Unit 3: Solved Numerical Problems

This section contains 25+ fully solved problems covering probability concepts, addition and multiplication rules, and probability distributions.

Section A: Basic Probability

Problem 1: Classical Probability

Question: A box contains 5 red, 3 blue, and 2 green balls. If one ball is drawn at random, find the probability of: a) Drawing a red ball b) Drawing a blue or green ball c) Not drawing a green ball

Click to reveal solution **Step 1:** Count total outcomes Total balls = 5 + 3 + 2 = 10 **Part (a):** P(Red) $$P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{5}{10} = 0.5$$ **Part (b):** P(Blue or Green) $$P(\text{Blue or Green}) = \frac{3 + 2}{10} = \frac{5}{10} = 0.5$$ **Part (c):** P(Not Green) $$P(\text{Not Green}) = 1 - P(\text{Green}) = 1 - \frac{2}{10} = \frac{8}{10} = 0.8$$ **Answers:** a) **0.5**, b) **0.5**, c) **0.8**

Problem 2: Probability with Playing Cards

Question: A card is drawn at random from a standard deck of 52 cards. Find the probability of: a) Drawing a King b) Drawing a Heart c) Drawing a King of Hearts d) Drawing a King or a Heart

Click to reveal solution **Given information:** - Total cards = 52 - Kings = 4 (one in each suit) - Hearts = 13 - King of Hearts = 1 **Part (a):** P(King) $$P(\text{King}) = \frac{4}{52} = \frac{1}{13} = 0.077$$ **Part (b):** P(Heart) $$P(\text{Heart}) = \frac{13}{52} = \frac{1}{4} = 0.25$$ **Part (c):** P(King of Hearts) $$P(\text{King of Hearts}) = \frac{1}{52} = 0.019$$ **Part (d):** P(King or Heart) - using addition rule $$P(\text{King or Heart}) = P(\text{King}) + P(\text{Heart}) - P(\text{King and Heart})$$ $$= \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} = 0.308$$ **Answers:** a) **1/13 or 0.077**, b) **1/4 or 0.25**, c) **1/52 or 0.019**, d) **4/13 or 0.308**

Problem 3: Probability with Dice

Question: Two fair dice are thrown. Find the probability of: a) Getting a sum of 7 b) Getting a sum greater than 9 c) Getting doubles (same number on both dice)

Click to reveal solution **Step 1:** Total outcomes when throwing 2 dice = 6 × 6 = 36 **Part (a):** Sum of 7 Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways $$P(\text{Sum} = 7) = \frac{6}{36} = \frac{1}{6} = 0.167$$ **Part (b):** Sum > 9 (i.e., sum = 10, 11, or 12) - Sum = 10: (4,6), (5,5), (6,4) = 3 ways - Sum = 11: (5,6), (6,5) = 2 ways - Sum = 12: (6,6) = 1 way - Total = 6 ways $$P(\text{Sum} > 9) = \frac{6}{36} = \frac{1}{6} = 0.167$$ **Part (c):** Doubles Favorable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 ways $$P(\text{Doubles}) = \frac{6}{36} = \frac{1}{6} = 0.167$$ **Answers:** a) **1/6**, b) **1/6**, c) **1/6**

Section B: Addition Rule of Probability

Problem 4: Addition Rule (Mutually Exclusive Events)

Question: In a government office, records show that 30% of employees have a Master’s degree, 25% have professional certifications, and these are mutually exclusive (no one has both). Find the probability that a randomly selected employee has either a Master’s degree or a professional certification.

