Comparing Binomial Distributions
A binomial distribution describes the number of successes in a fixed number of trials under the binomial conditions: each trial has two outcomes, the trials are independent, the number of trials is fixed, and the probability of success is the same on every trial. Once those conditions are appropriate, the parameters \(n\) and \(p\) determine the distribution. Comparing distributions with different values of \(n\) or \(p\) helps answer practical questions: does the count tend to be larger, does it vary more, and which counts are most plausible?
A probability table lists the possible counts and their probabilities. A probability plot represents the same information with a bar for each count, with bar height equal to its probability. The distribution’s location is where its probability mass tends to lie; its spread describes how dispersed the possible counts are around that location. The mean and standard deviation give numerical summaries of location and spread, but the individual probabilities show how the distribution is shaped.
When comparing distributions, change one parameter at a time if you want to isolate its effect. To study \(n\), hold \(p\) fixed; to study \(p\), hold \(n\) fixed. If both change, the mean and spread can change for more than one reason, so the comparison needs to account for both.
Changing the Number of Trials
For fixed \(p\), increasing \(n\) raises the mean \(np\): more trials create more opportunities for success. The variance \(np(1-p)\) also rises in direct proportion to \(n\), while the standard deviation rises by the square root of that factor. Thus, measured in numbers of successes, the distribution generally spreads out as \(n\) grows. At the same time, its relative spread compared with its mean becomes smaller when \(p>0\), because \( \sigma/\mu=\sqrt{(1-p)/(np)} \).
The possible range also expands from 0 through \(n\). This does not mean that the probability of every particular count increases. As the distribution shifts, a count that was near the center of the smaller distribution may become unusual in the larger one. The whole set of probabilities, rather than a single probability in isolation, is what reveals the change.
Worked Example: Increasing the Trial Count at Fixed Probability
Suppose a hypothetical process produces a success with probability \(0.5\) on each independent trial. Compare the number of successes in \(n=4\) trials with the number in \(n=8\) trials. The binomial conditions are assumed to hold: each process has a fixed number of trials, the outcomes are success or failure, and the trials are independent with the same success probability.
For \(n=4\), \(X\sim\operatorname{Bin}(4,0.5)\). For example, the probability of exactly two successes is $$ P(X=2)=\binom{4}{2}(0.5)^2(0.5)^2 =6(0.0625)=0.375. $$ The mean is \(4(0.5)=2\), and the standard deviation is \( \sqrt{4(0.5)(0.5)}=1 \) success.
For \(n=8\), let \(Y\sim\operatorname{Bin}(8,0.5)\). Its probability of exactly four successes is $$ P(Y=4)=\binom{8}{4}(0.5)^4(0.5)^4 =70(1/256)=0.2734. $$ Its mean is \(8(0.5)=4\), and its standard deviation is \( \sqrt{8(0.5)(0.5)}=\sqrt{2}=1.414 \) successes.
The table shows the full probability patterns. For \(n=8\), probabilities not listed in the \(n=4\) table are included as well.
| Success count | \(P(X=k)\), \(n=4,\ p=0.5\) | \(P(Y=k)\), \(n=8,\ p=0.5\) |
|---|---|---|
| 0 | 0.0625 | 0.0039 |
| 1 | 0.2500 | 0.0313 |
| 2 | 0.3750 | 0.1094 |
| 3 | 0.2500 | 0.2188 |
| 4 | 0.0625 | 0.2734 |
| 5 | — | 0.2188 |
| 6 | — | 0.1094 |
| 7 | — | 0.0313 |
| 8 | — | 0.0039 |
The distribution’s center moves from 2 to 4 successes, and its standard deviation increases from 1 to about 1.414 successes. The \(n=8\) distribution covers more possible counts, but its probabilities still cluster around its own mean. In particular, the probability at the old mean count, 2, is lower for \(n=8\) than for \(n=4\); the distribution has shifted rather than simply gaining probability at every count.
Changing the Success Probability
For fixed \(n\), increasing \(p\) raises the mean \(np\), so the distribution shifts toward larger success counts. The effect on spread is not simply “larger \(p\) means more spread.” The variance is \(np(1-p)\), which depends on both \(p\) and its complement \(1-p\). For fixed \(n\), this variance is greatest at \(p=0.5\). As \(p\) moves away from one half in either direction, the variance decreases. When \(p\) is near 0, most probability lies near 0 successes; when it is near 1, most lies near \(n\) successes.
The binomial formula also gives a useful symmetry. If \(X\sim\operatorname{Bin}(n,p)\), then the number of failures is \(n-X\). Each failure has probability \(1-p\), so \(n-X\) has a binomial distribution with parameters \(n\) and \(1-p\). Equivalently, the probability at count \(k\) when the success probability is \(p\) matches the probability at count \(n-k\) when it is \(1-p\):
The equality uses \(\binom{n}{k}=\binom{n}{n-k}\). This means the \(p\) and \(1-p\) distributions are mirror images across the midpoint \(n/2\): one puts high probabilities at low counts where the other puts high probabilities at corresponding high counts. They have the same standard deviation, since \(np(1-p)=n(1-p)p\).
