Probability Distributions and Reliability

Where the earlier statistics lesson described data you already had, this one predicts data you do not have yet. A probability distribution is a rule that assigns a probability to every possible outcome, and on the CELE the board leans on a short list of them: the binomial and Poisson laws for counts, the normal curve for measurements, and the reliability functions that turn a failure rate into the chance a component survives its mission. Each is formula-driven, so the trick is matching the situation to the right law and then carrying the arithmetic all the way to a clean number. This lesson does exactly that, worked example by worked example.

Discrete versus continuous distributions

The first fork is whether the quantity you are modeling is counted or measured. A discrete random variable takes separated values you can list, usually whole-number counts such as the number of defective bolts in a batch or the number of accidents in a month. Its probabilities live in a probability mass function (PMF), and every allowed value carries a lump of probability; the lumps sum to 1. A continuous random variable can take any value in a range, such as a concrete strength of 27.4 MPa or a beam deflection of 3.15 mm. Its probabilities live under a probability density curve, and probability is the area under that curve between two limits, so the chance of any single exact value is 0.

Distribution	Type	Use when	Mean	Variance
Binomial	discrete	fixed n independent trials, constant success probability p	n p	n p (1 - p)
Poisson	discrete	rare events counted over a fixed interval, average rate lambda	lambda	lambda
Normal	continuous	symmetric bell-shaped measurements	mu	sigma^2

Keep this table in your head. Half of solving a board item is naming the distribution before you reach for a formula, and the mean and variance columns let you sanity-check any answer in seconds.

The binomial distribution

Use the binomial law when you repeat the same trial a fixed number of times, each trial ends in one of two outcomes (call them success and failure), the probability of success p stays constant, and the trials are independent. The probability of exactly k successes in n trials is:

P(X = k) = C(n, k) p^k (1 - p)^(n - k)

where C(n, k) = n! / [k!(n - k)!] counts the ways to place the k successes among the n trials. The mean and variance are worth memorizing because they appear as their own questions: mean = n p and variance = n p (1 - p), so the standard deviation is sqrt(n p (1 - p)).

Worked example: A production line turns out bolts with a long-run defect rate of p = 0.05 (5 percent). In a random sample of n = 12 bolts, what is the probability that exactly 2 are defective? Here k = 2, so C(12, 2) = (12 x 11) / (2 x 1) = 66. Then p^2 = (0.05)^2 = 0.0025 and (1 - p)^(n - k) = (0.95)^10 = 0.5987. Multiply: P(X = 2) = 66 x 0.0025 x 0.5987 = 0.165 x 0.5987 = 0.0988. So the chance of exactly 2 defective bolts is about 0.0988, or 9.88 percent. As a check, the expected number of defects in the sample is mean = n p = 12(0.05) = 0.6 defective bolts, and the standard deviation is sqrt(12 x 0.05 x 0.95) = sqrt(0.57) = 0.755 defective bolts.

The Poisson distribution

The Poisson law counts how many times a rare event happens over a fixed interval of time, length, or area when the events occur independently at a steady average rate. That rate is the only parameter, lambda, and it equals the expected count over the interval. The probability of exactly k events is:

P(X = k) = e^(-lambda) lambda^k / k!

A signature of the Poisson distribution is that its mean and its variance are both equal to lambda, so its standard deviation is sqrt(lambda). If a problem tells you the average count over an interval and asks for the chance of a specific count, reach for Poisson.

Worked example: A stretch of provincial highway averages lambda = 3 accidents per month. What is the probability of exactly 5 accidents in a given month? Substitute k = 5: P(X = 5) = e^(-3) (3^5) / 5!. Now e^(-3) = 0.04979, 3^5 = 243, and 5! = 120, so P(X = 5) = 0.04979 x (243 / 120) = 0.04979 x 2.025 = 0.1008. The probability of exactly 5 accidents is about 0.1008, or 10.08 percent per month. The standard deviation of the monthly count is sqrt(lambda) = sqrt(3) = 1.73 accidents.

The normal distribution and z-scores

The normal (Gaussian) distribution is the bell-shaped curve that describes most continuous measurements in engineering: material strengths, survey errors, dimensions off a machine. It is symmetric about its mean mu, and its spread is set by its standard deviation sigma, so the mean, median, and mode all coincide at the peak. Because every normal curve has the same shape, you convert any value x to a standard normal score that counts how many standard deviations it sits from the mean:

z = (x - mu) / sigma

Once you have z, the probability of landing below x is the area to the left of z under the standard normal curve, written Phi(z) = P(Z <= z). You read Phi(z) from a z-table. A few anchor values cover most board items, and by symmetry Phi(-z) = 1 - Phi(z).

z	Phi(z) = P(Z <= z)
0.0	0.5000
0.5	0.6915
1.0	0.8413
1.5	0.9332
2.0	0.9772
2.5	0.9938

To find the area to the right of z, use 1 - Phi(z). To find the area between two values, subtract the two cumulative areas, Phi(z2) - Phi(z1).