Click to reveal solution **Step 1:** Identify given information - P(Master's) = 0.30 - P(Certification) = 0.25 - Events are mutually exclusive **Step 2:** Apply addition rule for mutually exclusive events $$P(A \cup B) = P(A) + P(B)$$ $$P(\text{Master's or Certification}) = 0.30 + 0.25 = 0.55$$ **Answer:** **0.55 or 55%**

Problem 5: Addition Rule (Non-Mutually Exclusive Events)

Question: In a survey of 200 citizens:

  • 120 read newspaper A
  • 80 read newspaper B
  • 50 read both newspapers

Find the probability that a randomly selected citizen reads: a) At least one newspaper b) Only newspaper A c) Neither newspaper

Solution:

Step 1: Convert to probabilities

  • P(A) = 120/200 = 0.60
  • P(B) = 80/200 = 0.40
  • P(A ∩ B) = 50/200 = 0.25

Part (a): P(A or B) \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) \(= 0.60 + 0.40 - 0.25 = 0.75\)

Part (b): P(Only A) \(P(\text{Only A}) = P(A) - P(A \cap B) = 0.60 - 0.25 = 0.35\)

Part (c): P(Neither) \(P(\text{Neither}) = 1 - P(A \cup B) = 1 - 0.75 = 0.25\)

Answers: a) 0.75, b) 0.35, c) 0.25

Problem 6: Addition Rule with Venn Diagram

Question: In a class of 100 students:

  • 50 study Economics
  • 40 study Statistics
  • 30 study both

Find: a) P(Economics only) b) P(Statistics only) c) P(at least one subject) d) P(exactly one subject)

Solution:

Using Venn Diagram approach:

Economics only = 50 - 30 = 20
Statistics only = 40 - 30 = 10
Both = 30
Neither = 100 - (20 + 10 + 30) = 40

Part (a): P(Economics only) = 20/100 = 0.20

Part (b): P(Statistics only) = 10/100 = 0.10

Part (c): P(at least one) = (20 + 30 + 10)/100 = 60/100 = 0.60

Part (d): P(exactly one) = (20 + 10)/100 = 30/100 = 0.30

Section C: Multiplication Rule and Conditional Probability

Problem 7: Independent Events

Question: A die is thrown twice. Find the probability of: a) Getting 6 on both throws b) Getting 6 on the first throw and not 6 on the second c) Getting at least one 6

Click to reveal solution **Note:** Throws are independent events. **Part (a):** P(6 and 6) $$P(6 \text{ and } 6) = P(6) \times P(6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} = 0.028$$ **Part (b):** P(6 first, not 6 second) $$= P(6) \times P(\text{not } 6) = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36} = 0.139$$ **Part (c):** P(at least one 6) Method 1: Using complement $$P(\text{at least one 6}) = 1 - P(\text{no 6})$$ $$= 1 - \frac{5}{6} \times \frac{5}{6} = 1 - \frac{25}{36} = \frac{11}{36} = 0.306$$ **Answers:** a) **1/36**, b) **5/36**, c) **11/36**

Problem 8: Dependent Events (Without Replacement)

Question: A bag contains 6 white and 4 black balls. Two balls are drawn without replacement. Find the probability of: a) Both white b) Both black c) One of each color

Click to reveal solution **Part (a):** P(Both White) $$P(\text{Both White}) = P(W_1) \times P(W_2|W_1)$$ $$= \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} = 0.333$$ **Part (b):** P(Both Black) $$P(\text{Both Black}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} = 0.133$$ **Part (c):** P(One of Each) Can happen two ways: WB or BW $$P(WB) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$$ $$P(BW) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$$ $$P(\text{One of each}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15} = 0.533$$ **Verification:** 1/3 + 2/15 + 8/15 = 5/15 + 2/15 + 8/15 = 15/15 = 1 ✓ **Answers:** a) **1/3**, b) **2/15**, c) **8/15**

Problem 9: Conditional Probability

Question: In a company, 60% of employees are male. Of the males, 30% have management positions. Of the females, 20% have management positions. Find: a) P(Manager | Male) b) P(Manager) c) P(Male | Manager)

