Three Ways to Answer a Probability Question
A probability question about a sum, average, or sample proportion can often be approached in more than one way. The key decision is not simply whether the sample size is large. First identify what the question requests: an exact probability, a useful guarantee about concentration, or an approximation to a particular tail probability. Then check what is known about the random variables.
An exact distribution calculation uses the probability law of the quantity in question. Law-of-large-numbers reasoning describes concentration around a mean as the sample size grows; with an inequality such as Chebyshev’s inequality, it can also give a finite-sample bound. A central limit theorem (CLT) gives an approximation to a standardized sum or sample mean when its assumptions are suitable. These answers are not interchangeable: an approximation is not an exact probability, and a bound is not an estimate of the probability itself.
A Decision Framework
Start by naming the random quantity and the event. For example, “at least 12 claims” concerns a count, while “the average claim amount is within $50 of its mean” concerns a sample mean. The event determines which distribution or result is relevant.
Identify the random variables, their dependence or independence, and the probability event. Note whether the question calls for an exact value, an approximation, or a guarantee.
If the quantity has a known distribution with a practical calculation, use it for an exact answer. For example, a sum of independent Bernoulli variables with the same success probability has a binomial distribution.
If the question is whether a sample average or proportion is close to its expected value, use the LLN for the long-run conclusion. If a numerical finite-sample guarantee is needed, check whether a bound such as Chebyshev’s inequality supplies one.
If an exact calculation is unavailable or impractical and the question asks for a tail probability or threshold, check whether a CLT applies to the sum or mean. State the result as approximate, not exact.
The phrase “large sample” alone does not decide the method. A binomial count can have an exact binomial law even when the sample size is large; a CLT approximation may be convenient, but it is still an approximation. Conversely, knowing only a mean and variance may support an LLN-based bound even when it does not specify an exact distribution.
Exact Calculations When the Distribution Is Known
An exact calculation is especially useful when the model directly identifies the distribution of the quantity and the requested probability can be evaluated without undue difficulty. Consider independent Bernoulli trials, each with success probability \(p\). If \(X\) is the number of successes in \(n\) trials, then \(X\sim\operatorname{Binomial}(n,p)\), with \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). For a modest \(n\), a tail probability can be obtained by adding the relevant binomial probabilities.
Worked Example: An Exact Probability for a Small Portfolio
A hypothetical portfolio has 5 independent policies. During a year, each policy has probability \(0.2\) of producing a specified type of claim. Let \(X\) be the number of policies producing that claim. Find the exact probability that at least 2 policies produce it.
Method and model. Each policy has the same claim probability \(p=0.2\), and the policy outcomes are independent. Therefore \(X\sim\operatorname{Binomial}(5,0.2)\). The requested event is \(X\geq 2\).
It is shorter to use the complement:
The probability of no claims of this type is \(P(X=0)=\binom{5}{0}(0.2)^0(0.8)^5=0.32768\). The probability of exactly one is \(P(X=1)=\binom{5}{1}(0.2)(0.8)^4=5(0.2)(0.4096)=0.4096\). Thus \(P(X\geq 2)=1-0.32768-0.4096=0.26272\). To four decimal places, the exact probability is 0.2627.
Check. The complement probability is \(0.32768+0.4096=0.73728\), and \(1-0.73728=0.26272\), as required. The common mistake is to treat a normal approximation as necessary merely because the question concerns a count; here the binomial model gives a direct exact calculation.
LLN Reasoning and Finite-Sample Bounds
For independent, identically distributed observations \(X_1,\ldots,X_n\) with finite mean \(\mu\) and finite variance \(\sigma^2\), the sample mean is \(\bar X=(X_1+\cdots+X_n)/n\). Its expected value is \(\mu\), and its variance is \(\sigma^2/n\). The weak law of large numbers says that for every \(\varepsilon>0\), \(P(|\bar X-\mu|\geq\varepsilon)\) tends to zero as \(n\) grows.
