Unit 2: Probability & Distributions

Course - Inferential Statistics
Component - LLT Assignment
Unit - Unit 2

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Question 1

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Question Statement

A security system requires users to create a 5-character access code using the letters A-Z (uppercase only). Each character in the code must be a letter, and letters cannot be repeated.

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Solution 1

1. Identify Parameters & Counting Principle:

   • Total available uppercase English letters: $n = 26$ ($A$ through $Z$).
   • Length of access code: $r = 5$.
   • Constraint: Characters cannot be repeated (sampling without replacement).
   • Order significance: In an access code/password, the sequence matters (e.g., $ABCDE \neq EDCBA$). Hence, this is a Permutation problem.

2. Permutation Formula: $P(n, r) = {}^n P_r = \frac{n!}{(n - r)!}$

3. Calculation: ${}^{26}P_5 = \frac{26!}{(26 - 5)!} = \frac{26!}{21!} = 26 \times 25 \times 24 \times 23 \times 22$

   • $26 \times 25 = 650$
   • $650 \times 24 = 15,600$
   • $15,600 \times 23 = 358,800$
   • $358,800 \times 22 = \mathbf{7,893,600}$

4. Conclusion: The security system can generate exactly $7,893,600$ unique 5-character access codes.

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Question 2

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Question Statement

You are organizing a speaker panel for a conference and need to select 4 speakers from a pool of 10 applicants. The order in which the speakers are selected does not matter.

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Solution 2

1. Identify Parameters & Counting Principle:

   • Total applicant pool size: $n = 10$.
   • Number of speakers to select: $r = 4$.
   • Constraint: "The order in which the speakers are selected does not matter."
   • Because sequence is irrelevant to panel membership, this is a Combination problem.

2. Combination Formula: $C(n, r) = \binom{n}{r} = \frac{n!}{r!(n - r)!}$

3. Calculation: $\binom{10}{4} = \frac{10!}{4!(10 - 4)!} = \frac{10!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$

   • Numerator: $10 \times 9 \times 8 \times 7 = 5,040$
   • Denominator: $4 \times 3 \times 2 \times 1 = 24$ $\binom{10}{4} = \frac{5,040}{24} = \mathbf{210}$

4. Conclusion: There are exactly $210$ distinct ways to select the 4-member speaker panel.

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Question 3

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Question Statement

A company produces three types of beverages: A, B, and C. The probability of a beverage being defective is as follows:The probability of a beverage being defective for A is 0.02, The probability of a beverage being defective for B is 0.05. The probability of a beverage being defective for C is 0.01, The conditional probabilities that a defective beverage comes from each type are : The probability that a defective beverage is of type A is 0.2, The probability that a defective beverage is of type B is 0.3, The probability that a defective beverage is of type C is 0.35. If a defective beverage is discovered, what is the most likely type of beverage (A, B, or C) that it is?

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Solution 3

Let $D$ denote the event that a beverage is defective, and $A, B, C$ denote the beverage types.

• Priors (Production Shares): $P(A) = 0.20, \; P(B) = 0.30, \; P(C) = 0.35$
• Likelihoods (Defect Rates): $P(D \mid A) = 0.02, \; P(D \mid B) = 0.05, \; P(D \mid C) = 0.01$

1. Joint Defect Probabilities $P(\text{Type} \cap D)$

• $P(A \cap D) = 0.20 \times 0.02 = \mathbf{0.0040}$
• $P(B \cap D) = 0.30 \times 0.05 = \mathbf{0.0150}$
• $P(C \cap D) = 0.35 \times 0.01 = \mathbf{0.0035}$

2. Total Probability of Defect $P(D)$

$P(D) = 0.0040 + 0.0150 + 0.0035 = \mathbf{0.0225}$

3. Posterior Probabilities (Bayes' Rule)

$P(\text{Type} \mid D) = \frac{P(\text{Type} \cap D)}{P(D)}$

• $P(A \mid D) = \frac{0.0040}{0.0225} \approx \mathbf{17.78\%}$
• $P(B \mid D) = \frac{0.0150}{0.0225} \approx \mathbf{66.67\%}$
• $P(C \mid D) = \frac{0.0035}{0.0225} \approx \mathbf{15.56\%}$

Conclusion: Comparing posterior probabilities ($66.67\% > 17.78\% > 15.56\%$), Type B is overwhelmingly the most likely defective beverage.

