NuclearNotes · Chapter 04 — Statistical Methods in Analytical Chemistry

1 Chapter overview

Every number you will ever measure in nuclear chemistry comes from counting — counts of disintegrations, counts of particles reaching a detector, counts in a minute. And counting is a statistical act. Count the same source twice for five minutes each and you will get two different numbers. That scatter is not sloppy technique; it is the physics of radioactive decay itself.

Each nucleus in your sample decays independently, with only a tiny probability of decaying in any given second. With millions of nuclei, the total number that happen to decay while you watch fluctuates randomly about a mean. This chapter gives you the mathematics of that fluctuation — the binomial and Poisson distributions — and shows how to extract the two things every chemist needs from a counting experiment: the best value and its uncertainty.

§2 Concepts — why decay is random, the probability rules you need, and what each distribution describes.
§3 Derivations — the binomial law from coin-toss logic, $\bar{r} = np$ , $\sigma^{2} = npq$ , the binomial → Poisson limit, $\sigma = \sqrt{N}$ for counting data, background-subtraction error propagation, and the optimal split of counting time.
§4 Examples — full numericals: rate ± error, counts needed for 1% precision, a χ² check of counter behaviour, background subtraction, optimal time division.
§6 PYQ bank — all 6 real questions (2022 and 2024), each solved.

📋 Exam record

This unit has appeared only in the 2022 and 2024 MSCH-102 papers, and only on three themes: the binomial distribution (derivation, asked twice), variance in terms of p (asked twice) and Poisson from binomial with its applicability conditions. Standard deviation of counting data and optimisation of counting experiments are in the syllabus but have never been asked — they are covered here in full from the reference books, marked with a "Syllabus, never asked" callout.

2 Core concepts

2.1 · Radioactivity is a statistical phenomenon

You cannot predict when a single nucleus will decay. There is no internal clock; the decay constant $\lambda$ is a probability per unit time — in a short interval $\Delta t$ , each undecayed nucleus has probability $\lambda\,\Delta t$ of decaying, independent of its history and of every other nucleus. This is why radioactivity is called a statistical phenomenon: the law $N = N_{0}e^{-\lambda t}$ predicts only the fraction of a large number of nuclei surviving, never the fate of one.

In a counting experiment you watch a vast number $n$ of nuclei, each with a tiny probability $p$ of producing a recorded count in your counting interval. The count you get is one random draw from a distribution — repeat the count and you get a different draw. The whole of this chapter is the statistics of those draws.

2.2 · The probability rules you need

An event is one possible outcome of a trial (a nucleus decays / does not decay in $\Delta t$ ). Its probability $P$ is a number between 0 and 1 measuring how likely it is. Two rules do all the work in this chapter:

Addition rule (mutually exclusive events — they cannot both happen): $P(A \cup B) = P(A) + P(B)$ . Example: the probability of getting either exactly 3 or exactly 4 counts is $P(3) + P(4)$ .
Multiplication rule (independent events — one does not affect the other): $P(A \cap B) = P(A)\,P(B)$ . Example: if each nucleus decays with probability $p$ in $\Delta t$ , the probability that three specified nuclei all decay is $p^{3}$ .

A sequence of $n$ independent trials, each with the same success probability $p$ , is called a set of Bernoulli trials — the coin-toss model. Radioactive counting is a Bernoulli process with enormous $n$ and tiny $p$ .

