Binomial Distribution vs Probability Distribution

A binomial distribution counts successes across a fixed number of trials that are binary, independent, and identical in success probability. Write it $X \sim B(n, p)$: $n$ trials, success probability $p$ on each one, and $X$ the number of successes, which can be any whole number from 0 to $n$. The probability of exactly $k$ successes is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $\binom{n}{k}$ (n choose k) counts the orders in which those $k$ successes could fall. The mean is $\mu_X = np$ and the standard deviation is $\sigma_X = \sqrt{np(1-p)}$.

Flip a fair coin 10 times, so $n = 10$ and $p = 0.5$. Then $P(X = 6) = \binom{10}{6}(0.5)^6(0.5)^4 = \frac{210}{1024} = 0.2051$. The mean is $\mu_X = 10(0.5) = 5$ heads and the standard deviation is $\sigma_X = \sqrt{10(0.5)(0.5)} = \sqrt{2.5} \approx 1.581$ heads.

Now the error that costs the most marks: "so the probability of 6 or more heads is 0.2051." It is not. That number is the probability of exactly 6. At least 6 means 6, 7, 8, 9 or 10, and adding those five probabilities gives 0.3770, nearly double. Circle the words at least, at most, more than and fewer than before any arithmetic starts, because the formula answers only the exactly question and every other question is assembled out of it.

Independence is the condition that breaks most often in practice. Deal 10 cards from a deck without replacement and the chance of a heart shifts with every card removed, so the count of hearts is not binomial. It is close enough to treat as binomial when the sample is a small fraction of the population, which is what the 10 percent condition checks.

The binomial distribution is topic 2.10 of Unit 2, Probability, Random Variables, and Probability Distributions.

A probability distribution says how the total probability of 1 is divided among the values a random variable can take. For a discrete variable it is a table or a formula giving $P(X = x)$ for every possible $x$, and two conditions define it: each probability satisfies $0 \le P(X = x) \le 1$, and $\sum P(X = x) = 1$. For a continuous variable there is no table. Probability comes from area under a density curve, and the area under the whole curve is 1.

Let $X$ be the number of people in a car passing a checkpoint, with $P(1) = 0.35$, $P(2) = 0.30$, $P(3) = 0.20$, and the last probability left blank. The sum condition fills it in: $P(4) = 1 - (0.35 + 0.30 + 0.20) = 0.15$. That is not a trick, it is the definition doing the work, and it is also why a table whose entries add to 0.97 is not a probability distribution at all.

The misreading to watch is "I rolled the die 60 times and got twelve 3s, so $P(X = 3) = 0.20$." That is a relative frequency from one sample, not the distribution. A probability distribution is what the chance process does over the long run and it is fixed before any data exist; a sample wobbles around it and settles closer as the number of trials grows.

The discrete and continuous cases part company over a single value. For the car table, $P(X \le 2) = 0.65$ while $P(X < 2) = 0.35$, because $X = 2$ carries real probability of its own. For a continuous variable, $P(X = x) = 0$ at every single $x$, so $P(X \le 2)$ and $P(X < 2)$ are equal. Reading a strict inequality as if it were the other kind is a genuine error for counts and harmless for measurements.

Probability distributions enter the course at topic 2.8, Introduction to Random Variables and Probability Distributions.