binom{10}{3} \cdot 0.6^3 \cdot 0.4^7$

  • Given choices check:

  • Choice 1: $\binom{10}{3} \cdot 0.6^3 \cdot (1 - 0.4)^7$ [Incorrect: $1 - 0.4 = 0.6$]

  • Choice 2: $\binom{10}{3} \cdot 0.6^3 \cdot (1 - 0.6)^7$ [Correct: $1 - 0.6 = 0.4$]

  • Choice 3: $720 \cdot 0.6^3 \cdot 0.4^{10-3}$ [Incorrect: $\binom{10}{3} = 120$, not 720]

  • Choice 4: $120 \cdot 0.6^3 \cdot 0.4^7$ [Correct: $\binom{10}{3} = 120$]

  • Choice 5: $120 \cdot 0.6^7 \cdot 0.4^3$ [Incorrect: exponents swapped]

• Exercise 12: Football Penalty Kicks

  • Setup: Conversion probability $p = 0.90$. $n = 7$ attempts. $X$ counts successful penalty goals.
  • a. Possible values for $X$: $X \in \{0, 1, 2, 3, 4, 5, 6, 7\}$. Binomially distributed because independent attempts with binary outcomes.
  • b. Tabulate probability distribution.
  • c. Probability of scoring exactly 4 penalties: $P(X = 4)$.
  • d. Probability of scoring at least 5 penalties: $P(X \ge 5)$.
  • e. Interpretation of expression:

$\binom{7}{1} \cdot 0.9^1 \cdot 0.1^6$

    Represents the probability of scoring exactly 1 penalty goal out of 7 attempts.

• Exercise 13: Rolling a Standard Die

  • Setup: Standard 6-sided die rolled 18 times ($n = 18$). Success is rolling a 3 ($p = \frac{1}{6}$). $X$ is the count of 3s.
  • a. Is $X$ binomially distributed? Yes, independent rolls with constant probability.
  • b. Probability of rolling exactly four 3s: $P(X = 4)$.
  • c. Probability of rolling at least ten 3s: $P(X \ge 10)$.
  • d. Probability of rolling at most two 3s: $P(X \le 2)$.
  • e. Minimum rolls $n$ required such that $P(X \ge 1) \ge 0.97$:

$1 - \left(\frac{5}{6}\right)^n \ge 0.97 \implies 0.03 \ge \left(\frac{5}{6}\right)^n$

$\frac{\ln(0.03)}{\ln\left(\frac{5}{6}\right)} \le n$

• Exercise 14: Rolling Two Dice

  • Setup: Two 6-sided dice thrown simultaneously 10 times ($n = 10$). Success is a sum of 7 ($p = \frac{6}{36} = \frac{1}{6}$). $X$ counts sum-of-7 throws.
  • a. Binomial check: Yes.
  • b. Probability sum is 7 exactly 3 times: $P(X = 3)$.
  • c. Probability sum is 7 at most 2 times: $P(X \le 2)$.
  • d. Minimum rolls $n$ so $P(X \ge 1) \ge 0.90$:

$1 - \left(\frac{5}{6}\right)^n \ge 0.90$

• Exercise 15: Multiple-Choice Test

  • Setup: 20 questions ($n = 20$), 4 choices per question ($p = 0.25$). Random guessing.
  • a. Passing grade requires $\ge 60\%$ correct ($k \ge 12$): $P(X \ge 12)$.
  • b. Grade "Sehr Gut" requires $\ge 90\%$ correct ($k \ge 18$): $P(X \ge 18)$.

• Exercise 16: Hunter Shooting Cans

  • Setup: Can hit probability $p = 0.54$ at 100 meters.
  • a. Minimum shots $n$ required so $P(X \ge 1) \ge 0.97$:

$1 - 0.46^n \ge 0.97 \implies 0.03 \ge 0.46^n \implies n \ge \frac{\ln(0.03)}{\ln(0.46)}$

• b. For $n = 24$ shots, calculate probability of landing between 7 and 10 hits: $P(7 \le X \le 10)$.

  • Exercise 17: Product Defect Inspection

• Setup: Defect rate $p = 0.13$, sample size $n = 12$. $X$ counts defective devices.

• a. Probability exactly 3 devices are defective: $P(X = 3)$.

• b. Expected value: $E(X) = 12 \cdot 0.13 = 1.56$. Contextual interpretation.

• c. Variance $V(X) = 12 \cdot 0.13 \cdot 0.87 = 1.3572$; Standard deviation \sigma = \sqrt{1.3572} \approx 1.165$.\n - d. Sample size n$required for$P(X \ge 1) \ge 0.95:\n\n1 - 0.87^n \ge 0.95 \implies 0.05 \ge 0.87^n \implies n \ge \frac{\ln(0.05)}{\ln(0.87)}\n\n - e. Probability that X$falls in$[\mu - \sigma, \mu + \sigma]$:$P(1.56 - 1.165 \le X \le 1.56 + 1.165) = P(0.395 \le X \le 2.725) = P(1 \le X \le 2).\n\n- Exercise 18: Penalty Goalkeeper Saves\n\n - Setup: Goalkeeper save probability p = 0.55$.$n = 17$shooters.$X counts saved penalties.\n - a. Expected value: E(X) = 17 \cdot 0.55 = 9.35. Interpret.\n - b. Variance V(X) = 17 \cdot 0.55 \cdot 0.45 = 4.2075$; Standard deviation$\sigma = \sqrt{4.2075} \approx 2.051$.