Click to reveal solution **Given:** - P(Male) = 0.60, P(Female) = 0.40 - P(Manager | Male) = 0.30 - P(Manager | Female) = 0.20 **Part (a):** Already given P(Manager | Male) = **0.30** **Part (b):** P(Manager) - using total probability $$P(\text{Manager}) = P(\text{Manager}|\text{Male}) \times P(\text{Male}) + P(\text{Manager}|\text{Female}) \times P(\text{Female})$$ $$= 0.30 \times 0.60 + 0.20 \times 0.40 = 0.18 + 0.08 = 0.26$$ **Part (c):** P(Male | Manager) - using Bayes' Theorem $$P(\text{Male}|\text{Manager}) = \frac{P(\text{Manager}|\text{Male}) \times P(\text{Male})}{P(\text{Manager})}$$ $$= \frac{0.30 \times 0.60}{0.26} = \frac{0.18}{0.26} = 0.692$$ **Answers:** a) **0.30**, b) **0.26**, c) **0.692**

Problem 10: Bayes’ Theorem

Question: A factory has three machines A, B, and C producing 50%, 30%, and 20% of total output respectively. The defective rates are 2%, 3%, and 4% respectively. If a randomly selected item is found defective, what is the probability it was produced by machine A?

Click to reveal solution **Step 1:** Define events - A, B, C: Item produced by machine A, B, C - D: Item is defective **Given:** - P(A) = 0.50, P(B) = 0.30, P(C) = 0.20 - P(D|A) = 0.02, P(D|B) = 0.03, P(D|C) = 0.04 **Step 2:** Calculate P(D) using total probability $$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$$ $$= 0.02(0.50) + 0.03(0.30) + 0.04(0.20)$$ $$= 0.01 + 0.009 + 0.008 = 0.027$$ **Step 3:** Apply Bayes' Theorem $$P(A|D) = \frac{P(D|A) \times P(A)}{P(D)} = \frac{0.02 \times 0.50}{0.027}$$ $$= \frac{0.01}{0.027} = 0.370$$ **Answer:** P(Machine A | Defective) = **0.370 or 37%**

Section D: Binomial Distribution

Problem 11: Basic Binomial Probability

Question: In a multiple-choice exam with 10 questions, each having 4 options, a student guesses all answers. Find the probability of: a) Getting exactly 3 correct b) Getting at least 2 correct c) Getting at most 1 correct

Click to reveal solution **Step 1:** Identify parameters - n = 10 (number of trials) - p = 1/4 = 0.25 (probability of success) - q = 1 - p = 0.75 **Binomial formula:** $$P(X = r) = \binom{n}{r} p^r q^{n-r}$$ **Part (a):** P(X = 3) $$P(X = 3) = \binom{10}{3} (0.25)^3 (0.75)^7$$ $$= 120 \times 0.015625 \times 0.1335 = 0.250$$ **Part (b):** P(X ≥ 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)] $$P(X = 0) = \binom{10}{0} (0.25)^0 (0.75)^{10} = 1 \times 1 \times 0.0563 = 0.0563$$ $$P(X = 1) = \binom{10}{1} (0.25)^1 (0.75)^9 = 10 \times 0.25 \times 0.0751 = 0.1877$$ $$P(X \geq 2) = 1 - (0.0563 + 0.1877) = 1 - 0.244 = 0.756$$ **Part (c):** P(X ≤ 1) $$P(X \leq 1) = P(X=0) + P(X=1) = 0.0563 + 0.1877 = 0.244$$ **Answers:** a) **0.250**, b) **0.756**, c) **0.244**

Problem 12: Binomial Distribution - Quality Control

Question: A machine produces items with 5% defective rate. If a sample of 15 items is selected, find: a) P(exactly 2 defectives) b) P(at most 1 defective) c) Expected number of defectives d) Standard deviation