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Question 4

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Question Statement

A community center offers three types of sports activities: Swimming (S), Volleyball (V), and Table Tennis (TT). The participation data is summarized in the table below, categorized by age groups.

	Swimming (S)	Volleyball (V)	Table Tennis (TT)
Adult	80	50	40
Teen	30	60	20
Child	40	30	70

i. What is the probability that a randomly chosen participant is enrolled in Volleyball (V)?
ii. What is the probability that a randomly chosen participant is a Child?
iii. What is the probability that a randomly chosen participant is an Adult enrolled in Swimming (S)?
iv. What is the probability that a randomly chosen participant is a Teen enrolled in Table Tennis (TT)?
v. What is the probability that a participant is an Adult given that the participant is enrolled in Volleyball (V)?

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Solution 4

Contingency Table with Marginal Totals

First, construct the full marginal table to determine the total sample space ($N$):

Age Group	Swimming (S)	Volleyball (V)	Table Tennis (TT)	Row Total
Adult ($A$)	80	50	40	170
Teen ($T$)	30	60	20	110
Child ($C$)	40	30	70	140
Column Total	150	140	130	Total $N = 420$

• Grand Total: $N = 170 + 110 + 140 = 150 + 140 + 130 = \mathbf{420}$.

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Subpart Evaluations

• i. Probability that a participant is enrolled in Volleyball (V): $P(V) = \frac{\text{Total enrolled in Volleyball}}{\text{Grand Total}} = \frac{140}{420} = \frac{1}{3} \approx \mathbf{0.3333}\text{ (}33.33\%\text{)}$

• ii. Probability that a participant is a Child: $P(\text{Child}) = \frac{\text{Total Children}}{\text{Grand Total}} = \frac{140}{420} = \frac{1}{3} \approx \mathbf{0.3333}\text{ (}33.33\%\text{)}$

• iii. Probability that a participant is an Adult enrolled in Swimming (S): This is the joint intersection $P(\text{Adult} \cap S)$: $P(\text{Adult} \cap S) = \frac{n(\text{Adult} \cap S)}{N} = \frac{80}{420} = \frac{4}{21} \approx \mathbf{0.1905}\text{ (}19.05\%\text{)}$

• iv. Probability that a participant is a Teen enrolled in Table Tennis (TT): This is the joint intersection $P(\text{Teen} \cap TT)$: $P(\text{Teen} \cap TT) = \frac{n(\text{Teen} \cap TT)}{N} = \frac{20}{420} = \frac{1}{21} \approx \mathbf{0.0476}\text{ (}4.76\%\text{)}$

• v. Probability that a participant is an Adult given that they are enrolled in Volleyball (V): This is a conditional probability $P(\text{Adult} \mid V)$: $P(\text{Adult} \mid V) = \frac{P(\text{Adult} \cap V)}{P(V)} = \frac{n(\text{Adult} \cap V)}{n(V)} = \frac{50}{140} = \frac{5}{14} \approx \mathbf{0.3571}\text{ (}35.71\%\text{)}$

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Question 5

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Question Statement

A traffic safety study examines the effect of seatbelt usage on accident outcomes. In the study, the results show the following: Drivers involved in accidents who were wearing seatbelts and had a minor injury: 15, Drivers involved in accidents who were wearing seatbelts and had a serious injury: 5, Drivers involved in accidents who were not wearing seatbelts and had a minor injury: 10, Drivers involved in accidents who were not wearing seatbelts and had a serious injury: 20.

