How Success Probability Changes Binomial Shape

What Changes When the Success Probability Changes?

A binomial model counts successes in a fixed number of independent trials, each with the same probability of success. If the number of trials stays fixed but the success probability rises, the count tends to move toward larger values. That shift affects the mean in a straightforward way, but the variance needs more care: it is largest when success and failure are equally likely, and it decreases if the probability moves farther away from one-half.

This distinction matters whenever a probability model is used to describe counts, including hypothetical insurance settings such as the number of policies in a group with a specified event during a year. A larger expected count does not automatically mean greater variability. The trial count must be held fixed when making this comparison; how changing the trial count affects shape is a separate question.

Definition: For a fixed positive integer $n$, let $X_p\sim\operatorname{Bin}(n,p)$ count successes in $n$ independent Bernoulli trials, each with success probability $p$, where $0\le p\le1$. Its support is $0,1,\ldots,n$. For $k=0,1,\ldots,n$, $$ P(X_p=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$ The mean and variance are $$ E[X_p]=np,\qquad \operatorname{Var}(X_p)=np(1-p). $$

The support remains the same as $p$ changes: the count can still be any integer from zero through $n$. What changes is the probability assigned to each possible count. Raising $p$ makes larger counts more likely overall, and the mean $np$ increases linearly. But the variance depends on both $p$ and $1-p$, so it first increases and then decreases as $p$ runs from zero to one.

How the Distribution Shifts

A precise way to compare the shift is to couple the trials. Imagine independent random variables $U_1,\ldots,U_n$, each uniformly distributed on $[0,1]$. For a given success probability $p$, call trial $i$ a success when $U_i\le p$. This gives a binomial count with parameters $n$ and $p$. If $p_1<p_2$, every value $U_i\le p_1$ also satisfies $U_i\le p_2$. Thus, under this construction, the count at $p_2$ is never below the count at $p_1$.

Theorem: With the number of trials $n$ fixed, if $p_1\le p_2$, then $X_{p_1}$ is stochastically no larger than $X_{p_2}$. In particular, for every integer threshold $k$, $$ P(X_{p_1}\ge k)\le P(X_{p_2}\ge k). $$ The expected count also increases from $np_1$ to $np_2$.

The coupling explains the direction of the shift; it does not say that every individual probability $P(X_p=k)$ must decrease as $p$ rises. For example, probabilities in the middle can rise and then fall as mass moves across the support. The reliable comparison is the one for cumulative upper-tail probabilities, not a claim about each separate point probability.

Why the Variance Has a Maximum at One-Half

The variance is $np(1-p)$. The factor $p(1-p)$ is a concave quadratic: completing the square makes its maximum clear.

p(1-p)=p-p^2=\frac14-\left(p-\frac12\right)^2\le\frac14.

Equality holds only when $p=\frac12$. Therefore the largest variance for fixed $n$ is $n/4$. The variance increases as $p$ moves from zero toward one-half and decreases as $p$ moves from one-half toward one. For two particular probabilities, the exact comparison is especially useful:

\operatorname{Var}(X_{p_2})-\operatorname{Var}(X_{p_1}) =n(p_2-p_1)(1-p_1-p_2).

If $p_2>p_1$, the sign depends on whether $p_1+p_2$ is below, equal to, or above one. A rise in $p$ therefore does not by itself determine whether variance rises. For instance, increasing $p$ from 0.2 to 0.4 increases variance, while increasing it from 0.6 to 0.8 decreases variance.

The binomial skewness gives a further description of shape when $0<p<1$:

\frac{1-2p}{\sqrt{np(1-p)}}.

Its sign is positive for $p<\frac12$, zero at $p=\frac12$, and negative for $p>\frac12$. This describes the direction of asymmetry: the distribution has a longer tail toward higher counts when $p$ is below one-half and toward lower counts when $p$ is above one-half. At one-half, the binomial probabilities are symmetric about $n/2$. Skewness describes asymmetry; it is not a measure of variance.

Worked Examples

Worked Example: Comparing the Full Pattern at Three Probabilities

A hypothetical system records whether each of five independent components passes a test. Each component has the same pass probability $p$. Let $X_p$ be the number that pass. Compare the distribution when $p=0.2$, $p=0.5$, and $p=0.8$, including the mean and variance.

Method. Use $X_p\sim\operatorname{Bin}(5,p)$, calculate the point probabilities from the binomial formula, and then use $E[X_p]=5p$ and $\operatorname{Var}(X_p)=5p(1-p)$. The table entries below are $P(X_p=k)$, in order for $k=0,1,2,3,4,5$.

$p$	Probabilities for $k=0,1,2,3,4,5$	Mean	Variance
0.2	0.32768, 0.40960, 0.20480, 0.05120, 0.00640, 0.00032	1	0.8
0.5	0.03125, 0.15625, 0.31250, 0.31250, 0.15625, 0.03125	2.5	1.25
0.8	0.00032, 0.00640, 0.05120, 0.20480, 0.40960, 0.32768	4	0.8

For example, when $p=0.2$, the probability of exactly two passes is \[ \binom52(0.2)^2(0.8)^3 =10(0.04)(0.512)=0.2048. \] The six probabilities in that row sum to $0.32768+0.40960+0.20480+0.05120+0.00640+0.00032=1$. The other two rows also sum to 1; the $p=0.8$ row reverses the $p=0.2$ row, and the $p=0.5$ row sums to $1$.

The means are $5(0.2)=1$, $5(0.5)=2.5$, and $5(0.8)=4$. The variances are $5(0.2)(0.8)=0.8$, $5(0.5)(0.5)=1.25$, and $5(0.8)(0.2)=0.8$. As $p$ rises, the distribution shifts from small counts toward large counts. The variance is largest at $p=0.5$, not at $p=0.8$.