A finite-sample guarantee can be obtained from Chebyshev’s inequality: for a random variable \(W\) with finite variance, \(P(|W-E[W]|\geq\varepsilon)\leq\operatorname{Var}(W)/\varepsilon^2\). Applying it to \(\bar X\) gives the bound below. This is a guarantee, not generally the exact probability or a close estimate.
The equality uses independence: when the observations are independent and identically distributed, \(\operatorname{Var}(\bar X)=\sigma^2/n\). If they are dependent, that variance formula need not hold, and this application cannot be made without further justification.
Worked Example: A Guaranteed Accuracy for a Claim Proportion
In a hypothetical group of 2,000 independent policies, each policy has probability \(0.4\) of producing a specified claim during the year. Let \(\hat p\) be the proportion of policies producing the claim. Give a guaranteed lower bound for the probability that \(\hat p\) is within \(0.04\) of \(0.4\).
Method and model. The claim indicators are independent Bernoulli variables with mean \(p=0.4\) and variance \(p(1-p)=0.4(0.6)=0.24\). The sample proportion has mean \(0.4\) and variance \(\operatorname{Var}(\hat p)=0.24/2000=0.00012\).
Apply Chebyshev’s inequality with \(\varepsilon=0.04\). Since \((0.04)^2=0.0016\),
Therefore \(P(|\hat p-0.4|<0.04)\geq 1-0.075=0.925\). The guaranteed lower bound is 0.9250, or 92.50%.
Check. The bound for being outside the stated range is \(0.075\); its complement is \(0.925\). The common mistake is to report \(0.075\) as the probability of being within \(0.04\). It bounds the probability of being at least \(0.04\) away, so the corresponding lower bound for being within the range is \(1-0.075\). This result does not claim that the true probability equals \(0.9250\).
When a CLT Approximation Is Appropriate
For independent, identically distributed observations with finite mean \(\mu\) and positive finite variance \(\sigma^2\), the CLT says that the distribution of \((\sum_{j=1}^n X_j-n\mu)/(\sigma\sqrt n)\) approaches the standard normal distribution as \(n\) grows. Equivalently, the standardized sample mean \(\sqrt n(\bar X-\mu)/\sigma\) approaches a standard normal distribution. The theorem motivates approximating certain probabilities; it does not make the original sum or mean exactly normal.
Before using this approximation, check that the observations have the dependence structure required by the result being used and that the mean and variance exist. Also consider whether the sample size and the shape of the individual distribution make a normal approximation credible for the event of interest. A far-tail probability or a highly skewed distribution may be less well approximated than a probability near the center.
Worked Example: A CLT Approximation for Total Daily Cost
A hypothetical service center records costs \(X_1,\ldots,X_{300}\) on 300 independent days. Assume each daily cost is uniformly distributed from 0 to 2 thousand dollars. Approximate the probability that the total cost exceeds 310 thousand dollars.
Method and model. A uniform distribution on \([0,2]\) has mean \(\mu=(0+2)/2=1\) thousand dollars and variance \(\sigma^2=(2-0)^2/12=1/3\) in squared thousands of dollars. Thus the total \(T=\sum_{j=1}^{300}X_j\) has mean \(300(1)=300\) thousand dollars and variance \(300(1/3)=100\), so its standard deviation is \(10\) thousand dollars.
The requested event is \(T>310\). Under the CLT, standardize the threshold using the total’s mean and standard deviation:
The normal approximation gives \(P(T>310)\approx P(Z>1)=1-\Phi(1)=1-0.8413=0.1587\), where \(Z\) is standard normal and \(\Phi\) is its cumulative distribution function. The estimated probability is 0.1587, to four decimal places.
Check. The threshold is one total standard deviation above the mean, so an upper-tail probability below \(0.5\) is sensible. The common mistake is to divide the threshold’s difference from the mean by the standard deviation of one day, \(\sqrt{1/3}\), rather than by the standard deviation \(10\) of the 300-day total. This is an approximation, not an exact probability.
