Discrete Distribution Problems & Calculators

Part 1: Step-by-Step Examples

This section provides detailed solutions to 20 problems covering various discrete probability distributions. Round answers as specified or to 4 decimal places if not mentioned.

Question 1: A quality control inspector finds that 5% of products are defective. If 10 products are selected at random, what is the probability that exactly 1 is defective?

Binomial Distribution Calculator

Note: If the calculator does not load, please ensure your browser allows third-party iframes or visit the site directly.

Solution to Question 1:

Type: Binomial Distribution

Formula: P(X=k) = C(n, k) * pk * (1-p)(n-k)

Parameters:

  • n (number of trials) = 10
  • p (probability of success, i.e., defective) = 0.05
  • k (number of successes, i.e., defectives) = 1
  • q (probability of failure) = 1 - p = 1 - 0.05 = 0.95

Calculation Steps:

  1. Calculate combinations: C(n, k) = C(10, 1) = 10! / (1! * (10-1)!) = 10.
  2. Calculate pk: (0.05)1 = 0.05.
  3. Calculate (1-p)(n-k): (0.95)(10-1) = (0.95)9 ≈ 0.630249399.
  4. Multiply the results: P(X=1) = 10 * 0.05 * 0.630249399 ≈ 0.3151246995.

Final Answer: The probability of selecting exactly 1 defective product is approximately 0.3151.

Question 2: For a binomial distribution with n=20 trials and probability of success p=0.4, calculate the expected number of successes E(X) and the variance Var(X).

Geometric Distribution Calculator

Note: If the calculator does not load, please ensure your browser allows third-party iframes or visit the site directly.

Solution to Question 2:

Type: Binomial Distribution

Formulas:

  • E(X) = n * p
  • Var(X) = n * p * (1-p)

Parameters:

  • n = 20
  • p = 0.4
  • q = 1 - p = 1 - 0.4 = 0.6

Calculation Steps:

  1. Calculate E(X): E(X) = 20 * 0.4 = 8.
  2. Calculate Var(X): Var(X) = 20 * 0.4 * 0.6 = 8 * 0.6 = 4.8.

Final Answer: The expected number of successes E(X) is 8, and the variance Var(X) is 4.8.

Question 3: A student takes a 3-question multiple-choice quiz. Each question has 4 options, only one of which is correct. The student guesses randomly. What is the probability the student gets at least 2 questions correct (P(X ≥ 2))?

Poisson Distribution Calculator

Note: If the calculator does not load, please ensure your browser allows third-party iframes or visit the site directly.

Solution to Question 3:

Type: Binomial Distribution

Formula: P(X=k) = C(n, k) * pk * (1-p)(n-k)

We need P(X ≥ 2) = P(X=2) + P(X=3).

Parameters:

  • n (number of questions) = 3
  • p (probability of guessing correctly) = 1/4 = 0.25
  • q (probability of guessing incorrectly) = 1 - 0.25 = 0.75

Calculation Steps for P(X=2):

  1. C(3, 2) = 3.
  2. p2 = (0.25)2 = 0.0625.
  3. q(3-2) = (0.75)1 = 0.75.
  4. P(X=2) = 3 * 0.0625 * 0.75 = 0.140625.

Calculation Steps for P(X=3):

  1. C(3, 3) = 1.
  2. p3 = (0.25)3 = 0.015625.
  3. q(3-3) = (0.75)0 = 1.
  4. P(X=3) = 1 * 0.015625 * 1 = 0.015625.

Final Calculation:

  1. P(X ≥ 2) = P(X=2) + P(X=3) = 0.140625 + 0.015625 = 0.15625.

Final Answer: The probability of getting at least 2 questions correct is 0.15625.

Question 4: A basketball player has a 60% chance of making a free throw. What is the probability that the player makes their first successful free throw on the 4th attempt?

Discrete Uniform Distribution Calculator

Note: If the calculator does not load, please ensure your browser allows third-party iframes or visit the site directly.