2.3 · The three distributions of this chapter

Distribution	Describes	Parameters	Mean	Variance
Binomial	number of successes $r$ in a fixed number $n$ of trials	$n, p$	$np$	$npq$
Poisson	number of rare, random events $r$ in a fixed interval of time/space	$\mu$ (mean count)	$\mu$	$\mu$
Normal (Gaussian)	large-sample limit of both; the familiar bell curve	$\mu, \sigma$	$\mu$	$\sigma^{2}$

Binomial is the exact, fundamental law: it answers "in $n$ trials with success probability $p$ each, what is the chance of exactly $r$ successes?"
Poisson is the binomial law in the limit $n \to \infty$ , $p \to 0$ with $np = \mu$ held fixed — exactly the situation of a counting experiment (huge number of nuclei, tiny per-nucleus probability). Its signature: variance equals mean, $\sigma^{2} = \mu$ .
Normal is what both become when the mean is large (roughly $\mu \gtrsim 20\text{–}30$ ): the distribution turns symmetric and bell-shaped with $\sigma = \sqrt{\mu}$ . In practice, counting data with more than a few dozen counts is treated as normal — which is why " $\pm\sqrt{N}$ error bars" work.

🔑 The one line to remember

For counting data the total count $N$ estimates the Poisson mean, so the standard deviation of the count is the square root of the count: $\sigma_{N} = \sqrt{N}$ , and the relative standard deviation is $1/\sqrt{N}$ . Want 1% precision? You need $N = 10{,}000$ counts. Want 0.1%? $N = 10^{6}$ .

2.4 · Where the syllabus topics sit

Counting statistics — this whole chapter: treating a count as a random variable with a distribution, not as an exact number.
Probability and binomial distribution — §2.2 and the binomial derivation in §3.
Radioactivity as a statistical phenomenon — §2.1: decay is random per nucleus, predictable only in the large-number average.
Standard deviation of counting data — $\sigma = \sqrt{N}$ , relative error $1/\sqrt{N}$ , background subtraction.
Poisson distribution — derived as the binomial limit; the working distribution of counting experiments.
Optimisation of counting experiments — how to split a fixed total time between sample and background, and preset-count versus preset-time.

3 Key derivations

3.1 · The binomial distribution law

🔥 Exam favourite — asked 2022(a) and 2024(a)

Learn this derivation cold, with every symbol defined. It is the single most repeated question of this unit.

Set up the trials. Consider $n$ independent, identical trials. In each trial the probability of "success" is $p$ and of "failure" is $q = 1 - p$ . (In counting: a trial is one nucleus in the counting interval; "success" is its decay being recorded.)

One particular sequence. Fix attention on one specific order of outcomes — say the first $r$ trials succeed and the remaining $n-r$ fail. The trials are independent, so by the multiplication rule the probability of this exact sequence is $p^{r}\,q^{\,n-r}$ . Every specific sequence with $r$ successes and $n-r$ failures has this same probability.

Count the sequences. How many distinct sequences contain exactly $r$ successes among $n$ trials? Choose which $r$ of the $n$ positions hold the successes: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ ways. These sequences are mutually exclusive.

Add them up. By the addition rule, the probability of exactly $r$ successes in any order is the number of sequences times the probability of each:

(1)

P(r) = \binom{n}{r}\,p^{r}\,q^{\,n-r}, \qquad q = 1-p

Meaning of every symbol: $P(r)$ — probability of obtaining exactly $r$ successes; $n$ — total number of independent trials (fixed); $r$ — number of successes observed, $r = 0, 1, \ldots, n$ ; $p$ — probability of success in a single trial (same for all trials); $q = 1-p$ — probability of failure in a single trial; $\binom{n}{r}$ — binomial coefficient, the number of ways to choose the $r$ successful trials out of $n$ .

3.2 · Mean of the binomial distribution: $\bar{r} = np$

🔥 Exam favourite — asked 2024(b)

A short, clean proof. Examiners expect the combinatorial identity $r\binom{n}{r} = n\binom{n-1}{r-1}$ used explicitly.

Write the mean as a sum. By definition, $\bar{r} = \sum_{r=0}^{n} r\,P(r) = \sum_{r=0}^{n} r\binom{n}{r}p^{r}q^{n-r}$ . The $r = 0$ term is zero.