Click to reveal solution **Parameters:** n = 15, p = 0.05, q = 0.95 **Part (a):** P(X = 2) $$P(X = 2) = \binom{15}{2} (0.05)^2 (0.95)^{13}$$ $$= 105 \times 0.0025 \times 0.5133 = 0.135$$ **Part (b):** P(X ≤ 1) $$P(X = 0) = \binom{15}{0} (0.05)^0 (0.95)^{15} = 0.4633$$ $$P(X = 1) = \binom{15}{1} (0.05)^1 (0.95)^{14} = 15 \times 0.05 \times 0.4877 = 0.3658$$ $$P(X \leq 1) = 0.4633 + 0.3658 = 0.8291$$ **Part (c):** Expected value $$E(X) = np = 15 \times 0.05 = 0.75$$ **Part (d):** Standard deviation $$\sigma = \sqrt{npq} = \sqrt{15 \times 0.05 \times 0.95} = \sqrt{0.7125} = 0.844$$ **Answers:** a) **0.135**, b) **0.829**, c) **0.75**, d) **0.844**

Problem 13: Binomial - Election Polling

Question: In an election, 40% of voters favor candidate A. If 20 voters are randomly selected, find: a) P(exactly 8 favor A) b) P(between 6 and 10 inclusive favor A) c) Mean and variance

Click to reveal solution **Parameters:** n = 20, p = 0.40, q = 0.60 **Part (a):** P(X = 8) $$P(X = 8) = \binom{20}{8} (0.40)^8 (0.60)^{12}$$ $$= 125970 \times 0.000655 \times 0.00218 = 0.180$$ **Part (b):** P(6 ≤ X ≤ 10) This requires calculating P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) | X | Calculation | P(X) | | --- | ------------------------------------ | ----- | | 6 | $\binom{20}{6}(0.4)^6(0.6)^{14}$ | 0.124 | | 7 | $\binom{20}{7}(0.4)^7(0.6)^{13}$ | 0.166 | | 8 | Already calculated | 0.180 | | 9 | $\binom{20}{9}(0.4)^9(0.6)^{11}$ | 0.160 | | 10 | $\binom{20}{10}(0.4)^{10}(0.6)^{10}$ | 0.117 | $$P(6 \leq X \leq 10) = 0.124 + 0.166 + 0.180 + 0.160 + 0.117 = 0.747$$ **Part (c):** Mean and Variance $$\mu = np = 20 \times 0.40 = 8$$ $$\sigma^2 = npq = 20 \times 0.40 \times 0.60 = 4.8$$ **Answers:** a) **0.180**, b) **0.747**, c) Mean = **8**, Variance = **4.8**

Section E: Normal Distribution

Problem 14: Standard Normal Distribution (Z-Score)

Question: For a standard normal distribution, find: a) P(Z < 1.5) b) P(Z > 2.0) c) P(-1.5 < Z < 1.5) d) P(Z < -0.5)

Click to reveal solution Using standard normal table (Z-table): **Part (a):** P(Z < 1.5) From table: **P(Z < 1.5) = 0.9332** **Part (b):** P(Z > 2.0) $$P(Z > 2.0) = 1 - P(Z < 2.0) = 1 - 0.9772 = 0.0228$$ **Part (c):** P(-1.5 < Z < 1.5) $$P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)$$ $$= 0.9332 - 0.0668 = 0.8664$$ **Part (d):** P(Z < -0.5) By symmetry: P(Z < -0.5) = P(Z > 0.5) = 1 - 0.6915 = **0.3085** **Answers:** a) **0.9332**, b) **0.0228**, c) **0.8664**, d) **0.3085**

Problem 15: Converting Raw Scores to Z-Scores

Question: Exam scores are normally distributed with mean μ = 70 and standard deviation σ = 10. Find: a) P(score > 85) b) P(score < 60) c) P(65 < score < 80) d) Score at 90th percentile

