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Binomial and Poisson Distributions

The Binomial distribution answers: "In $n$ independent trials (each with probability $p$ of success), what is the probability of exactly $x$ successes?" Requirements: fixed number of trials, only two outcomes per trial (success/failure), same $p$ each time, and trials are independent. The Poisson distribution models how many times an event occurs in a fixed interval when the average rate is $\lambda$. Use it when events happen randomly over time or space (calls per hour, defects per meter of pipe, accidents per month). Mean and variance of Binomial: $\mu = np$, $\sigma^2 = np(1-p)$. For Poisson: $\mu = \sigma^2 = \lambda$.

$$\text{Binomial: }P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}, \quad \mu=np$$
$$\text{Poisson: }P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}, \quad \mu=\lambda$$

Binomial Defective Items

A batch has 5% defective items. If 10 items are selected, find the probability exactly 2 are defective.

$$P(X=2)=C(10,2)(0.05)^2(0.95)^8=0.0746$$

Final answer: 0.0746.

At Least One Success

A machine has 0.90 probability of passing inspection. If 4 machines are inspected, find the probability at least one fails.

At least one fails is the complement of all pass.

$$P=1-(0.90)^4=0.3439$$

Final answer: 0.3439.

Poisson Arrivals

Calls arrive at an average of 3 per hour. Find the probability of exactly 5 calls in one hour.

$$P(X=5)=\frac{e^{-3}3^5}{5!}=0.1008$$

Final answer: 0.1008.

Binomial Mean and Variance

A fair coin is flipped 20 times. Find the expected number of heads and the standard deviation.

$n=20$, $p=0.5$, $q=0.5$.

$$\mu=np=20(0.5)=10$$
$$\sigma^2=np(1-p)=20(0.5)(0.5)=5, \quad \sigma=\sqrt{5}\approx2.24$$

Final answer: Expected heads = 10, σ ≈ 2.24.

Binomial: P(X ≥ 3)

The probability a concrete cylinder passes a strength test is 0.8. If 5 cylinders are tested, find P(X ≥ 3).

Find $P(X\geq3)=P(X=3)+P(X=4)+P(X=5)$, or use the complement: $1-P(X\leq2)$.

$$P(X=3)=C(5,3)(0.8)^3(0.2)^2=10(0.512)(0.04)=0.2048$$
$$P(X=4)=C(5,4)(0.8)^4(0.2)^1=5(0.4096)(0.2)=0.4096$$
$$P(X=5)=(0.8)^5=0.3277$$
$$P(X\geq3)=0.2048+0.4096+0.3277=0.9421$$

Final answer: 0.9421.

Poisson: Probability of Zero Events

Accidents occur at a construction site at a rate of 2 per month. What is the probability of no accidents in a given month?

$\lambda=2$, $x=0$. Recall $e^{-2}\approx0.1353$.

$$P(X=0)=\frac{e^{-2}(2)^0}{0!}=e^{-2}=0.1353$$

Final answer: 0.1353 or about 13.5% chance of a zero-accident month.

Poisson: Changing the Interval

Defects occur at a rate of 4 per 100 meters of cable. Find the probability of exactly 2 defects in 50 meters.

Rescale the rate: for 50 m, $\lambda=4\times(50/100)=2$.

$$P(X=2)=\frac{e^{-2}(2)^2}{2!}=\frac{0.1353(4)}{2}=0.2707$$

Final answer: 0.2707.

Quality Control with Binomial

A lot has 10% defective items. A sample of 8 is inspected. Find P(exactly 1 defective) and P(at most 1 defective).

$n=8$, $p=0.10$, $q=0.90$.

$$P(X=0)=C(8,0)(0.1)^0(0.9)^8=0.4305$$
$$P(X=1)=C(8,1)(0.1)^1(0.9)^7=8(0.1)(0.4783)=0.3826$$
$$P(X\leq1)=0.4305+0.3826=0.8131$$

Final answer: P(exactly 1) = 0.3826; P(at most 1) = 0.8131.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q662

MSTE - Statistics and Probability / Probability / Engr. Janclyde Espinosa (Clidez)

A contractor estimated that one of his two bricklayers would take 9 hours to build a certain wall and the other 10 hours. However, from experience he knew that if they worked together 10 fewer bricks get laid per hour. Since he was in a hurry, he put both men on the job and found it took exactly 5 hours to build the wall. How many bricks did it contain?