Worked example: Rebar yield strength is normally distributed with mean mu = 420 MPa and standard deviation sigma = 20 MPa. A bar is rejected if its yield strength falls below 390 MPa. What fraction of bars is rejected? First standardize the cutoff: z = (390 - 420) / 20 = -30 / 20 = -1.5. The rejected fraction is the area to the left of z = -1.5, which is P(Z < -1.5) = 1 - Phi(1.5) = 1 - 0.9332 = 0.0668. So about 0.0668, or 6.68 percent of the bars are rejected.

Expected value and variance

For a discrete random variable, the expected value E[X] is the long-run average outcome, found by weighting each value by its probability:

E[X] = sum of (x times P(x))

The variance measures spread around that mean, Var(X) = sum of ((x - E[X])^2 times P(x)), and the standard deviation is its square root. Expected value is the engine behind decision problems: the option with the higher expected payoff is the better long-run bet.

Worked example: A contractor bidding on a project will earn a profit of PHP 200,000 with probability 0.6 if the bid wins on schedule, but will absorb a loss of PHP 50,000 with probability 0.4 if delays trigger penalties. The expected profit is E[X] = (200,000)(0.6) + (-50,000)(0.4) = 120,000 - 20,000 = PHP 100,000. Because the expected value is positive, the bid is worth taking on the numbers alone.

Reliability as a probability of survival

Reliability engineering is applied probability. The reliability R(t) of a component is the probability that it performs its intended function without failure for a stated period t under stated conditions. Its complement is the probability of failure, F(t) = 1 - R(t). For the very common case of a constant failure rate lambda (failures per unit time), the reliability follows the exponential model:

R(t) = e^(-lambda t)

The reciprocal of the failure rate is the mean time between failures, MTBF = 1 / lambda, the average life you can expect from the part. A larger lambda means a shorter life and a faster drop in reliability.

Worked example: A pump has a constant failure rate lambda = 0.02 failures per year. What is its reliability (probability of survival) over a 5-year mission, and what is its MTBF? The reliability is R(5) = e^(-(0.02)(5)) = e^(-0.10) = 0.9048, so there is about a 90.48 percent chance the pump survives 5 years. Its mean time between failures is MTBF = 1 / 0.02 = 50 years.

Series and parallel system reliability

Real systems are built from many components, and how they are wired decides the system reliability. In a series system every component must work for the system to work, so the reliabilities multiply:

R_series = R1 x R2 x ... x Rn

A series chain is always less reliable than its weakest link, because multiplying probabilities below 1 only drives the product down. In a parallel (redundant) system the whole works as long as at least one path works, so you multiply the failure probabilities and subtract from 1:

R_parallel = 1 - (1 - R1)(1 - R2) ... (1 - Rn)

Redundancy always raises reliability above the best single unit. Many boards combine the two: a parallel block sits inside a series chain, so you collapse the parallel block to a single equivalent reliability first, then multiply it down the series.

Worked example: For the diagram above, component A has R = 0.95 and sits in series with a parallel block of two redundant units B and C, each with R = 0.80. Collapse the parallel block first: R_block = 1 - (1 - 0.80)(1 - 0.80) = 1 - (0.20)(0.20) = 1 - 0.04 = 0.96. Now the block is in series with A, so R_system = R_A x R_block = 0.95 x 0.96 = 0.912. The overall system reliability is 0.912, or 91.2 percent. Notice the redundant block (0.96) is more reliable than a single unit B alone (0.80), which is exactly why designers pay for redundancy.

Exam-day strategy

• Name the distribution before you compute: fixed number of trials with a constant success probability means binomial, an average count of rare events over an interval means Poisson, and a bell-shaped measurement means normal.
• For the binomial, lock mean = n p and variance = n p (1 - p); the standard deviation is the square root of the variance, so an answer that skips the square root is the planted trap.
• For the Poisson, remember the mean and variance are both lambda, so the standard deviation is sqrt(lambda); if a choice sets the variance equal to lambda^2, it is wrong.
• Standardize with z = (x - mu) / sigma, then decide left tail Phi(z), right tail 1 - Phi(z), or between Phi(z2) - Phi(z1); use Phi(-z) = 1 - Phi(z) so one z-table row covers both signs.
• Keep the wiring straight: series reliabilities multiply and always drop below the weakest link, while parallel uses R = 1 - product of (1 - Ri) and always rises above the best single unit.
• On a mixed system, collapse each parallel block to one equivalent reliability first, then multiply the equivalents down the series chain; never multiply parallel reliabilities directly.