Why Unequal Bernoulli Trials Need Their Own Calculation
Many counts are built from independent yes-or-no outcomes whose success probabilities are not all the same. For example, a hypothetical portfolio might contain policies with different probabilities of a claim, or a quality-control program might test components with different defect rates. The total number of successes is still a sum of indicator variables, but it is not generally binomial: the binomial model requires both independence and a common success probability.
A normal approximation can still be useful. The essential calculation is to add the individual means and variances, then use those totals to standardize a continuity-corrected count event. The approximation is aligned with the SOA Exam P (Probability) syllabus.
Building the Normal Approximation
The variance formula depends on independence. Each indicator has \(E[I_i]=p_i\) and \(\operatorname{Var}(I_i)=p_i(1-p_i)\). The expectation of a sum is the sum of expectations whether or not the terms are independent; for variance, independence ensures the covariances are zero. Thus the exact mean and variance of \(S\) are \(\mu\) and \(V\) above.
When a normal approximation is reasonable, model \(S\) approximately by a normal random variable \(Y\sim N(\mu,V)\). Here \(V\) is the variance, so the standard deviation is \(\sigma=\sqrt{V}\). The normal distribution is continuous whereas \(S\) takes integer values. A continuity correction places the boundary halfway between adjacent integers before standardizing.
The corresponding standardized value for a real endpoint \(c\) is \(z=(c-\mu)/\sqrt V\). If \(\Phi(z)\) denotes the standard normal cumulative distribution function, a corrected lower-tail approximation is \(\Phi(z)\); an upper-tail approximation is \(1-\Phi(z)\). Integer interval endpoints must also be corrected: \(P(a\le S\le b)\) is approximated using the continuous interval from \(a-0.5\) to \(b+0.5\).
Why the Summed-Variance Normal Model Can Work
The normal approximation is justified as a limit when the total variance grows. Center each indicator by writing \(I_i-p_i\). These centered variables are independent, have mean zero, and their variances sum to \(V\). A short Lyapunov-condition argument explains the role of \(V\): since \(|I_i-p_i|\le 1\), its absolute cube is no greater than its square. Therefore,
As \(V\) grows, this upper bound tends to zero. The Lyapunov central limit theorem then implies that \((S-\mu)/\sqrt V\) approaches a standard normal distribution under such a sequence of independent-trial models. This explains why the summed variance, rather than the number of trials alone, is central to the approximation. It does not guarantee high accuracy for every finite problem: a small variance, substantial skewness, or an event far into a tail can still make the estimate poor. There is no universal cutoff that makes every finite approximation reliable.
Worked Example: An Upper Tail Across Three Risk Groups
A hypothetical screening program checks 40 independent items. Ten have success probability \(0.20\), fifteen have probability \(0.50\), and fifteen have probability \(0.80\). Approximate the probability that at least 25 items succeed. Use a normal approximation and give the probability to four decimals.
Method. The unequal probabilities rule out a single binomial model. Sum the individual means and variances by group. The screening outcomes are assumed independent, as stated.
Mean. The three groups contribute \(10(0.20)=2.00\), \(15(0.50)=7.50\), and \(15(0.80)=12.00\). Hence \(\mu=2.00+7.50+12.00=21.50\).
Variance. Their contributions are \(10(0.20)(0.80)=1.60\), \(15(0.50)(0.50)=3.75\), and \(15(0.80)(0.20)=2.40\). Thus \(V=1.60+3.75+2.40=7.75\), and \(\sigma=\sqrt{7.75}=2.7839\).
The event “at least 25” is \(S\ge25\). Its continuity-corrected boundary is \(24.5\), so \(z=(24.5-21.5)/2.7839=1.0776\). Therefore, \(P(S\ge25)\approx1-\Phi(1.0776)=1-0.8594=0.1406\).
Check and trap avoided. The result is an upper-tail probability and is below one-half because the threshold is above the mean. Using \(25.5\) instead of \(24.5\) would shift the boundary in the wrong direction and understate the upper tail.
Worked Example: A Corrected Interval Probability
A hypothetical audit records whether each of 50 independent transactions meets a standard. Twenty transactions have success probability \(0.10\), fifteen have probability \(0.40\), and fifteen have probability \(0.70\). Approximate the probability that between 16 and 21 transactions, inclusive, meet the standard. Give the answer to four decimals.
Method. Calculate the summed mean and variance, then convert the inclusive integer interval to a continuous interval using half-unit corrections.
