UNIT II

 Unit–II : Probability Distributions for Discrete and Continuous Random Variables Probability  distributions  for  discrete  random  variables  – Bernoulli’s,  Binomial,  Geometric and Poisson distributions – applications 

Lecture Notes: Probability Distributions for Discrete and Continuous Random Variables

Introduction to Probability Distributions

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

• Discrete Random Variables: These take specific, countable values (e.g., number of cars on a bridge, cracks in a concrete slab).
• Continuous Random Variables: These take any value within a range (e.g., rainfall in mm, compressive strength of concrete).

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Discrete Probability Distributions

1. Bernoulli Distribution

• Definition: A distribution for a single trial with only two outcomes: success ($p$) or failure ($1-p$).

• Formula:

  $$P(X = x) = \begin{cases} p, & \text{if } x = 1 \, (\text{success}) \\ 1-p, & \text{if } x = 0 \, (\text{failure}) \end{cases}$$

  Here, $X \in \{0, 1\}$.

• Example in Civil Engineering: A soil test is conducted to check whether it is suitable ($X=1$) or unsuitable ($X=0$) for construction.

• Numerical Example:

  • Probability that a soil sample is suitable ($p$) = 0.8.
  • Probability of being unsuitable = $1-p = 0.2$.

  Outcome probabilities:

  $$P(X=1) = 0.8, \quad P(X=0) = 0.2$$

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2. Binomial Distribution

• Definition: Models the number of successes in $n$ independent Bernoulli trials.

• Formula:

  $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, 2, \dots, n$$

  Where:

  • $n$: Number of trials.
  • $k$: Number of successes.
  • $p$: Probability of success.

• Example in Civil Engineering: Estimating the number of defective parts in a batch of $n$ parts manufactured.

• Numerical Example:

  • Probability of a defective part ($p$) = 0.1.
  • Batch size ($n$) = 5.
  • Calculate $P(X=2)$, the probability of 2 defective parts: $$P(X=2) = \binom{5}{2} (0.1)^2 (0.9)^3 = 10 \cdot 0.01 \cdot 0.729 = 0.0729$$

  Result: Probability of 2 defective parts is $7.29\%$.

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3. Geometric Distribution

• Definition: Models the number of trials needed to get the first success.

• Formula:

  $$P(X = k) = (1-p)^{k-1} p, \quad k = 1, 2, 3, \dots$$

  Where:

  • $k$: Number of trials.
  • $p$: Probability of success.

• Example in Civil Engineering: Number of soil tests required to find the first suitable site for construction.

• Numerical Example:

  • Probability that a site is suitable ($p$) = 0.3.
  • Find $P(X=3)$, the probability that the 3rd test is the first success: $$P(X=3) = (1-0.3)^2 \cdot 0.3 = 0.7^2 \cdot 0.3 = 0.147$$

  Result: Probability is $14.7\%$.

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4. Poisson Distribution

• Definition: Models the number of events occurring in a fixed interval of time or space.

• Formula:

  $$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots$$

  Where:

  • $\lambda$: Average rate of occurrence.
  • $k$: Number of events.

• Example in Civil Engineering: Number of vehicles crossing a bridge in an hour.

• Numerical Example:

  • Average number of vehicles per hour ($\lambda$) = 5.
  • Find $P(X=3)$, the probability of 3 vehicles crossing: $$P(X=3) = \frac{5^3 e^{-5}}{3!} = \frac{125 \cdot 0.0067}{6} \approx 0.14$$

  Result: Probability is $14\%$.

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Continuous Probability Distributions

1. Uniform Distribution

• Definition: Models variables uniformly distributed over an interval $[a, b]$.

• Formula:

  $$f(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{otherwise} \end{cases}$$

• Example in Civil Engineering: The thickness of a concrete slab uniformly varying between 10 cm and 20 cm.

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2. Normal Distribution

• Definition: Models natural phenomena, with a symmetric bell-shaped curve.

• Formula:

  $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

  Where:

  • $\mu$: Mean.
  • $\sigma^2$: Variance.

• Example in Civil Engineering: Variations in compressive strength of concrete.

• Numerical Example:

  • Compressive strength ($X$) is normally distributed with $\mu = 40$ MPa and $\sigma = 5$ MPa.
  • Find $P(35 \leq X \leq 45)$:

    • Standardize $X$ to $Z$:

      $$Z = \frac{X - \mu}{\sigma}$$

      For $X=35$, $Z = \frac{35-40}{5} = -1$.
      For $X=45$, $Z = \frac{45-40}{5} = 1$.