Worked Example: Changing \(p\) With the Trial Count Fixed
A hypothetical set of 10 independent trials records success with probability \(p\). Compare \(p=0.2\) with \(p=0.5\). The trial count is fixed, and in each case the probability of success is constant across trials, so the binomial model applies.
At \(p=0.2\), the mean is \(10(0.2)=2\) successes and the standard deviation is \( \sqrt{10(0.2)(0.8)}=\sqrt{1.6}=1.265 \) successes. For example, $$ P(X=2)=\binom{10}{2}(0.2)^2(0.8)^8 =45(0.04)(0.16777216)=0.3020. $$
At \(p=0.5\), the mean is \(10(0.5)=5\) successes and the standard deviation is \( \sqrt{10(0.5)(0.5)}=\sqrt{2.5}=1.581 \) successes. For exactly two successes, $$ P(X=2)=\binom{10}{2}(0.5)^2(0.5)^8 =45(1/1024)=0.04395. $$
| Success count \(k\) | \(P(X=k)\), \(n=10,\ p=0.2\) | \(P(X=k)\), \(n=10,\ p=0.5\) |
|---|---|---|
| 0 | 0.1074 | 0.0010 |
| 2 | 0.3020 | 0.0439 |
| 5 | 0.0264 | 0.2461 |
| 8 | 0.0001 | 0.0439 |
| 10 | 0.0000* | 0.0010 |
The asterisked probability is \(0.2^{10}=0.0000001024\), which rounds to 0.0000 at four decimal places but is not zero. The table shows the shift toward higher counts when \(p\) rises from 0.2 to 0.5. The standard deviation also rises, because the probability moves toward the value that maximizes \(p(1-p)\). A higher success probability does not make every count more likely: the probability of exactly two successes falls from 0.3020 to 0.04395.
A Probability Can Rise or Fall as Parameters Change
It is tempting to expect every probability to move in the same direction as the mean, but an individual event may behave differently. A helpful example is the probability of at least one success. With fixed \(p\), this probability rises as \(n\) increases because there are more opportunities for a success. The complement rule gives a direct calculation: “at least one” is the complement of “zero successes.”
Worked Example: More Trials and the Chance of at Least One Success
Suppose a hypothetical trial has success probability \(0.1\), independently from trial to trial. Compare five trials with ten trials. The same success probability and independence assumption hold in both cases; only the fixed trial count changes.
With five trials, the probability of no successes is \(0.9^5=0.59049\), so $$ P(X\geq 1)=1-P(X=0)=1-0.9^5=0.40951. $$ With ten trials, the corresponding probability is $$ P(Y\geq 1)=1-P(Y=0)=1-0.9^{10} =1-0.34867844=0.65132156. $$
Thus the chance of at least one success rises from about 0.4095 to 0.6513 when the number of independent trials doubles. The mean count rises from \(5(0.1)=0.5\) to \(10(0.1)=1\) success, but the event probability is not itself the mean: it is a probability computed from the distribution.
How to Compare Distributions Clearly
A careful comparison separates three questions: where the distribution is centered, how much it spreads in the units of the count, and how its probability mass is shaped. A table or probability plot is useful because two distributions with similar means or standard deviations can still assign different probabilities to a particular event.
State which parameter changes and which is held fixed. Verify that each set of trials still satisfies the binomial conditions.
Calculate \(np\) and \(\sqrt{np(1-p)}\) for each distribution, reporting the mean and standard deviation in units of successes.
Use the probability mass function, a table, or a probability plot to compare the relevant counts or event. Do not infer a particular event probability from the mean alone.
For example, when \(n\) is fixed and \(p\) changes from 0.2 to 0.8, the symmetry result says the distributions are mirror images, not identical at each count. Their means are \(0.2n\) and \(0.8n\), while their standard deviations are equal. This is a useful check when reading a table or plot.
Common Mistakes
- Assuming a larger \(n\) means every count is more likely. Increasing \(n\) changes the range and shifts the distribution. Individual probabilities can fall, as the probability of exactly two successes did when \(n\) rose from 4 to 8 at \(p=0.5\).
- Assuming spread always increases as \(p\) increases. For fixed \(n\), spread is greatest at \(p=0.5\). It decreases as \(p\) moves away from one half toward either endpoint.
- Confusing standard deviation with a range of possible values. The standard deviation describes typical spread around the mean; it is not a boundary. A binomial variable can take any integer count from 0 through \(n\), even if some such counts have very small probability.
- Comparing only means. Means describe centers, not the probabilities of all events. Use the binomial formula or a probability table when the question concerns a particular count or tail.
Check Your Understanding
Use the binomial formulas and the comparison ideas above to answer these questions.
- For a fixed success probability \(p=0.3\), what happens to the mean and standard deviation when the trial count increases from 20 to 40?
- With \(n=12\), which value of \(p\) gives the largest standard deviation: 0.2, 0.5, or 0.8? Explain.
- For \(n=8\), explain how the distribution with \(p=0.7\) relates to the distribution with \(p=0.3\).
- Why can the probability of exactly two successes decrease even when the mean number of successes increases?
- For 6 independent trials with success probability 0.15, write an expression for the probability of at least one success. What binomial condition is needed for using it?