i) What are the odds in favor of a driver sustaining a minor injury while wearing a seatbelt compared to sustaining a serious injury?
ii) What are the odds against a driver sustaining a minor injury while not wearing a seatbelt?
iii) What is the probability of a driver sustaining a minor injury while wearing a seatbelt?
iv) Calculate the odds ratio of sustaining a minor injury while wearing a seatbelt compared to the odds of sustaining a minor injury without a seatbelt.
v) What does this odds ratio indicate about the effectiveness of wearing a seatbelt in minimizing injury severity during a traffic accident?

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Solution 5

Summary Contingency Table

Status	Minor Injury ($M$)	Serious Injury ($S$)	Total
Wearing Seatbelt ($W$)	15	5	20
Not Wearing Seatbelt ($NW$)	10	20	30
Total	25	25	50

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Subpart Evaluations

• i) Odds in favor of sustaining a minor injury while wearing a seatbelt compared to sustaining a serious injury: $\text{Odds}_{\text{wearing}} = \frac{n(\text{Minor} \cap \text{Wearing})}{n(\text{Serious} \cap \text{Wearing})} = \frac{15}{5} = \frac{3}{1} = \mathbf{3.0}\quad (\text{or } 3:1)$

• ii) Odds against sustaining a minor injury while not wearing a seatbelt:

  • Odds in favor of minor injury without seatbelt: $\frac{n(\text{Minor} \cap NW)}{n(\text{Serious} \cap NW)} = \frac{10}{20} = 0.5$ (or $1:2$).
  • Odds against minor injury without seatbelt: $\text{Odds against} = \frac{n(\text{Serious} \cap NW)}{n(\text{Minor} \cap NW)} = \frac{20}{10} = \frac{2}{1} = \mathbf{2.0}\quad (\text{or } 2:1)$

• iii) Probability of a driver sustaining a minor injury while wearing a seatbelt: $P(\text{Minor} \mid \text{Wearing}) = \frac{n(\text{Minor} \cap \text{Wearing})}{n(\text{Wearing})} = \frac{15}{15 + 5} = \frac{15}{20} = \mathbf{0.75}\text{ (}75\%\text{)}$

• iv) Odds Ratio (OR) of minor injury wearing vs not wearing seatbelt: $\text{Odds Ratio} = \frac{\text{Odds of Minor Injury (Wearing)}}{\text{Odds of Minor Injury (Not Wearing)}} = \frac{15 / 5}{10 / 20} = \frac{3.0}{0.5} = \mathbf{6.0}$ (Cross-product shortcut: $\frac{15 \times 20}{5 \times 10} = \frac{300}{50} = \mathbf{6.0}$)

• v) Practical interpretation of the Odds Ratio:

  • An Odds Ratio of $6.0$ indicates that the odds of sustaining only a minor injury (rather than a serious injury) are $6$ times higher for drivers wearing seatbelts compared to drivers not wearing seatbelts.
  • This provides robust statistical evidence that wearing a seatbelt is highly effective in significantly reducing injury severity during vehicular accidents.

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Question 6

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Question Statement

In a volleyball training camp, data was collected on the serving accuracy of a player. From 20 serves, the player successfully hit the target 14 times. Considering the same player, what is the probability that out of 60 serves

Questions:
i. Calculate the probability of hitting the target for five serves.
ii. Calculate the probability of hitting the target for 45 serves.
iii. determine the probability of hitting at most 55 serves will hit the target.
iv. Calculate the probability of hitting at least 2 serves

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Solution 6

Step 1: Model Identification & Parameters

• Observed Accuracy from Training Camp: $p = P(\text{Hit Target}) = \frac{14}{20} = \mathbf{0.70}\quad (70\%)$ $q = 1 - p = 1 - 0.70 = \mathbf{0.30}\quad (30\%)$
• Number of Independent Trials: $n = \mathbf{60\text{ serves}}$
• Distribution Model: Since each serve has a binary outcome (success/failure), constant success probability $p = 0.70$, and mutually independent trials, the random variable $X$ (number of targets hit) follows a Binomial Distribution: $X \sim B(n = 60, \, p = 0.70)$