Use the identity $r\binom{n}{r} = n\binom{n-1}{r-1}$ (check: $r\cdot n!/[r!(n-r)!] = n\cdot (n-1)!/[(r-1)!(n-r)!]$ . Pull the constant $n$ out and split one power of $p$ :

$\bar{r} = np\sum_{r=1}^{n}\binom{n-1}{r-1}\,p^{\,r-1}\,q^{\,(n-1)-(r-1)}$

Recognise the binomial expansion. Put $s = r-1$ ; the sum is $\sum_{s=0}^{n-1}\binom{n-1}{s}p^{s}q^{(n-1)-s} = (p+q)^{n-1}$ . Since $p + q = 1$ , the sum equals 1.

(2)

\bar{r} = np

3.3 · Variance of the binomial distribution: $\sigma^{2} = npq$

🔥 Exam favourite — asked 2022(b) and 2024(c)

The question asks for variance "in terms of p" — the expected final line is $\sigma^{2} = np(1-p)$ . Derive it via $E[r(r-1)]$ ; it is far shorter than expanding $(r-\bar{r})^{2}$ directly.

Variance in terms of moments. $\sigma^{2} = \overline{(r-\bar{r})^{2}} = \overline{r^{2}} - \bar{r}^{2}$ , where $\overline{r^{2}} = \sum r^{2}P(r)$ . Write $r^{2} = r(r-1) + r$ , so $\overline{r^{2}} = \overline{r(r-1)} + \bar{r}$ .

Evaluate $\overline{r(r-1)}$ . Using $r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2}$ :

$\overline{r(r-1)} = \sum_{r=2}^{n}r(r-1)\binom{n}{r}p^{r}q^{n-r} = n(n-1)p^{2}\sum_{s=0}^{n-2}\binom{n-2}{s}p^{s}q^{(n-2)-s}$

The sum is $(p+q)^{n-2} = 1$ , so $\overline{r(r-1)} = n(n-1)p^{2}$ .

Assemble. $\overline{r^{2}} = n(n-1)p^{2} + np$ , and $\bar{r}^{2} = n^{2}p^{2}$ . Hence $\sigma^{2} = n(n-1)p^{2} + np - n^{2}p^{2} = np - np^{2} = np(1-p).$

(3)

\sigma^{2} = npq = np(1-p), \qquad \sigma = \sqrt{npq}

3.4 · The Poisson distribution as the limit of the binomial

🔥 Exam favourite — asked 2022(c)

Do this limit carefully, one factor at a time. The examiner wants to see why each piece tends to its limit.

Take the binomial law and fix the mean. Start from $P(r) = \frac{n!}{r!(n-r)!}\,p^{r}(1-p)^{\,n-r}$ . Now let the number of trials grow without bound, $n \to \infty$ , while the per-trial probability shrinks, $p \to 0$ , in such a way that the mean stays fixed: $np = \mu$ (constant). Substitute $p = \mu/n$ :

$P(r) = \frac{n(n-1)\cdots(n-r+1)}{r!}\left(\frac{\mu}{n}\right)^{r}\left(1-\frac{\mu}{n}\right)^{n-r}$

Regroup into three factors:

$P(r) = \frac{\mu^{r}}{r!}\times \underbrace{\frac{n(n-1)\cdots(n-r+1)}{n^{r}}}_{\text{(A)}}\times \underbrace{\left(1-\frac{\mu}{n}\right)^{n}}_{\text{(B)}}\times \underbrace{\left(1-\frac{\mu}{n}\right)^{-r}}_{\text{(C)}}\,,$

where $\left(1-\frac{\mu}{n}\right)^{n-r} = \left(1-\frac{\mu}{n}\right)^{n}\left(1-\frac{\mu}{n}\right)^{-r}$ was used.

Take the limit term by term. (A) has $r$ factors, each of the form $(n-k)/n = 1 - k/n \to 1$ , so (A) → 1. (C) has a fixed exponent $-r$ , and $(1-\mu/n) \to 1$ , so (C) → 1. (B) is the classic exponential limit: $\lim_{n\to\infty}(1-\mu/n)^{n} = e^{-\mu}$ .

