Random variables and distributions

For a discrete random variable, a probability mass function assigns probability directly to each possible value. The masses are nonnegative and sum to one. A Bernoulli variable is one for success and zero for failure. If we count successes in a fixed number of independent Bernoulli trials with the same success probability, the count has a binomial distribution. Changing these assumptions can change the distribution even if the average success rate stays the same.

The binomial coefficient counts which trial positions contain the successes. Factorial notation means $n!=n(n-1)\cdots1$, with $0!=1$. For four trials and two successes, the coefficient is $4!/(2!2!)=6$. At $p=0.25$, the two-success probability is therefore $6(0.25)^2(0.75)^2=0.2109375$. The factor six groups all successful-position arrangements; omit it when the question specifies just one exact ordered sequence.

A continuous density assigns probability through area over an interval, not through its height at a single point. Its integral over the full line is one, but its height can exceed one without violating probability rules. For example, a uniform density over an interval of length one half has height two. The units of density are inverse units of the measured quantity, so multiplying by an interval width produces a dimensionless probability.

For an absolutely continuous variable, an individual exact value has probability zero even though some value is observed. Practical measurements have finite resolution, so intervals are usually the operational events. A normal density is useful for many additive-noise models, but it is not a universal law of data. Heavy tails, bounded quantities, multimodality, and dependent observations all require attention before a familiar bell curve is selected.

Expectation is a probability-weighted average, not necessarily a likely or even possible outcome. A binomial count can have a noninteger mean despite taking only integer values. Variance measures the expected squared deviation from that mean; standard deviation returns to the original units. For independent Bernoulli trials the binomial mean is $np$ and variance is $np(1-p)$. Those compact formulas encode the independence assumption.

Expectation is linear even when variables are dependent. Variance needs covariance terms unless dependence vanishes in the relevant way: strongly correlated failures can produce much larger count variability than the binomial model predicts. In infrastructure engineering, the tail probability of a bad burst can matter more than average failures. In machine learning, an average loss similarly cannot reveal every subgroup error. Read summaries as answers to specific questions rather than complete portraits.