The mean is \(20(0.10)+15(0.40)+15(0.70)=2.00+6.00+10.50=18.50\). The variance is \(20(0.10)(0.90)+15(0.40)(0.60)+15(0.70)(0.30)=1.80+3.60+3.15=8.55\). Thus \(\sigma=\sqrt{8.55}=2.9240\).
The inclusive event \(16\le S\le21\) becomes \(15.5<Y<21.5\). The standardized endpoints are \(z_L=(15.5-18.5)/2.9240=-1.0260\) and \(z_U=(21.5-18.5)/2.9240=1.0260\). Consequently,
Check and trap avoided. The interval is centered on the mean, so its corrected standardized endpoints are symmetric. Omitting the half-unit corrections would instead use 16 and 21 as the continuous boundaries and would calculate a different probability.
Worked Example: A Strict Count Threshold
A hypothetical reliability study combines 60 independent component tests: 25 have success probability \(0.20\), 20 have probability \(0.50\), and 15 have probability \(0.80\). Approximate the probability that more than 30 tests succeed. Give the answer to four decimals.
Method. First translate the strict inequality into an integer event. “More than 30” means \(S\ge31\), not \(S\ge30\).
The mean is \(25(0.20)+20(0.50)+15(0.80)=5.00+10.00+12.00=27.00\). The variance is \(25(0.20)(0.80)+20(0.50)(0.50)+15(0.80)(0.20)=4.00+5.00+2.40=11.40\), giving \(\sigma=\sqrt{11.40}=3.3764\).
For \(S\ge31\), the corrected boundary is \(30.5\). Therefore \(z=(30.5-27.0)/3.3764=1.0366\), and \(P(S>30)=P(S\ge31)\approx1-\Phi(1.0366)=1-0.8500=0.1500\).
Check and trap avoided. The estimated probability is less than one-half because the cutoff exceeds the mean. Treating “more than 30” as “at least 30” would incorrectly include the count 30 and use a corrected boundary of 29.5.
Choosing and Interpreting the Approximation
The calculation requires the indicators to be independent, but it does not require identical success probabilities. The individual \(p_i\) values must be used in both sums. Replacing them with their average and then using a binomial variance is not generally valid: a binomial variance would be \(n\bar p(1-\bar p)\), whereas the appropriate variance here is \(\sum_i p_i(1-p_i)\). These expressions need not agree.
Before using the normal model, check that the total variance is large enough to support a useful approximation and consider whether the event is close to a boundary or far into a tail. The earlier material on skewness and boundary effects is also relevant: highly uneven probabilities can produce asymmetry, and the normal model may represent extreme tails poorly. When the variance is small or the approximation is questionable, the normal answer should not be mistaken for an exact probability.
A reliable workflow is to identify the independent indicators, calculate \(\mu\) and \(V\) from the individual probabilities, translate the integer event, apply its continuity correction, and standardize using \(\sqrt V\). Keeping these steps separate helps prevent the two common errors: treating unequal trials as binomial and correcting the wrong endpoint.
Write \(S=\sum I_i\), record each success probability \(p_i\), and verify independence.
Find \(\mu=\sum p_i\), \(V=\sum p_i(1-p_i)\), and \(\sigma=\sqrt V\).
Translate strict or inclusive integer language carefully, then move each relevant boundary by one-half.
Use the normal CDF with the corrected endpoint or endpoints, and interpret the result as an approximation.
Check Your Understanding
For each problem, assume the trials are independent. Use continuity-corrected normal approximations and give probabilities to four decimals.
- A device test contains six trials with success probability \(0.25\) and six with probability \(0.75\). Approximate the probability of at least 8 successes. (Answer: 0.1587)
- A review covers 40 independent files: 10 have success probability \(0.20\), 20 have probability \(0.40\), and 10 have probability \(0.60\). Approximate the probability of between 14 and 18 successes, inclusive. (Answer: 0.6006)
- A survey combines 50 independent responses: 20 have success probability \(0.10\), 20 have probability \(0.50\), and 10 have probability \(0.80\). Approximate the probability of at most 17 successes. (Answer: 0.1942)
- A hypothetical test program has 80 independent trials: 20 have success probability \(0.15\), 30 have probability \(0.45\), and 30 have probability \(0.75\). Approximate the probability of between 35 and 44 successes, inclusive. (Answer: 0.7908)
- A production check includes 100 independent items: 40 have success probability \(0.20\), 30 have probability \(0.50\), and 30 have probability \(0.80\). Approximate the probability of at least 53 successes. (Answer: 0.1017)