Solution to Question 4:

Type: Geometric Distribution

Formula: P(X=k) = (1-p)(k-1) * p

Parameters:

  • p (probability of success, making the shot) = 0.60
  • k (attempt number of first success) = 4
  • q (probability of failure) = 1 - p = 1 - 0.60 = 0.40

Calculation Steps:

  1. Calculate (1-p)(k-1): (0.40)(4-1) = (0.40)3 = 0.064.
  2. Multiply by p: P(X=4) = 0.064 * 0.60 = 0.0384.

Final Answer: The probability that the first successful free throw is on the 4th attempt is 0.0384.

Question 5: If the probability of finding a rare gem in a mine is 0.02 per ton of ore processed, what is the expected number of tons of ore that need to be processed to find the first gem?

Type: Geometric Distribution

Formula: E(X) = 1 / p

Parameters:

  • p (probability of success, finding a gem) = 0.02

Calculation Steps:

  1. E(X) = 1 / 0.02 = 50.

Final Answer: The expected number of tons of ore to process to find the first gem is 50 tons.

Question 6: A salesperson has a 10% chance of closing a deal with each client. What is the probability that they close their first deal within the first 3 clients (P(X ≤ 3))?

Type: Geometric Distribution

Formula for P(X ≤ k): P(X ≤ k) = 1 - (1-p)k

Alternatively, P(X ≤ 3) = P(X=1) + P(X=2) + P(X=3)

Parameters:

  • p (probability of success, closing a deal) = 0.10
  • k = 3
  • q (probability of failure) = 1 - p = 0.90

Calculation Steps (using P(X ≤ k) formula):

  1. Calculate (1-p)k: (0.90)3 = 0.729.
  2. Subtract from 1: P(X ≤ 3) = 1 - 0.729 = 0.271.

Calculation Steps (summing individual probabilities):

  1. P(X=1) = (0.90)0 * 0.10 = 1 * 0.10 = 0.10.
  2. P(X=2) = (0.90)1 * 0.10 = 0.90 * 0.10 = 0.09.
  3. P(X=3) = (0.90)2 * 0.10 = 0.81 * 0.10 = 0.081.
  4. P(X ≤ 3) = 0.10 + 0.09 + 0.081 = 0.271.

Final Answer: The probability of closing the first deal within the first 3 clients is 0.271.

Question 7: An urn contains 7 red balls and 5 blue balls. If 4 balls are drawn without replacement, what is the probability that exactly 2 of them are red?

Type: Hypergeometric Distribution

Formula: P(X=k) = [C(K, k) * C(N-K, n-k)] / C(N, n)

Parameters:

  • N (total number of balls) = 7 (red) + 5 (blue) = 12
  • K (total number of "success" items in population, i.e., red balls) = 7
  • n (number of items drawn, sample size) = 4
  • k (number of "success" items in sample, i.e., red balls drawn) = 2
  • N-K (total number of "failure" items, i.e., blue balls) = 5
  • n-k (number of "failure" items in sample, i.e., blue balls drawn) = 4 - 2 = 2

Calculation Steps:

  1. C(K, k) = C(7, 2) = 7! / (2! * 5!) = (7*6)/(2*1) = 21.
  2. C(N-K, n-k) = C(5, 2) = 5! / (2! * 3!) = (5*4)/(2*1) = 10.
  3. C(N, n) = C(12, 4) = 12! / (4! * 8!) = (12*11*10*9)/(4*3*2*1) = 495.
  4. P(X=2) = (21 * 10) / 495 = 210 / 495.
  5. Simplify: 210 / 495 = 42 / 99 = 14 / 33 ≈ 0.424242.

Final Answer: The probability of drawing exactly 2 red balls is 210/495 ≈ 0.4242.

Question 8: A small pond has 10 fish, 4 of which are tagged. If a fisherman catches 3 fish, what is the probability that all 3 caught fish are tagged?

Type: Hypergeometric Distribution

Formula: P(X=k) = [C(K, k) * C(N-K, n-k)] / C(N, n)

Parameters:

  • N (total fish) = 10
  • K (total tagged fish) = 4
  • n (fish caught) = 3
  • k (tagged fish caught) = 3
  • N-K (untagged fish) = 10 - 4 = 6
  • n-k (untagged fish caught) = 3 - 3 = 0

Calculation Steps:

  1. C(K, k) = C(4, 3) = 4! / (3! * 1!) = 4.
  2. C(N-K, n-k) = C(6, 0) = 1.
  3. C(N, n) = C(10, 3) = 10! / (3! * 7!) = (10*9*8)/(3*2*1) = 120.
  4. P(X=3) = (4 * 1) / 120 = 4 / 120.
  5. Simplify: 4 / 120 = 1 / 30 ≈ 0.033333.