Why the CLT Formula Has This Scale
The scale in the CLT comes from the mean and variance of a sum, not from a rule of thumb. If \(T=\sum_{j=1}^n X_j\) for independent, identically distributed variables with mean \(\mu\) and variance \(\sigma^2\), linearity of expectation gives \(E[T]=n\mu\). Independence makes the variance of the sum equal to the sum of the variances, so \(\operatorname{Var}(T)=n\sigma^2\). Its standard deviation is therefore \(\sigma\sqrt n\), not \(n\sigma\).
Consequently, the standardized sum subtracts \(n\mu\) and divides by \(\sigma\sqrt n\). For the sample mean \(\bar X=T/n\), the corresponding standard deviation is \(\sigma/\sqrt n\), because dividing a random variable by \(n\) divides its standard deviation by \(n\). This justifies the standard CLT scales:
The two expressions are equal because \(\bar X=T/n\). This derivation also explains why a CLT calculation must use the standard deviation of the quantity named in the question: a sum and a sample mean have different standard deviations.
Choosing the Method Without Overstating the Result
| Question’s aim | Useful method | What the answer means |
|---|---|---|
| A probability from a tractable, specified distribution | Exact distribution calculation | The probability under the stated model, without a normal approximation |
| A guarantee that a sample mean or proportion is close to its expectation | LLN reasoning; use a finite-sample inequality when needed | A convergence statement or a bound, not necessarily the exact probability |
| An approximate tail probability for a sum or mean when the exact calculation is impractical | CLT, if its assumptions and approximation quality are reasonable | An estimate based on a normal distribution |
A familiar model does not automatically make every calculation worthwhile: an exact sum of many probabilities can be cumbersome, and an approximation may be the practical choice if the question permits it. Conversely, a large sample does not erase an available exact distribution. Match the method to both the requested result and the quality of the assumptions.
Common Mistakes
- Calling a bound an exact probability. Chebyshev’s inequality supplies an upper bound on a deviation probability. Its bound may be conservative.
- Treating the LLN as a numerical answer. The LLN says deviations become unlikely as the sample size increases; by itself, it does not give the exact probability for a particular \(n\).
- Calling a CLT result exact. A normal approximation should be reported with language such as “approximately,” even when the calculation is straightforward.
- Ignoring the model. A binomial calculation requires independent trials with a common success probability. A CLT or variance bound also has conditions; check independence and the relevant finite moments rather than assuming them.
- Using “large \(n\)” as the only decision rule. First check for a tractable exact law and identify whether the question asks for a value, an approximation, or a guarantee.
Check Your Understanding
For each problem, choose or apply a method that matches the model and the requested result.
- A hypothetical sensor independently records an alarm on each of 6 trials with probability \(0.3\). Find the exact probability of at least one alarm. (Answer: \(1-0.7^6=0.8824\), exact.)
- For 100 independent Bernoulli observations with success probability \(0.5\), use Chebyshev’s inequality to give a lower bound for \(P(|\hat p-0.5|<0.1)\). (Answer: at least \(0.7500\).)
- Independent daily amounts are uniformly distributed on \([0,2]\) thousand dollars. For 100 days, use the CLT to approximate the probability that the total exceeds 110 thousand dollars. (Answer: \(z=(110-100)/\sqrt{100/3}=\sqrt 3\), so the approximate probability is \(0.0416\).)
- A hypothetical group has 1,600 independent policies, each with claim probability \(0.2\). Use a finite-sample concentration bound to give a lower bound for the probability that the claim proportion is within \(0.02\) of \(0.2\). (Answer: Chebyshev gives \(1-[0.2(0.8)/1600]/(0.02)^2=0.7500\) as a lower bound.)
- Suppose 25 independent observations are normally distributed with mean 50 and variance 100. Find the probability that their sample mean exceeds 54, and identify the method. (Answer: the sample mean is exactly normal with standard deviation \(10/\sqrt{25}=2\); \(P(\bar X>54)=P(Z>2)=0.0228\), an exact normal calculation rather than a CLT approximation.)