    • Use standard normal tables:

      $$P(-1 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -1)$$ $$P(Z \leq 1) = 0.8413, \quad P(Z \leq -1) = 0.1587$$ $$P(-1 \leq Z \leq 1) = 0.8413 - 0.1587 = 0.6826$$

  Result: Probability is $68.26\%$.

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Applications in Civil Engineering

1. Bernoulli Distribution: Checking soil suitability in a binary pass/fail test.
2. Binomial Distribution: Predicting the number of defective items in a batch.
3. Geometric Distribution: Determining the number of tests to find a viable construction site.
4. Poisson Distribution: Estimating vehicle counts or defects over a time period.
5. Normal Distribution: Modeling variability in material properties like tensile strength or elasticity.

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Conclusion

Understanding probability distributions is critical in civil engineering for risk assessment, design, and decision-making under uncertainty. These distributions allow engineers to quantify variability and make informed decisions.

UNIT 1 Conditional Probability

 

Lecture Notes: Conditional Probability

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Definition

Conditional probability measures the likelihood of an event occurring given that another event has already occurred. It helps analyze dependencies between events.

Formula

If $A$ and $B$ are two events, and $P(B) > 0$, the conditional probability of $A$ given $B$ is:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

Where:

• $P(A|B)$: Probability of $A$ given $B$
• $P(A \cap B)$: Probability of both $A$ and $B$ occurring
• $P(B)$: Probability of $B$ occurring

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Solved Problems

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Problem 1

Question: In a class of 50 students, 30 study Mathematics, and 20 study both Mathematics and Physics. If a student studies Physics, what is the probability that they also study Mathematics?
Solution:

$$P(\text{Mathematics | Physics}) = \frac{P(\text{Mathematics} \cap \text{Physics})}{P(\text{Physics})}$$ $$P(\text{Physics}) = \frac{20}{50} = 0.4,\ P(\text{Mathematics} \cap \text{Physics}) = \frac{20}{50} = 0.4$$ $$P(\text{Mathematics | Physics}) = \frac{0.4}{0.4} = 1$$

Answer: $P(\text{Mathematics | Physics}) = 1$

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Problem 2

Question: A die is rolled. If the outcome is an even number, what is the probability that it is a 4?
Solution:

$$P(4 | \text{Even}) = \frac{P(4 \cap \text{Even})}{P(\text{Even})}$$

• $P(4 \cap \text{Even}) = P(4) = \frac{1}{6}$
• $P(\text{Even}) = \frac{3}{6} = \frac{1}{2}$

$$P(4 | \text{Even}) = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$

Answer: $P(4 | \text{Even}) = \frac{1}{3}$

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Problem 3

Question: A box contains 5 red, 3 green, and 2 blue balls. If a ball is randomly picked and found to be red, what is the probability that it is one of the 3 red balls numbered 1, 2, or 3?
Solution:

$$P(\text{Numbered Red | Red}) = \frac{P(\text{Numbered Red} \cap \text{Red})}{P(\text{Red})}$$

• $P(\text{Red}) = \frac{5}{10} = 0.5$
• $P(\text{Numbered Red} \cap \text{Red}) = \frac{3}{10} = 0.3$

$$P(\text{Numbered Red | Red}) = \frac{0.3}{0.5} = 0.6$$

Answer: $P(\text{Numbered Red | Red}) = 0.6$

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Problem 4

Question: A student is selected randomly. The probability that they study both Chemistry and Biology is 0.2, and the probability that they study Biology is 0.5. What is the probability that a student studies Chemistry, given that they study Biology?
Solution:

$$P(\text{Chemistry | Biology}) = \frac{P(\text{Chemistry} \cap \text{Biology})}{P(\text{Biology})}$$ $$P(\text{Chemistry | Biology}) = \frac{0.2}{0.5} = 0.4$$

Answer: $P(\text{Chemistry | Biology}) = 0.4$

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Problem 5

Question: In a company, 60% of employees are male, and 70% of males are engineers. What is the probability that a randomly chosen engineer is male?
Solution: Let $M$ = Male, $E$ = Engineer.

$$P(M|E) = \frac{P(M \cap E)}{P(E)}$$

• $P(M) = 0.6, P(E|M) = 0.7$

$$P(M \cap E) = P(M) \cdot P(E|M) = 0.6 \cdot 0.7 = 0.42$$

Assume $P(E) = 0.5$ (given context or additional data).

$$P(M|E) = \frac{0.42}{0.5} = 0.84$$

Answer: $P(M|E) = 0.84$

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Summary

1. Conditional probability measures the dependency of events.
2. Formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
3. Practice problems enhance understanding of real-life applications.

UNIT I Bayes Theorem

 

Bayes' Theorem for Civil Engineering Students with a Rural Background