(4)

P(r) = \frac{\mu^{r}\,e^{-\mu}}{r!}\qquad r = 0, 1, 2, \ldots

Conditions for Poisson to apply to a counting experiment (the second half of the 2022 question): the limit above demands (i) a very large number of nuclei, $n \to \infty$ ; (ii) a very small probability of any one nucleus being counted in the interval, $p \to 0$ ; (iii) a finite, constant mean count $\mu = np$ . In laboratory language: disintegrations must be random, independent events occurring at a constant average rate — the source must not decay appreciably during the count, geometry and detector efficiency must stay fixed, and dead-time losses must be negligible. Then $\mu$ is estimated by the observed mean count, and — the Poisson signature — variance equals mean.

3.5 · Mean and variance of the Poisson distribution

Mean. $\bar{r} = \sum_{r=0}^{\infty} r\,\frac{\mu^{r}e^{-\mu}}{r!} = \mu e^{-\mu}\sum_{r=1}^{\infty}\frac{\mu^{r-1}}{(r-1)!} = \mu e^{-\mu}\cdot e^{\mu} = \mu.$ (The series is $e^{\mu}$ .)

Variance. $\overline{r(r-1)} = \mu^{2}e^{-\mu}\sum_{r=2}^{\infty}\mu^{r-2}/(r-2)! = \mu^{2}$ , so $\overline{r^{2}} = \mu^{2} + \mu$ and $\sigma^{2} = \overline{r^{2}} - \bar{r}^{2} = \mu.$

(5)

\text{Poisson:}\quad \bar{r} = \mu,\qquad \sigma^{2} = \mu,\qquad \sigma = \sqrt{\mu}

3.6 · Standard deviation of counting data: $\sigma = \sqrt{N}$

Syllabus, never asked

Never appeared in 2020–2024, but it is explicitly in the syllabus and it is the most-used result of the chapter in real analytical work. Examiners can lift it straight from Friedlander or Arnikar at any time.

Identify the Poisson mean with the observed count. A counting experiment satisfies the Poisson conditions (§3.4), so the counts follow $P(r) = \mu^{r}e^{-\mu}/r!$ with $\sigma^{2} = \mu$ . The best (maximum-likelihood) estimate of the unknown true mean $\mu$ from a single measurement is the observed total count $N$ itself.

Read off the standard deviation. Replacing $\mu$ by its estimate $N$ :

(6)

\sigma_{N} = \sqrt{N}, \qquad \frac{\sigma_{N}}{N} = \frac{1}{\sqrt{N}}

So a count of $N = 10{,}000$ carries $\sigma = 100$ , i.e. ±1% relative. For a count rate $R = N/t$ measured over time $t$ , dividing by the exact time gives $\sigma_{R} = \sqrt{N}/t = \sqrt{R/t}$ . Two consequences examiners love: (a) the absolute error grows as $\sqrt{N}$ while the relative error shrinks as $1/\sqrt{N}$ ; (b) to halve the relative error you must quadruple the counts (or the counting time).

3.7 · Background subtraction and error propagation

Syllabus, never asked

Straight from the reference books. Any numerical on "net count rate" needs this.

A real measurement gives gross counts $N_{s}$ in time $t_{s}$ ; a separate background count gives $N_{b}$ in time $t_{b}$ . The net (sample-only) rate is $R_{n} = R_{s} - R_{b}$ with $R_{s} = N_{s}/t_{s}$ , $R_{b} = N_{b}/t_{b}$ . The two counts are independent Poisson variables, and for a sum or difference of independent quantities variances add (never standard deviations):

Net counts. $N_{n} = N_{s} - N_{b}$ , so $\sigma_{n}^{2} = \sigma_{s}^{2} + \sigma_{b}^{2} = N_{s} + N_{b}$ . Note the background increases the error even though it is subtracted from the value.