Step 2: Distribution Baseline Properties

• Expected Value (Mean): $\mu = n \cdot p = 60 \times 0.70 = \mathbf{42.0\text{ serves}}$
• Standard Deviation: $\sigma = \sqrt{n \cdot p \cdot q} = \sqrt{60 \times 0.70 \times 0.30} = \sqrt{12.60} \approx \mathbf{3.55\text{ serves}}$

Step 3: Binomial Probability Mass Function (PMF)

$P(X = k) = \binom{n}{k} p^k q^{n - k} = \binom{60}{k} (0.70)^k (0.30)^{60 - k}$

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Subpart Evaluations

• i. Probability of hitting the target for exactly 5 serves ($k = 5$): $P(X = 5) = \binom{60}{5} (0.70)^5 (0.30)^{55}$
  • $\binom{60}{5} = \frac{60!}{5! \times 55!} = \mathbf{5,461,512}$
  • $(0.70)^5 = \mathbf{0.16807}$
  • $(0.30)^{55} \approx \mathbf{1.7445 \times 10^{-29}}$ $P(X = 5) \approx 5,461,512 \times 0.16807 \times (1.7445 \times 10^{-29}) \approx \mathbf{1.60 \times 10^{-23}}\quad (\approx \mathbf{0.00\%})$
  • Understanding: Because the expected average is $\mu = 42$ serves, hitting only $5$ serves lies over $10$ standard deviations below the mean ($Z \approx -10.4$), making this outcome virtually impossible under normal conditions.

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• ii. Probability of hitting the target for exactly 45 serves ($k = 45$): $P(X = 45) = \binom{60}{45} (0.70)^{45} (0.30)^{15}$
  • $\binom{60}{45} = \binom{60}{15} = \mathbf{53,194,089,192,720}\quad (\approx 5.319 \times 10^{13})$
  • $(0.70)^{45} \approx \mathbf{1.0701 \times 10^{-7}}$
  • $(0.30)^{15} \approx \mathbf{1.4349 \times 10^{-8}}$ $P(X = 45) \approx 5.3194 \times 10^{13} \times 1.0701 \times 10^{-7} \times 1.4349 \times 10^{-8} \approx \mathbf{0.0817}\quad (\mathbf{8.17\%})$
  • Understanding: Since $45$ is close to the expected center of $\mu = 42$ ($Z \approx +0.85$), it represents a highly realistic single-game performance that occurs roughly once every 12 matches.

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• iii. Probability of hitting at most 55 serves ($X \le 55$): Direct summation of $56$ terms ($k = 0$ to $55$) is inefficient. We apply the Complement Rule: $P(X \le 55) = 1 - P(X \ge 56) = 1 - [P(X=56) + P(X=57) + P(X=58) + P(X=59) + P(X=60)]$

  Upper Tail Probabilities Breakdown:

  $k$	Binomial Term $\binom{60}{k} (0.70)^k (0.30)^{60 - k}$	Exact Probability
  56	$\binom{60}{56} (0.70)^{56} (0.30)^4$	$0.0000083574$
  57	$\binom{60}{57} (0.70)^{57} (0.30)^3$	$0.0000013685$
  58	$\binom{60}{58} (0.70)^{58} (0.30)^2$	$0.0000001652$
  59	$\binom{60}{59} (0.70)^{59} (0.30)^1$	$0.0000000131$
  60	$\binom{60}{60} (0.70)^{60} (0.30)^0$	$0.0000000005$
  Total $\sum P(X \ge 56)$		$0.0000099046$

  • Final Subtraction: $P(X \le 55) = 1 - 0.0000099046 = \mathbf{0.9999900954}\quad (\approx \mathbf{99.999\%})$
  • Understanding: Hitting $56$ or more targets out of $60$ demands an extraordinary accuracy rate of $\ge 93.3\%$. The player is essentially $99.999\%$ certain to hit at most $55$ serves.