Final Answer: The probability that all 3 caught fish are tagged is 4/120 or 1/30 ≈ 0.0333.

Question 9: A batch of 50 components contains 5 defective ones. If a sample of 10 components is selected, what is the expected number of defective components in the sample?

Type: Hypergeometric Distribution

Formula for Expected Value: E(X) = n * (K / N)

Parameters:

  • N (total components in batch) = 50
  • K (total defective components in batch) = 5
  • n (sample size) = 10

Calculation Steps:

  1. Calculate the proportion of defectives in the population: K / N = 5 / 50 = 0.1.
  2. Multiply by sample size: E(X) = 10 * 0.1 = 1.

Final Answer: The expected number of defective components in the sample is 1.

Question 10: A call center receives an average of 4 calls per minute. What is the probability of receiving exactly 2 calls in a given minute? (Use e ≈ 2.71828)

Type: Poisson Distribution

Formula: P(X=k) = (e-λ * λk) / k!

Parameters:

  • λ (average rate of events) = 4 calls per minute
  • k (number of events) = 2 calls
  • e (Euler's number) ≈ 2.71828

Calculation Steps:

  1. Calculate e-λ: e-4 ≈ (2.71828)-4 ≈ 0.0183156.
  2. Calculate λk: 42 = 16.
  3. Calculate k!: 2! = 2 * 1 = 2.
  4. P(X=2) = (0.0183156 * 16) / 2 = 0.2930496 / 2 = 0.1465248.

Final Answer: The probability of receiving exactly 2 calls in a given minute is approximately 0.1465.

Question 11: The average number of typos on a page of a certain book is 0.8. What is the probability that a randomly selected page has no typos (P(X=0))? (Use e ≈ 2.71828)

Type: Poisson Distribution

Formula: P(X=k) = (e-λ * λk) / k!

Parameters:

  • λ (average number of typos) = 0.8
  • k (number of typos) = 0

Calculation Steps:

  1. Calculate e-λ: e-0.8 ≈ (2.71828)-0.8 ≈ 0.449329.
  2. Calculate λk: (0.8)0 = 1.
  3. Calculate k!: 0! = 1.
  4. P(X=0) = (0.449329 * 1) / 1 = 0.449329.

Final Answer: The probability that a page has no typos is approximately 0.4493.

Question 12: A shop sells an average of 2 specialty cakes per day. What is the probability that it sells at most 1 specialty cake on a given day (P(X ≤ 1))? (Use e ≈ 2.71828)

Type: Poisson Distribution

Formula: P(X=k) = (e-λ * λk) / k!

We need P(X ≤ 1) = P(X=0) + P(X=1).

Parameters:

  • λ (average cakes sold) = 2

Calculation for P(X=0):

  1. e-2 ≈ (2.71828)-2 ≈ 0.135335.
  2. λ0 = 20 = 1.
  3. 0! = 1.
  4. P(X=0) = (0.135335 * 1) / 1 = 0.135335.

Calculation for P(X=1):

  1. e-2 ≈ 0.135335.
  2. λ1 = 21 = 2.
  3. 1! = 1.
  4. P(X=1) = (0.135335 * 2) / 1 = 0.27067.

Final Calculation:

  1. P(X ≤ 1) = P(X=0) + P(X=1) = 0.135335 + 0.27067 = 0.406005.

Final Answer: The probability of selling at most 1 specialty cake is approximately 0.4060.

Question 13: If a website receives an average of 120 hits per hour, what is the probability of receiving exactly 3 hits in a 1-minute interval? (Use e ≈ 2.71828)

Type: Poisson Distribution

Formula: P(X=k) = (e-λ * λk) / k!