Net rate. $R_{n} = N_{s}/t_{s} - N_{b}/t_{b}$ ; the times are exact, so $\sigma_{R_{n}}^{2} = \sigma_{s}^{2}/t_{s}^{2} + \sigma_{b}^{2}/t_{b}^{2} = N_{s}/t_{s}^{2} + N_{b}/t_{b}^{2}.$

(7)

\sigma_{R_{n}}^{2} = \frac{R_{s}}{t_{s}} + \frac{R_{b}}{t_{b}} \qquad \bigl(R_{n} = R_{s} - R_{b}\bigr)

using $N_{s}/t_{s}^{2} = R_{s}/t_{s}$ . If background is negligible, this collapses to $\sigma_{R} = \sqrt{R/t}$ .

3.8 · Optimisation: how to split the counting time

Syllabus, never asked

The syllabus says "optimisation of counting experiments". This is the standard textbook result (Evans; Friedlander, Kennedy and Macias): with total time $T$ fixed, do not split it equally — give more time to whichever count is noisier.

Minimise the net-rate variance. From (7), with $t_{s} + t_{b} = T$ fixed, minimise $f(t_{s}) = R_{s}/t_{s} + R_{b}/(T - t_{s})$ . Differentiate and set to zero:

$\frac{df}{dt_{s}} = -\frac{R_{s}}{t_{s}^{2}} + \frac{R_{b}}{(T-t_{s})^{2}} = 0 \;\;\Longrightarrow\;\; \frac{R_{s}}{t_{s}^{2}} = \frac{R_{b}}{t_{b}^{2}}.$

Take square roots. $\sqrt{R_{s}}/t_{s} = \sqrt{R_{b}}/t_{b}$ , i.e.

(8)

\frac{t_{s}}{t_{b}} = \sqrt{\frac{R_{s}}{R_{b}}}

(The second derivative is positive, so this is a minimum.) In words: split the time in proportion to the square roots of the rates — spend longer on the noisier (higher-rate) count. If sample and background rates are equal, split equally; if the sample is much hotter than background, most of the time goes to the sample. A useful companion choice is preset-count versus preset-time: in preset-time you fix $t$ and let the counts vary (Poisson); in preset-count you fix $N$ and measure the time needed. The relative precision of the rate is $\sigma_{R}/R = 1/\sqrt{N}$ either way — preset-count simply guarantees the precision you asked for, while preset-time is operationally simpler.

4 Worked examples

Example 1 — Rate and its error from one count

Given. A sample gives $N = 2500$ counts in $t = 5$ min.

Solution. The count rate is $R = N/t = 2500/5 = 500$ cpm. The standard deviation of the count is $\sigma_{N} = \sqrt{N} = \sqrt{2500} = 50$ . The time is exact, so the error in the rate is $\sigma_{R} = \sigma_{N}/t = 50/5 = 10$ cpm. Result: $R = 500 \pm 10$ cpm, a relative error of $10/500 = 2\%$ . (≈68% of repeat counts would fall in 490–510 cpm; ≈95% in $\pm 2\sigma$ , i.e. 480–520 cpm.)

Example 2 — How many counts for 1% precision?

Given. You need the count rate to ±1% relative.

Solution. Relative error $= 1/\sqrt{N}$ . Set $1/\sqrt{N} = 0.01 \;\Rightarrow\; \sqrt{N} = 100 \;\Rightarrow\; N = 10{,}000$ counts. At 500 cpm that needs $10{,}000/500 = 20$ min of counting. For ±0.5% you need $1/\sqrt{N} = 0.005 \Rightarrow N = 40{,}000$ counts — four times the counts (80 min): halving the error always quadruples the counting time. Preset-count shortcut: set the scaler to stop at $N = 10{,}000$ ; if it stops after 25.0 min, $R = 10{,}000/25.0 = 400$ cpm with $\sigma_{R}/R = 1/\sqrt{10{,}000} = 1\%$ , i.e. $400 \pm 4$ cpm — the precision is guaranteed by the preset.