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• iv. Probability of hitting at least 2 serves ($X \ge 2$): Using the Complement Rule: $P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X = 0) + P(X = 1)]$
  • Probability of 0 hits ($k = 0$): $P(X = 0) = \binom{60}{0} (0.70)^0 (0.30)^{60} = (0.30)^{60} \approx \mathbf{4.24 \times 10^{-32}} \approx \mathbf{0}$
  • Probability of 1 hit ($k = 1$): $P(X = 1) = \binom{60}{1} (0.70)^1 (0.30)^{59} = 60 \times 0.70 \times (0.30)^{59} \approx \mathbf{5.93 \times 10^{-30}} \approx \mathbf{0}$
  • Sum of Lower Tail: $P(X \le 1) = P(X = 0) + P(X = 1) \approx 5.98 \times 10^{-30} \approx \mathbf{0}$
  • Final Probability: $P(X \ge 2) = 1 - 0 = \mathbf{1.0000}\quad (\mathbf{100\%\text{ certainty}})$
  • Understanding: For a player with 70% accuracy, missing 59 or 60 consecutive serves is statistically impossible under normal play. Hitting at least 2 targets is guaranteed.

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Summary of Results for Question 6

Question Part	Condition	Formula Representation	Evaluated Result	Practical Interpretation
i. Exactly 5 serves	$P(X = 5)$	$\binom{60}{5}(0.7)^5(0.3)^{55}$	$\approx \mathbf{1.60 \times 10^{-23}}$	Astronomically rare ($> 10\sigma$ below mean).
ii. Exactly 45 serves	$P(X = 45)$	$\binom{60}{45}(0.7)^{45}(0.3)^{15}$	$\mathbf{0.0817\text{ (}8.17\%\text{)}}$	Highly realistic single-game score near $\mu = 42$.
iii. At most 55 serves	$P(X \le 55)$	$1 - \sum_{k=56}^{60} P(X = k)$	$\mathbf{0.99999\text{ (}99.999\%\text{)}}$	Almost guaranteed upper ceiling for the player.
iv. At least 2 serves	$P(X \ge 2)$	$1 - [P(0) + P(1)]$	$\mathbf{1.0000\text{ (}100\%\text{)}}$	Absolute certainty; missing $\ge 59$ serves is impossible.

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Question 7

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Question Statement

A local coffee shop receives an average of 15 customer arrivals per hour. The shop manager wants to determine the probabilities associated with customer arrivals to optimize staffing and inventory.

i) What is the likelihood of receiving exactly 12 customer arrivals in a particular hour?
ii) What is the probability of receiving no more than 3 customer arrivals during the same hour?
iii) Additionally, the manager is interested in the overall probability of receiving more than 5 customer arrivals during the busiest 2-hour period of the day.

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Solution 7

Step 1: Model Identification

Arrivals occur independently over a continuous time interval at a known average rate $\lambda = 15$ arrivals/hour. Therefore, the number of arrivals follows a Poisson Distribution:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

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Subpart Evaluations

• i) Likelihood of exactly 12 arrivals in a particular 1-hour period ($\lambda = 15, k = 12$): $P(X = 12) = \frac{e^{-15} \cdot 15^{12}}{12!}$

  • $e^{-15} \approx 3.05902 \times 10^{-7}$
  • $15^{12} \approx 1.29746 \times 10^{14}$
  • $12! = 479,001,600$ $P(X = 12) = \frac{(3.05902 \times 10^{-7}) \times (1.29746 \times 10^{14})}{479,001,600} \approx \mathbf{0.0829}\text{ (}8.29\%\text{)}$