Parameters:

  • Original λ = 120 hits per hour.
  • We need λ for a 1-minute interval. Since 1 hour = 60 minutes:
  • Adjusted λ = 120 hits / 60 minutes = 2 hits per minute.
  • k (number of hits in 1 minute) = 3.

Calculation Steps (using adjusted λ = 2):

  1. Calculate e-λ: e-2 ≈ 0.135335.
  2. Calculate λk: 23 = 8.
  3. Calculate k!: 3! = 3 * 2 * 1 = 6.
  4. P(X=3) = (0.135335 * 8) / 6 = 1.08268 / 6 = 0.180446.

Final Answer: The probability of receiving exactly 3 hits in a 1-minute interval is approximately 0.1804.

Question 14: A manufacturer knows that 10% of their light bulbs are defective. If they ship a box of 6 bulbs, what is the probability that 2 or fewer bulbs are defective (P(X ≤ 2))?

Type: Binomial Distribution

Formula: P(X=k) = C(n, k) * pk * (1-p)(n-k)

We need P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2).

Parameters:

  • n = 6 (bulbs in a box)
  • p = 0.10 (probability of a bulb being defective)
  • q = 1 - p = 0.90

Calculation for P(X=0):

  1. C(6, 0) = 1.
  2. p0 = (0.10)0 = 1.
  3. q6 = (0.90)6 ≈ 0.531441.
  4. P(X=0) = 1 * 1 * 0.531441 = 0.531441.

Calculation for P(X=1):

  1. C(6, 1) = 6.
  2. p1 = (0.10)1 = 0.10.
  3. q5 = (0.90)5 ≈ 0.59049.
  4. P(X=1) = 6 * 0.10 * 0.59049 = 0.354294.

Calculation for P(X=2):

  1. C(6, 2) = 15.
  2. p2 = (0.10)2 = 0.01.
  3. q4 = (0.90)4 ≈ 0.6561.
  4. P(X=2) = 15 * 0.01 * 0.6561 = 0.098415.

Final Calculation:

  1. P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) = 0.531441 + 0.354294 + 0.098415 = 0.98415.

Final Answer: The probability that 2 or fewer bulbs are defective is 0.98415.

Question 15: A computer program has a 5% chance of crashing on any given run. What is the probability that the program runs successfully more than 3 times before its first crash (i.e., the first crash occurs on the 4th run or later, P(X > 3))?

Type: Geometric Distribution

Let "success" be a crash. So p = 0.05. X is the run number of the first crash.

P(X > k) means the first k trials were failures (no crash).

Formula for P(X > k): P(X > k) = (1-p)k

Parameters:

  • p (probability of a crash) = 0.05
  • k = 3 (we want the first crash to be *after* the 3rd run)
  • q (probability of no crash) = 1 - p = 0.95

Calculation Steps:

  1. Calculate (1-p)k: (0.95)3 = 0.95 * 0.95 * 0.95 = 0.857375.

Final Answer: The probability that the first crash occurs after the 3rd run is 0.857375.

Question 16: A bag contains 5 green marbles, 4 red marbles, and 3 blue marbles. If 3 marbles are drawn at random without replacement, what is the probability of drawing 1 green, 1 red, and 1 blue marble?

Type: Hypergeometric Distribution (extended concept)

This is a direct application of combinations for multiple groups.

Formula: P(event) = (Ways to choose 1G * Ways to choose 1R * Ways to choose 1B) / (Total ways to choose 3 marbles)

Parameters:

  • Total marbles N = 5 (G) + 4 (R) + 3 (B) = 12
  • Number of Green (KG) = 5, want kG = 1
  • Number of Red (KR) = 4, want kR = 1
  • Number of Blue (KB) = 3, want kB = 1
  • Sample size n = 3

Calculation Steps:

  1. Ways to choose 1 green from 5: C(5, 1) = 5.
  2. Ways to choose 1 red from 4: C(4, 1) = 4.
  3. Ways to choose 1 blue from 3: C(3, 1) = 3.
  4. Total ways to form the desired sample: 5 * 4 * 3 = 60.
  5. Total ways to choose any 3 marbles from 12: C(12, 3) = 12! / (3! * 9!) = (12*11*10)/(3*2*1) = 2 * 11 * 10 = 220.
  6. Probability = 60 / 220.
  7. Simplify: 60 / 220 = 6 / 22 = 3 / 11 ≈ 0.272727.