Click to reveal solution **Z-score formula:** $$Z = \frac{X - \mu}{\sigma} = \frac{X - 70}{10}$$ **Part (a):** P(X > 85) $$Z = \frac{85 - 70}{10} = 1.5$$ $$P(X > 85) = P(Z > 1.5) = 1 - 0.9332 = 0.0668$$ **Part (b):** P(X < 60) $$Z = \frac{60 - 70}{10} = -1.0$$ $$P(X < 60) = P(Z < -1.0) = 0.1587$$ **Part (c):** P(65 < X < 80) $$Z_1 = \frac{65 - 70}{10} = -0.5$$ $$Z_2 = \frac{80 - 70}{10} = 1.0$$ $$P(65 < X < 80) = P(-0.5 < Z < 1.0)$$ $$= P(Z < 1.0) - P(Z < -0.5) = 0.8413 - 0.3085 = 0.5328$$ **Part (d):** 90th percentile - From table, P(Z < 1.28) = 0.90 - Therefore Z = 1.28 $$X = \mu + Z\sigma = 70 + 1.28(10) = 70 + 12.8 = 82.8$$ **Answers:** a) **0.0668**, b) **0.1587**, c) **0.5328**, d) **82.8**

Problem 16: Normal Distribution - Salary Analysis

Question: Salaries of government employees are normally distributed with mean Rs. 45,000 and SD Rs. 8,000. Find: a) Probability of salary between Rs. 40,000 and Rs. 55,000 b) Percentage of employees earning more than Rs. 60,000 c) Salary above which top 5% of employees earn

Click to reveal solution **Given:** μ = 45,000, σ = 8,000 **Part (a):** P(40,000 < X < 55,000) $$Z_1 = \frac{40000 - 45000}{8000} = -0.625$$ $$Z_2 = \frac{55000 - 45000}{8000} = 1.25$$ $$P(-0.625 < Z < 1.25) = P(Z < 1.25) - P(Z < -0.625)$$ $$= 0.8944 - 0.2660 = 0.6284$$ **Part (b):** P(X > 60,000) $$Z = \frac{60000 - 45000}{8000} = 1.875$$ $$P(Z > 1.875) = 1 - 0.9696 = 0.0304 = 3.04\%$$ **Part (c):** Top 5% threshold - P(Z > z*) = 0.05, so P(Z < z*) = 0.95 - From table: z\* = 1.645 $$X = 45000 + 1.645(8000) = 45000 + 13160 = 58,160$$ **Answers:** a) **62.84%**, b) **3.04%**, c) **Rs. 58,160**

Problem 17: Normal Distribution - Quality Control

Question: Weights of packaged items are normally distributed with mean 500g and SD 15g. Quality standards require weights between 475g and 525g. Find: a) Probability an item meets the standard b) Expected number of substandard items in 1000 items c) If mean shifts to 495g (SD same), what proportion would be substandard?

Click to reveal solution **Part (a):** P(475 < X < 525) with μ = 500, σ = 15 $$Z_1 = \frac{475 - 500}{15} = -1.67$$ $$Z_2 = \frac{525 - 500}{15} = 1.67$$ $$P(-1.67 < Z < 1.67) = 0.9525 - 0.0475 = 0.9050$$ **Part (b):** Expected substandard items $$P(\text{substandard}) = 1 - 0.9050 = 0.0950$$ $$\text{Expected} = 1000 \times 0.0950 = 95 \text{ items}$$ **Part (c):** With μ = 495, σ = 15 $$Z_1 = \frac{475 - 495}{15} = -1.33$$ $$Z_2 = \frac{525 - 495}{15} = 2.00$$ $$P(-1.33 < Z < 2.00) = 0.9772 - 0.0918 = 0.8854$$ $$P(\text{substandard}) = 1 - 0.8854 = 0.1146 = 11.46\%$$ **Answers:** a) **90.50%**, b) **95 items**, c) **11.46%**

Problem 18: Finding Mean or SD from Given Probabilities

Question: In a normally distributed population:

  • 10% of values are below 50
  • 5% of values are above 80

Find the mean and standard deviation.