Example 3 — χ² check: is the counter behaving statistically?

Given. Ten successive 1-min background counts: 52, 41, 49, 58, 44, 50, 39, 55, 47, 51. Is the scatter consistent with pure statistical fluctuation?

Solution. Mean $\bar{x} = 486/10 = 48.6$ . For Poisson data the test statistic $\chi^{2} = \sum_{i}(x_{i}-\bar{x})^{2}/\bar{x}$ should be ≈ the degrees of freedom, $n-1 = 9$ . Deviations squared: $3.4^{2} + 7.6^{2} + 0.4^{2} + 9.4^{2} + 4.6^{2} + 1.4^{2} + 9.6^{2} + 6.4^{2} + 1.6^{2} + 2.4^{2} = 322.4$ , so $\chi^{2} = 322.4/48.6 = 6.63$ . The 95% acceptance band for $\chi^{2}_{9}$ is about 2.7–19.0, and 6.63 lies comfortably inside — the counter is behaving; the scatter is statistical. (A value far above ~19 would signal extra, non-statistical noise — drifting high voltage, for instance.)

Example 4 — Background subtraction with error propagation

Given. Gross count: $N_{s} = 3600$ in $t_{s} = 10$ min. Background: $N_{b} = 900$ in $t_{b} = 10$ min.

Solution. Rates: $R_{s} = 360$ cpm, $R_{b} = 90$ cpm, net $R_{n} = 360 - 90 = 270$ cpm. Variances add for a difference: $\sigma_{R_{n}}^{2} = R_{s}/t_{s} + R_{b}/t_{b} = 360/10 + 90/10 = 36 + 9 = 45$ , so $\sigma_{R_{n}} = \sqrt{45} = 6.71$ cpm. Result: $R_{n} = 270 \pm 6.7$ cpm. Check via net counts: $N_{n} = 2700$ , $\sigma_{n} = \sqrt{3600+900} = 67.1$ , and $67.1/10 = 6.71$ cpm ✓. Note the background contributes 9 of the 45 units of variance — subtracting background raises the error.

Example 5 — Optimal split of counting time

Given. Sample rate $R_{s} = 400$ cpm, background $R_{b} = 100$ cpm, total time $T = 30$ min. Compare the optimal split with an equal split.

Solution. Optimal: $t_{s}/t_{b} = \sqrt{R_{s}/R_{b}} = \sqrt{4} = 2$ , so $t_{s} = 20$ min, $t_{b} = 10$ min. Then $\sigma^{2} = 400/20 + 100/10 = 20 + 10 = 30$ , $\sigma = 5.48$ cpm. Equal split (15/15): $\sigma^{2} = 400/15 + 100/15 = 33.33$ , $\sigma = 5.77$ cpm. Net rate $R_{n} = 300$ cpm, so $300 \pm 5.5$ cpm optimal vs $300 \pm 5.8$ cpm equal-split. The gain is modest here but grows when the rates differ more — and it costs nothing.

5 Figures

Fig. 1. The binomial distribution with fixed mean

np = 5

n

grows and

p

shrinks: at

n = 100, p = 0.05

the bars already hug the dashed Poisson curve — the §3.4 limit made visible.

Fig. 2. Relative standard deviation

1/\sqrt{N}

against total counts on log–log axes — a straight line of slope −½. Memorise the three landmarks: 100 counts → 10%, 10⁴ counts → 1%, 10⁶ counts → 0.1%.

Fig. 3. Variance of the net rate versus the fraction of a fixed total time spent on the sample. The minimum sits at

t_{s}/t_{b} = \sqrt{R_{s}/R_{b}}

(dashed line, §3.8) — an equal split is measurably worse.

6 PYQ bank

Every question below was asked in a Burdwan M.Sc. MSCH-102 final paper — nothing is invented. This unit appears only in 2022 and 2024.