• ii) Probability of receiving no more than 3 customer arrivals during the same hour ($X \le 3$): $P(X \le 3) = \sum_{k=0}^{3} \frac{e^{-15} \cdot 15^k}{k!} = e^{-15} \left( \frac{15^0}{0!} + \frac{15^1}{1!} + \frac{15^2}{2!} + \frac{15^3}{3!} \right)$ $= e^{-15} \left( 1 + 15 + \frac{225}{2} + \frac{3375}{6} \right) = e^{-15} (1 + 15 + 112.5 + 562.5) = e^{-15} \times 691$ $P(X \le 3) \approx (3.05902 \times 10^{-7}) \times 691 \approx \mathbf{0.000211}\text{ (}0.0211\%\text{)}$

• iii) Overall probability of receiving more than 5 arrivals during a 2-hour period: For a 2-hour time frame ($t = 2$), the adjusted arrival parameter is: $\lambda_{2\text{hr}} = 15 \times 2 = 30\text{ arrivals}$ Let $Y$ be the number of arrivals in 2 hours: $Y \sim \text{Poisson}(\lambda = 30)$. $P(Y > 5) = 1 - P(Y \le 5) = 1 - \sum_{k=0}^{5} \frac{e^{-30} \cdot 30^k}{k!}$

  • $\sum_{k=0}^{5} \frac{e^{-30} \cdot 30^k}{k!} = e^{-30} \left( 1 + 30 + 450 + 4500 + 33750 + 202500 \right) = e^{-30} \times 241231$
  • $e^{-30} \approx 9.3576 \times 10^{-14}$
  • $P(Y \le 5) \approx (9.3576 \times 10^{-14}) \times 241231 \approx 2.26 \times 10^{-8}$ $P(Y > 5) = 1 - 2.26 \times 10^{-8} \approx \mathbf{0.99999998} \approx \mathbf{1.0000}\text{ (}100\%\text{)}$

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Question 8

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Question Statement

The yearly salaries of employees at a tech company are assumed to be normally distributed with a mean of $75,000 and a variance of $4,000,000.

I) What is the probability that a randomly selected employee has a salary of more than $78,000?
II) What is the probability that a randomly selected employee has a salary of less than $73,000?

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Solution 8

• Parameters: $X \sim N(\mu = 75,000, \, \sigma^2 = 4,000,000 \implies \sigma = \sqrt{4,000,000} = \mathbf{2,000})$
• Standard Normal Transformation: $Z = \frac{X - \mu}{\sigma} = \frac{X - 75,000}{2,000}$

Step-by-Step Probability Calculations

• I) Probability that a salary is more than $78,000 ($P(X > 78,000)$): $Z = \frac{78,000 - 75,000}{2,000} = \mathbf{+1.50}$ $P(X > 78,000) = P(Z > +1.50) = 1 - \Phi(1.50) = 1 - 0.9332 = \mathbf{0.0668}\quad (\mathbf{6.68\%})$

• II) Probability that a salary is less than $73,000 ($P(X < 73,000)$): $Z = \frac{73,000 - 75,000}{2,000} = \mathbf{-1.00}$ $P(X < 73,000) = P(Z < -1.00) = \Phi(-1.00) = 1 - \Phi(1.00) = 1 - 0.8413 = \mathbf{0.1587}\quad (\mathbf{15.87\%})$

Summary Table

Question Condition	$Z$-Score Calculation	$Z$-Table Area Lookup	Final Probability
i) More than $78,000	$Z = \frac{78,000 - 75,000}{2,000} = +1.50$	$1 - 0.9332$	$\mathbf{0.0668}\text{ (}6.68\%\text{)}$
ii) Less than $73,000	$Z = \frac{73,000 - 75,000}{2,000} = -1.00$	$1 - 0.8413$	$\mathbf{0.1587}\text{ (}15.87\%\text{)}$

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