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Probability Distributions

A probability distribution assigns probabilities to possible outcomes. The expected value is the long-run weighted average.

$$E(X)=\sum xp(x), \qquad Var(X)=E(X^2)-[E(X)]^2$$

Expected Value and Variance

Let $X=0,1,2$ with probabilities $0.20,0.50,0.30$. Find $E(X)$ and $Var(X)$.

$$E(X)=0(0.20)+1(0.50)+2(0.30)=1.10$$
$$E(X^2)=0^2(0.20)+1^2(0.50)+2^2(0.30)=1.70$$
$$Var(X)=1.70-(1.10)^2=0.49$$

Final answer: $E(X)=1.10$ and $Var(X)=0.49$.

Problem: Binomial Exact Count

A concrete cylinder has an 80% chance of passing a strength check. If 5 cylinders are tested, find the probability that exactly 4 pass.

$$P(X=4)=\binom{5}{4}(0.80)^4(0.20)=0.4096$$

Answer: The probability is 0.4096.

Problem: Poisson Arrival Count

An inspection desk receives an average of 3 requests per hour. Find the probability of exactly 2 requests in one hour.

$$P(X=2)=e^{-3}\frac{3^2}{2!}=0.2240$$

Answer: The probability is about 0.224.

Problem: Normal Distribution z-Score

Survey errors are normally distributed with mean 0 mm and standard deviation 5 mm. Find the z-score for an error of 8 mm.

$$z=\frac{x-\mu}{\sigma}=\frac{8-0}{5}=1.60$$

Answer: The z-score is 1.60.

Problem: Expected Value of a Discrete Variable

A repair cost is P1,000 with probability 0.30 and P4,000 with probability 0.70. Find the expected repair cost.

$$E(X)=1000(0.30)+4000(0.70)=3100$$

Answer: The expected repair cost is P3,100.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w238

MSTE - Statistics and Probability / Probability / MSTE April 2025

The time taken by a toll taker to collect the toll from the vehicles crossing a bridge is an exponential distribution with a mean of 23 seconds. The probability that a random vehicle will be processed in 25 seconds or more is:

  1. 0.225
  2. 0.552
  3. 0.337
  4. 0.773
For an exponential distribution with density $f(x) = \lambda e^{-\lambda x}$, the right-tail probability is
$P(x \ge a) = e^{-\lambda a}$

The rate is the reciprocal of the mean:
$\lambda = \dfrac{1}{23}$
$P(x \ge 25) = e^{-(1/23)(25)}$
$\boxed{P = 0.337}$

Question Bank: w277

MSTE - Statistics and Probability / Probability / MSTE April 2025

The Design Basis Earthquake (DBE) has a return period of 475 years with 10% probability of exceedance in how many years?

  1. 100 years
  2. 85 years
  3. 72 years
  4. 50 years
Use the exponential probability distribution, where the rate is the reciprocal of the return period.

The probability of no exceedance in $a$ years is $P = e^{-\lambda a}$, so the probability of exceedance is $1 - e^{-\lambda a}$:
$0.10 = 1 - e^{-(1/475)a}$
$e^{-a/475} = 0.90$
$\boxed{a = 50.046 \approx 50\ \text{years}}$