2022 · MSCH-102asked 2×

Q(a). Derive the binomial distribution law mentioning the meaning of all symbols used in the context of statistical data analysis.

Solution. See §3.1. Consider $n$ independent identical trials; in each, success has probability $p$ and failure $q = 1-p$ . One particular sequence with $r$ successes then $n-r$ failures has probability $p^{r}q^{\,n-r}$ by the multiplication rule. There are $\binom{n}{r}$ such mutually exclusive sequences, so by the addition rule

$P(r) = \binom{n}{r}\,p^{r}\,q^{\,n-r},\qquad q = 1-p.$

Symbols: $P(r)$ probability of exactly $r$ successes; $n$ number of trials (fixed); $r = 0,\dots,n$ successes observed; $p$ per-trial success probability; $q$ per-trial failure probability; $\binom{n}{r} = n!/[r!(n-r)!]$ the number of ways to place the $r$ successes. (Repeated as 2024(a).)

2022 · MSCH-102asked 2×

Q(b). Write down the equation of "Variance" in terms of "probability of success (p)".

Solution. For the binomial distribution the variance is (derived in §3.3)

$\sigma^{2} = np(1-p) = npq,$

where $n$ is the number of trials and $q = 1-p$ the failure probability. Derivation sketch: $\sigma^{2} = \overline{r^{2}}-\bar{r}^{2}$ ; with $\overline{r(r-1)} = n(n-1)p^{2}$ and $\bar{r} = np$ , one gets $\sigma^{2} = n(n-1)p^{2}+np-n^{2}p^{2} = np(1-p)$ . (Repeated as 2024(c).)

2022 · MSCH-102asked 1×

Q(c). Starting from binomial distribution, derive the expression for "Poisson distribution". Mention the condition that a radioactive counting experiment must satisfy so that Poisson distribution may be applied.

Solution. From $P(r) = \binom{n}{r}p^{r}(1-p)^{n-r}$ , let $n \to \infty$ , $p \to 0$ with $np = \mu$ fixed, and put $p = \mu/n$ . Regrouping (full steps in §3.4),

$P(r) = \frac{\mu^{r}}{r!}\cdot\frac{n(n-1)\cdots(n-r+1)}{n^{r}}\cdot\left(1-\frac{\mu}{n}\right)^{n}\cdot\left(1-\frac{\mu}{n}\right)^{-r}\xrightarrow[n\to\infty]{}\frac{\mu^{r}e^{-\mu}}{r!}.$

The first and third factors tend to 1, the middle one to $e^{-\mu}$ . Applicability conditions: the number of nuclei must be very large, the probability of any one nucleus being counted in the interval very small, with finite constant mean $\mu = np$ — i.e. disintegrations must be random, independent events at a constant average rate (no appreciable decay during counting, fixed geometry and efficiency, negligible dead-time).

2024 · MSCH-102asked 2×

Q(a). Derive binomial distribution law mentioning the meaning of all symbols used in the context of statistical data analysis.

Solution. Identical to 2022(a) above: $P(r) = \binom{n}{r}p^{r}q^{\,n-r}$ with $q = 1-p$ , from "one sequence has probability $p^{r}q^{n-r}$ (multiplication rule)" × " $\binom{n}{r}$ mutually exclusive sequences (addition rule)". Symbol meanings: $P(r)$ probability of exactly $r$ successes; $n$ trials; $r$ successes, $0 \le r \le n$ ; $p$ success probability per trial; $q$ failure probability per trial; $\binom{n}{r}$ ways to choose the successful trials. Full derivation in §3.1.

2024 · MSCH-102asked 1×

Q(b). Prove that $\bar{r} = np$ where the symbols carry usual meaning in the context of statistical data analysis.