  1. 900
  2. 800
  3. 1000
  4. 1100
Let the wall contain $N$ bricks. Individual rates are $N/9$ and $N/10$ bricks/hr. Together they lay 10 fewer bricks per hour than the sum of rates, and finish in 5 hr:
$5\left(\frac{N}{9}+\frac{N}{10}-10\right)=N$
$5\left(\frac{19N}{90}-10\right)=N$
$\frac{19N}{18}-50=N$
$\boxed{N=900}$

Question Bank: t18

MSTE - Statistics and Probability / Probability / Civil Engineering Refresher

A manufacturer finds that 1 out of 2000 chips is defective. How many chips must be ordered so the probability of at least one defect is 45%?

  1. 1194
  2. 1195
  3. 2000
  4. 900
$P(\text{at least one}) = 1 - \left(1 - \frac{1}{2000}\right)^n = 0.45$.
$(0.9995)^n = 0.55$
$n = \frac{\ln 0.55}{\ln 0.9995} = \frac{-0.5978}{-0.0005001}$
$\boxed{\approx 1194}$

Question Bank: t19

MSTE - Statistics and Probability / Probability / Civil Engineering Refresher

The probability that both stages of a missile function correctly is 0.95. If the first stage has a 0.98 probability of success, find the probability the second functions correctly given the first did.

  1. 0.97
  2. 0.95
  3. 0.98
  4. 0.93
Conditional probability: $P(\text{both}) = P(1)\cdot P(2\,|\,1)$
$0.95 = 0.98 \cdot P(2\,|\,1)$
$P(2\,|\,1) = \frac{0.95}{0.98}$
$\boxed{= 0.97}$

Question Bank: t2169

MSTE - Statistics and Probability / Probability / Besavilla CE Pre-Board Math & Surveying

The probability of a component failing in one year due to excessive temperature is 0.05, due to excessive vibration is 0.04 and due to excessive humidity is 0.02. Determine the probability that during one year period, a component fails due to excessive temperature and excessive vibration.

  1. 0.045
  2. 0.085
  3. 0.002
  4. 0.012
  5. 0.112
Assuming the two causes are independent, multiply the probabilities for excessive temperature and excessive vibration.
$P(T\cap V)=P(T)P(V)$
$P(T\cap V)=(0.05)(0.04)$
$\boxed{P(T\cap V)=0.002}$

Question Bank: w19

MSTE - Statistics and Probability / Probability / MSTE May 2019

Samsung, a computer chip manufacturer, has found that only 1 out of 2000 chips is defective. A certain company ordered a shipment of chips. How many chips will the company order before the probability that at least one chip is defective is 45%?

  1. 1246
  2. 1428
  3. 1086
  4. 1194
$P(\text{at least one defective}) = 1 - q^n$, where $q = \frac{1999}{2000}$.
$0.45 = 1 - \left(\frac{1999}{2000}\right)^{n}$
$n = \frac{\ln 0.55}{\ln(1999/2000)} \approx 1{,}195$
$\boxed{n \approx 1194\text{ chips}}$

Question Bank: w80

MSTE - Statistics and Probability / Probability / MSTE November 2019

A box contains 5 defective and 195 non-defective cell phones. A quality control engineer selects 2 cell phones at random without replacement. What is the probability that exactly 1 is defective?

  1. 0.049
  2. 0.043
  3. 0.038
  4. 0.027
Exactly one defective occurs as (defective then good) or (good then defective):
$P = \dfrac{5}{200}\cdot\dfrac{195}{199} + \dfrac{195}{200}\cdot\dfrac{5}{199} = 0.049$
$\boxed{P = 0.049}$

Question Bank: w81

MSTE - Statistics and Probability / Probability / MSTE November 2019

Smith and Jones, both 50% marksmen, decide to fight a duel in which they exchange alternate shots until one is hit. What are the odds in favor of the man who shoots first?

  1. 2/3
  2. 1/4
  3. 1/3
  4. 1/2
Let the first shooter win on his 1st, 2nd, 3rd, … shot. Each shot hits with probability $\tfrac{1}{2}$, and reaching a later shot requires both having missed (probability $\tfrac{1}{4}$ per round):
$P = \tfrac{1}{2} + \tfrac{1}{2}\!\left(\tfrac{1}{4}\right) + \tfrac{1}{2}\!\left(\tfrac{1}{4}\right)^2 + \cdots$
Infinite geometric series with $a_1 = \tfrac{1}{2}$, $r = \tfrac{1}{4}$:
$P = \dfrac{a_1}{1 - r} = \dfrac{1/2}{1 - 1/4} = \dfrac{2}{3}$
$\boxed{\tfrac{2}{3}}$

Question Bank: w202

MSTE - Statistics and Probability / Probability / MSTE May 2022

A test consists of 6 multiple choice questions. To pass the test, a student must answer at least 4 questions correctly. Each question has 3 possible answers, of which only one is correct. If a student guesses on each question, what is the probability that the student will pass the test?