Solution:

Step 1: Convert to Z-scores

  • P(X < 50) = 0.10 → Z = -1.28
  • P(X > 80) = 0.05 → Z = 1.645

Step 2: Set up equations \(50 = \mu - 1.28\sigma \quad ...(1)\) \(80 = \mu + 1.645\sigma \quad ...(2)\)

Step 3: Solve equations Subtracting (1) from (2): \(30 = 2.925\sigma\) \(\sigma = 10.26\)

Substituting in (1): \(50 = \mu - 1.28(10.26)\) \(\mu = 50 + 13.13 = 63.13\)

Answers: μ = 63.13, σ = 10.26

Problem 19: Normal Approximation to Binomial

Question: In a large city, 40% of households have broadband internet. For a random sample of 200 households, find the probability that: a) Between 70 and 90 have broadband b) More than 85 have broadband

Click to reveal solution **Step 1:** Check if normal approximation is appropriate - np = 200 × 0.40 = 80 ≥ 5 ✓ - nq = 200 × 0.60 = 120 ≥ 5 ✓ **Step 2:** Calculate mean and SD $$\mu = np = 80$$ $$\sigma = \sqrt{npq} = \sqrt{200 \times 0.4 \times 0.6} = \sqrt{48} = 6.93$$ **Part (a):** P(70 < X < 90) with continuity correction P(69.5 < X < 90.5) $$Z_1 = \frac{69.5 - 80}{6.93} = -1.51$$ $$Z_2 = \frac{90.5 - 80}{6.93} = 1.51$$ $$P(-1.51 < Z < 1.51) = 0.9345 - 0.0655 = 0.8690$$ **Part (b):** P(X > 85) with continuity correction P(X > 85.5) $$Z = \frac{85.5 - 80}{6.93} = 0.79$$ $$P(Z > 0.79) = 1 - 0.7852 = 0.2148$$ **Answers:** a) **0.869 or 86.9%**, b) **0.215 or 21.5%**

Problem 20: Sampling Distribution of Mean

Question: A population has mean μ = 100 and SD σ = 20. For samples of size n = 64, find: a) Mean and standard error of sampling distribution b) P(sample mean > 103) c) P(97 < sample mean < 102)

Click to reveal solution **Step 1:** Calculate standard error $$SE = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{64}} = \frac{20}{8} = 2.5$$ **Part (a):** Mean of sampling distribution = μ = **100** Standard error = **2.5** **Part (b):** P(X̄ > 103) $$Z = \frac{103 - 100}{2.5} = 1.2$$ $$P(Z > 1.2) = 1 - 0.8849 = 0.1151$$ **Part (c):** P(97 < X̄ < 102) $$Z_1 = \frac{97 - 100}{2.5} = -1.2$$ $$Z_2 = \frac{102 - 100}{2.5} = 0.8$$ $$P(-1.2 < Z < 0.8) = 0.7881 - 0.1151 = 0.6730$$ **Answers:** a) Mean = **100**, SE = **2.5**, b) **0.1151**, c) **0.6730**

Practice Problems

  1. A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find P(both same color).

  2. In a survey, P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2. Find P(A B) and P(B A).

  3. A coin is biased with P(Head) = 0.6. If tossed 8 times, find P(exactly 5 heads).

  4. Heights are normally distributed with μ = 170cm, σ = 10cm. Find P(height between 160 and 180cm).

  5. In a binomial distribution with n = 100 and p = 0.3, use normal approximation to find P(25 ≤ X ≤ 35).

Summary of Key Formulas

Distribution Formula/Property
Classical Probability $P(A) = \frac{n(A)}{n(S)}$
Addition Rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Multiplication Rule $P(A \cap B) = P(A) \times P(B \mid A)$
Bayes’ Theorem $P(A \mid B) = \frac{P(B \mid A) \times P(A)}{P(B)}$
Binomial P(X=r) $\binom{n}{r} p^r q^{n-r}$
Binomial Mean $\mu = np$
Binomial SD $\sigma = \sqrt{npq}$
Z-score $Z = \frac{X - \mu}{\sigma}$
Standard Error $SE = \frac{\sigma}{\sqrt{n}}$