• c. Expected value if $X$ represents non-saved penalties ($p = 0.45$): E(X) = 17 \cdot 0.45 = 7.65$.\n - d. Probability P(\mu - \sigma \le X \le \mu + \sigma).\n - e. Probability P(X < \mu - 2\sigma).\n\n- Exercise 19: Olympic Marksman\n\n - Setup: 10.0 hit probability p = 0.68$, total shots$n = 500$.$X counts 10.0 hits.\n - a. Probability of hitting 10.0 exactly 362 times: P(X = 362).\n - b. Expected value: E(X) = 500 \cdot 0.68 = 340. Interpret.\n - c. Variance V(X) = 500 \cdot 0.68 \cdot 0.32 = 108.8$; Standard deviation$\sigma = \sqrt{108.8} \approx 10.43$.

• d. Probability $P(\mu - \sigma \le X \le \mu + \sigma)$.

• e. Probability $P(X > \mu + 2\sigma)$.

  • Exercise 20: Comparative Multi-Sided Dice Rolls

• Setup: All dice rolled $n = 40$ times.

• a. Standard 6-sided die ($p = \frac{1}{6}$):

$E(X) = 40 \cdot \frac{1}{6} \approx 6.67$

$\sigma = \sqrt{40 \cdot \frac{1}{6} \cdot \frac{5}{6}} \approx 2.36$

• b. 20-sided die ($p = \frac{1}{20}$):

$E(X) = 40 \cdot \frac{1}{20} = 2.0$

$\sigma = \sqrt{40 \cdot \frac{1}{20} \cdot \frac{19}{20}} \approx 1.38$

    Comparison: As the number of sides increases, success probability $p$ decreases, leading to a lower expected value and lower absolute standard deviation.

• c. 50-sided die ($p = \frac{1}{50}$):

$E(X) = 40 \cdot \frac{1}{50} = 0.8$

$\sigma = \sqrt{40 \cdot \frac{1}{50} \cdot \frac{49}{50}} \approx 0.885$

• Exercise 21: Roulette Betting Strategies

  • Setup: 20 consecutive bets ($n = 20$).

  • Scenarios:

  • (i) Red: $p = \frac{18}{37}$

  • (ii) 2nd Dozen (13–24): $p = \frac{12}{37}$

  • (iii) Single Number 7: $p = \frac{1}{37}$

  • a. Expected values: $E(X_{\text{Red}}) = 20 \cdot \frac{18}{37} \approx 9.73$; $E(X_{\text{Dozen}}) = 20 \cdot \frac{12}{37} \approx 6.49$; E(X_7) = 20 \cdot \frac{1}{37} \approx 0.54$.\n - b. Standard deviations: \sigma_{\text{Red}} = \sqrt{20 \cdot \frac{18}{37} \cdot \frac{19}{37}} \approx 2.23$;$\sigma_{\text{Dozen}} = \sqrt{20 \cdot \frac{12}{37} \cdot \frac{25}{37}} \approx 2.09$;$\sigma_7 = \sqrt{20 \cdot \frac{1}{37} \cdot \frac{36}{37}} \approx 0.73$.

  • c. Probability of winning exactly 13 times ($P(X = 13)$) evaluated for each option.

• Exercise 22: Train Delay Monitoring

  • Setup: Delay probability $p_{\text{delayed}} = 0.05$, punctuality probability $p_{\text{punctual}} = 0.95$. Sample size $n = 400$ trains.
  • a. Expected value for punctual trains: $E(X) = 400 \cdot 0.95 = 380$. Interpret.
  • b. Standard deviation: \sigma = \sqrt{400 \cdot 0.95 \cdot 0.05} = \sqrt{19} \approx 4.36$.\n - c. Probability at least 370 trains are punctual: P(X \ge 370).\n - d. For delayed trains (X$counts delays):$E(X) = 400 \cdot 0.05 = 20$;$\sigma = \sqrt{19} \approx 4.36$; Probability at least 30 trains delayed:$P(X \ge 30).\n\n- Exercise 23: Porcelain Plate Shipments\n\n - Setup: Package breakage probability p = 0.03.\n - a. Inspection of n = 190 packages. Probabilities:\n\n - (i) Exactly 12 broken: P(X = 12)\n - (ii) At most 20 broken: P(X \le 20)\n - (iii) At least 100 broken: P(X \ge 100)\n - (iv) Between 10 and 20 broken: P(10 \le X \le 20)\n\n - b. Expected value: E(X) = 190 \cdot 0.03 = 5.7$; Standard deviation:$\sigma = \sqrt{190 \cdot 0.03 \cdot 0.97} \approx 2.35$.
  • c. Minimum package count $n$ required so $P(X \ge 1) \ge 0.99$:

$1 - 0.97^n \ge 0.99 \implies 0.01 \ge 0.97^n \implies n \ge \frac{\ln(0.01)}{\ln(0.97)}$

• d. Interval probability $P(\mu - \sigma \le X \le \mu + \sigma)$.