  1. 0.2001
  2. 0.3001
  3. 0.0209
  4. 0.1001
This is binomial with $n=6$, $p=1/3$, and $q=2/3$. The passing probability is $$P(X\ge4)=\binom64p^4q^2+\binom65p^5q+\binom66p^6.$$ The printed terms are $0.0823$, $0.0165$, and $0.0014$, so $$P=0.0823+0.0165+0.0014=0.1001.$$

Question Bank: w249

MSTE - Statistics and Probability / Statistics / MSTE April 2025

A binomial probability distribution involves 60 trials with a probability of failure of 0.26. Find the standard deviation of this probability distribution.

  1. 15.6
  2. 3.4
  3. 6.66
  4. 11.5
For a binomial distribution:
Mean, $\mu = np$
Standard deviation, $\sigma = \sqrt{npq}$

From the problem:
$n = 60$
$p = 1 - 0.26 = 0.74$
$q = 0.26$

$\sigma = \sqrt{60(0.74)(0.26)} = 3.3976$
$\boxed{\sigma \approx 3.4}$

Question Bank: w264

MSTE - Statistics and Probability / Probability / MSTE April 2025

In an NBA championship series, the team which wins four games out of seven will be the winner. Suppose that team A has probability 0.55 of winning over the team B and both teams A and B face each other in the championship games, what is the probability that team A will win the series?

  1. 0.185
  2. 0.608
  3. 0.292
  4. 0.590
Team A wins the series by clinching the 4th win in game 4, 5, 6, or 7. For a clinch in game $x$, A must win exactly 3 of the first $x-1$ games and then win game $x$.

$P = P_4 + P_5 + P_6 + P_7$
$= 0.55^{4} + 0.55^{3}(0.45^{1})({}_{4}C_{3})(0.55) + 0.55^{3}(0.45^{2})({}_{5}C_{3})(0.55) + 0.55^{3}(0.45^{3})({}_{6}C_{3})(0.55)$

Compactly,
$P = \sum_{x=3}^{6}\left[0.55^{3}(0.45^{\,x-3})({}_{x}C_{3})(0.55)\right]$
$\boxed{P = 0.6083}$

Question Bank: w267

MSTE - Statistics and Probability / Probability / MSTE April 2025

Samuel Pepys wrote Isaac Newton to ask which of three events is more likely: that a person get (a) at least 1 six when 6 dice are rolled, (b) at least two sixes when 12 dice are rolled, or (c) at least 3 sixes when 18 dice are rolled. What is the answer?

  1. Pepys will do better with the 18-dice wager than 6 or 12
  2. Pepys will do better with the 12-dice wager than 6 or 18
  3. Pepys will do better with the 6-dice wager than 12 or 18
  4. None of the above
Each case is binomial with $p = 1/6$. Compute the complement of "fewer than the required number of sixes".

(a) At least 1 six in 6 dice:
$P = 1 - (5/6)^{6} = 0.665$

(b) At least 2 sixes in 12 dice:
$P = 1 - (5/6)^{12} - (1/6)^{1}(5/6)^{11}({}_{12}C_{1}) = 0.619$

(c) At least 3 sixes in 18 dice:
$P = 1 - (5/6)^{18} - (1/6)^{1}(5/6)^{17}({}_{18}C_{1}) - (1/6)^{2}(5/6)^{16}({}_{18}C_{2}) = 0.597$

The 6-dice wager has the highest probability.
$\boxed{\text{Pepys will do better with the 6 dice wager}}$

Question Bank: w278

MSTE - Statistics and Probability / Probability / MSTE April 2025

A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%. Suppose that the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be 3 shipments containing at least one defective device?

  1. 0.1620
  2. 0.1062
  3. 0.1602
  4. 0.1206
Solve in two binomial stages.

For each shipment (probability that at least one of 20 tested devices is defective):
$P_1 = 1 - 0.97^{20} = 0.4562$

For 10 shipments (exactly 3 of them "flagged"):
$P = p^{x}q^{\,n-x}\,{}_{n}C_{x}$
$P = P_1^{3}(1 - P_1)^{7}({}_{10}C_{3})$
$\boxed{P = 0.1602}$

Question Bank: w299

MSTE - Statistics and Probability / Probability / MSTE April 2025

In a family of five children, what is the chance that there are three boys and two girls?

  1. 1/32
  2. 9/24
  3. 5/16
  4. 3/8
Binomial probability with $p = q = 1/2$ for each birth:
$P = p^{x}q^{\,n-x}\,{}_{n}C_{x}$
$P = \left(\tfrac{1}{2}\right)^{3}\left(\tfrac{1}{2}\right)^{2}({}_{5}C_{3})$
$P = \dfrac{10}{32}$
$\boxed{P = \tfrac{5}{16}}$