Final Answer: The probability of drawing 1 green, 1 red, and 1 blue marble is 60/220 or 3/11 ≈ 0.2727.

Question 17: The number of accidents in a factory per week follows a Poisson distribution with an average of 1.5. What is the probability of having at least 1 accident in a week (P(X ≥ 1))? (Use e ≈ 2.71828)

Type: Poisson Distribution

It's easier to calculate P(X ≥ 1) = 1 - P(X=0).

Formula for P(X=0): P(X=0) = e-λ

Parameters:

  • λ (average accidents) = 1.5

Calculation Steps:

  1. Calculate P(X=0): e-1.5 ≈ (2.71828)-1.5 ≈ 0.22313.
  2. Calculate P(X ≥ 1) = 1 - P(X=0) = 1 - 0.22313 = 0.77687.

Final Answer: The probability of having at least 1 accident in a week is approximately 0.7769.

Question 18: A survey found that 70% of college students own a laptop. If 5 students are randomly selected, what is the probability that exactly 4 of them own a laptop? (Identify the distribution)

Distribution Identification:

  • Fixed number of trials (n=5 students).
  • Each trial is independent.
  • Two outcomes for each trial (owns laptop / doesn't own laptop).
  • Probability of "success" (owning a laptop) is constant (p=0.70).
  • This is a Binomial Distribution.

Formula: P(X=k) = C(n, k) * pk * (1-p)(n-k)

Parameters:

  • n = 5
  • p = 0.70
  • k = 4
  • q = 1 - p = 0.30

Calculation Steps:

  1. C(n, k) = C(5, 4) = 5.
  2. pk = (0.70)4 = 0.2401.
  3. (1-p)(n-k) = (0.30)(5-4) = (0.30)1 = 0.30.
  4. P(X=4) = 5 * 0.2401 * 0.30 = 0.36015.

Final Answer: The probability that exactly 4 of the 5 students own a laptop is 0.36015.

Question 19: An archer hits the bullseye with a probability of 0.2 on any given shot. Shots are independent. What is the probability that her first bullseye occurs on her 5th shot? (Identify the distribution)

Distribution Identification:

  • Sequence of independent Bernoulli trials.
  • Looking for the trial number of the *first* success (first bullseye).
  • Probability of success is constant (p=0.2).
  • This is a Geometric Distribution.

Formula: P(X=k) = (1-p)(k-1) * p

Parameters:

  • p (probability of bullseye) = 0.2
  • k (shot number of first bullseye) = 5
  • q (probability of miss) = 1 - p = 0.8

Calculation Steps:

  1. Calculate (1-p)(k-1): (0.8)(5-1) = (0.8)4 = 0.4096.
  2. Multiply by p: P(X=5) = 0.4096 * 0.2 = 0.08192.

Final Answer: The probability that her first bullseye occurs on the 5th shot is 0.08192.

Question 20: A Geiger counter records an average of 10 radioactive decays per minute from a sample. What is the probability of observing exactly 7 decays in a particular minute? (Identify the distribution)

Distribution Identification:

  • Counting the number of events (decays) occurring in a fixed interval of time (1 minute).
  • Events occur independently.
  • The average rate of events is known and constant (λ=10).
  • This is a Poisson Distribution.

Formula: P(X=k) = (e-λ * λk) / k!

Parameters:

  • λ = 10 decays per minute
  • k = 7 decays
  • e ≈ 2.71828

Calculation Steps:

  1. e-λ: e-10 ≈ (2.71828)-10 ≈ 0.0000454.
  2. λk: 107 = 10,000,000.
  3. k!: 7! = 5040.
  4. P(X=7) = (0.0000454 * 10000000) / 5040 = 454 / 5040 ≈ 0.090079.

Final Answer: The probability of observing exactly 7 decays in that minute is approximately 0.0901.

Part 2: Interactive Practice Quiz

Test your knowledge with these 10 interactive questions. Enter your answer and click "Submit".

Practice Problems