Check. The endpoint probabilities agree with the model: $P(X_p=0)=(1-p)^5$ and $P(X_p=5)=p^5$. The table’s reversal between $p=0.2$ and $p=0.8$ is consistent with exchanging success and failure. Common mistake avoided: a higher mean does not imply a higher variance.

Worked Example: Mean Rises While Variance Falls

In a hypothetical quality review, eight independent items are inspected, and each item independently has probability $p$ of meeting a specified standard. Compare the count $X_p$ when $p=0.4$ and when $p=0.7$. State the change in mean and variance and interpret the skewness.

Method. Both counts have $n=8$; substitute each success probability into the binomial mean and variance formulas.

For $p=0.4$, the mean is $8(0.4)=3.2$, and the variance is $8(0.4)(0.6)=1.92$. For $p=0.7$, the mean is $8(0.7)=5.6$, and the variance is $8(0.7)(0.3)=1.68$. Thus the expected count increases by $5.6-3.2=2.4$, while the variance decreases by $1.68-1.92=-0.24$.

For the shape, the skewness at $p=0.4$ is \[ \frac{1-2(0.4)}{\sqrt{1.92}} =\frac{0.2}{1.3856\ldots} \approx 0.1443. \] At $p=0.7$, it is \[ \frac{1-2(0.7)}{\sqrt{1.68}} =\frac{-0.4}{1.2961\ldots} \approx -0.3086. \] The first distribution has positive skewness and the second negative skewness.

Check. Using the variance difference formula gives $8(0.7-0.4)(1-0.4-0.7)=8(0.3)(-0.1)=-0.24$, matching the direct calculation. Common mistake avoided: increasing $p$ raises the mean here, but it does not raise the variance.

Worked Example: Comparing an Upper-Tail Probability

A hypothetical test is run on six independent units, each with the same probability $p$ of triggering an alert. Let $X_p$ count the alerts. Compare the probability of at least five alerts when $p=0.25$ and when $p=0.75$, and compare the means and variances.

Method. Add the exact probabilities at $k=5$ and $k=6$, since “at least five” means $X_p=5$ or $X_p=6$.

For $p=0.25$, \[ \begin{aligned} P(X_{0.25}\ge5) &=\binom65(0.25)^5(0.75)+(0.25)^6\\ &=6\left(\frac1{1024}\right)\left(\frac34\right)+\frac1{4096}\\ &=\frac{18}{4096}+\frac1{4096} =\frac{19}{4096}\approx 0.0046. \end{aligned} \] For $p=0.75$, \[ \begin{aligned} P(X_{0.75}\ge5) &=\binom65(0.75)^5(0.25)+(0.75)^6\\ &=\frac{1458}{4096}+\frac{729}{4096} =\frac{2187}{4096}\approx 0.5349. \end{aligned} \] The means are $6(0.25)=1.5$ and $6(0.75)=4.5$. The variances are $6(0.25)(0.75)=1.125$ for both probabilities.

Check. Since $0.25<0.75$, the upper-tail probability must be no smaller at $p=0.75$, as the coupling argument predicts; the computed values satisfy this. Equal variances are consistent with the symmetry of $p(1-p)$ around one-half. Common mistake avoided: “at least five” includes both five and six; omitting the $k=6$ term understates each probability.

Common Exam Traps

Assuming variance always increases with $p$. The variance is $np(1-p)$, not $np$. The latter is the mean. For fixed $n$, variance peaks at $p=\frac12$.
Comparing variances without checking where the probabilities lie. When $p$ increases, the variance difference has the sign of $1-p_1-p_2$. A comparison that crosses one-half needs an actual calculation or this difference formula.
Claiming every point probability moves in the same direction. Increasing $p$ shifts the distribution to the right in the stochastic-order sense, but a particular $P(X_p=k)$ may increase or decrease.
Changing the trial count at the same time. To isolate the effect of $p$, hold $n$ fixed. If $n$ also changes, differences in mean and variance have more than one cause.
Confusing spread with skewness. Variance measures spread around the mean; the sign of skewness indicates the direction of asymmetry. Neither quantity alone describes every feature of the probability pattern.

Key takeaway: With fixed $n$, increasing $p$ moves the binomial count toward larger values and increases its mean $np$. Its variance $np(1-p)$ is largest at $p=\frac12$, so a higher success probability can accompany a lower variance.

Check Your Understanding

For each problem, hold the stated trial count fixed when comparing probabilities. Round probability answers to four decimal places where needed.

A hypothetical sample has $n=7$ independent trials. Find the mean and variance for $p=0.2$ and $p=0.8$. What changes, and what stays the same? (Answer: Means $1.4$ and $5.6$; both variances are $1.12$.)
For a binomial count with $n=12$, compare the mean and variance at $p=0.25$ and $p=0.5$. (Answer: Means $3$ and $6$; variances $2.25$ and $3$, respectively.)
For $X_p\sim\operatorname{Bin}(5,p)$, compare $P(X_p\ge4)$ at $p=0.2$ and $p=0.6$. (Answer: $0.0067$ at $p=0.2$ and $0.3370$ at $p=0.6$.)
With $n=9$, $p$ increases from $0.4$ to $0.7$. Calculate both means and variances, and decide whether the variance increases or decreases. (Answer: Means $3.6$ and $6.3$; variances $2.16$ and $1.89$, so the variance decreases by $0.27$.)
For $X_p\sim\operatorname{Bin}(4,p)$, compare $P(X_p\ge2)$ at $p=0.3$ and $p=0.6$. Show that the upper-tail probability is larger for the higher success probability. (Answer: $0.3483$ at $p=0.3$ and $0.8208$ at $p=0.6$.)

What You'll Learn