Solution. $\bar{r} = \sum_{r=0}^{n} r\binom{n}{r}p^{r}q^{n-r} = np\sum_{r=1}^{n}\binom{n-1}{r-1}p^{r-1}q^{(n-1)-(r-1)},$

using $r\binom{n}{r} = n\binom{n-1}{r-1}$ . With $s = r-1$ the sum is $\sum_{s=0}^{n-1}\binom{n-1}{s}p^{s}q^{(n-1)-s} = (p+q)^{n-1} = 1$ , since $p+q = 1$ . Hence $\bar{r} = np$ : the mean number of successes is the number of trials times the per-trial success probability. Full steps in §3.2.

2024 · MSCH-102asked 2×

Q(c). Hence, derive the equation of 'Variance' in terms of 'probability of success (p)'.

Solution. "Hence" points back to the binomial setup of Q(a)–(b). Using $\sigma^{2} = \overline{r^{2}}-\bar{r}^{2}$ with $\overline{r^{2}} = \overline{r(r-1)}+\bar{r}$ , and $\overline{r(r-1)} = n(n-1)p^{2}$ (via $r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2}$ , together with $\bar{r} = np$ from Q(b):

$\sigma^{2} = \bigl[n(n-1)p^{2} + np\bigr] - n^{2}p^{2} = np - np^{2} = np(1-p).$

Full derivation in §3.3. (Same demand as 2022(b).)

7 Exam Q&A

Q1. Why is radioactivity called a statistical phenomenon?

A. The moment of decay of any single nucleus is unpredictable — the decay constant $\lambda$ is only a probability per unit time ( $\lambda\Delta t$ in $\Delta t$ ). Only for a huge number of nuclei does the fraction decaying become predictable, via $N = N_{0}e^{-\lambda t}$ .

Q2. State the addition and multiplication rules of probability, with the condition for each.

A. Addition: $P(A \cup B) = P(A)+P(B)$ for mutually exclusive events. Multiplication: $P(A \cap B) = P(A)\,P(B)$ for independent events. The binomial law uses multiplication for one fixed sequence of outcomes and addition over the $\binom{n}{r}$ mutually exclusive sequences.

Q3. In $P(r) = \binom{n}{r}p^{r}q^{n-r}$ , what is $q$ , and why must $p+q = 1$ ?

A. $q = 1-p$ is the probability of failure in one trial. Success and failure are the only two, mutually exclusive and exhaustive outcomes of a trial, so their probabilities sum to 1 — which is also what makes the binomial probabilities sum to 1 via $(p+q)^{n} = 1$ .

Q4. What is special about the variance of a Poisson distribution?

A. It equals the mean: $\sigma^{2} = \bar{r} = \mu$ , so $\sigma = \sqrt{\mu}$ . This is the entire basis of counting statistics — one measurement $N$ gives both the value and its error, $\sigma = \sqrt{N}$ .

Q5. A scaler records 900 counts. Quote the result with its standard deviation and relative error.

A. $\sigma = \sqrt{900} = 30$ , so $900 \pm 30$ counts; relative error $30/900 = 3.3\%$ (equivalently $1/\sqrt{900}$ ).

Q6. When subtracting background, why do variances add rather than standard deviations?

A. For independent random variables, the variance of a sum or difference is the sum of the variances — cross terms average to zero because the fluctuations are uncorrelated. So $\sigma_{n}^{2} = N_{s}+N_{b}$ : the background measurement adds error even though its value is subtracted.

Q7. State the optimal time-split rule. When is an equal split optimal?

A. $t_{s}/t_{b} = \sqrt{R_{s}/R_{b}}$ : split the total time in proportion to the square roots of the rates — more time for the noisier count. Equal split is optimal only when $R_{s} = R_{b}$ .

Q8. Preset-count versus preset-time: which guarantees the precision you want, and what is the relative error in each?

A. Preset-count (fix $N$ , measure $t$ ) guarantees the precision, because $\sigma_{R}/R = 1/\sqrt{N}$ is set by your chosen $N$ . Preset-time (fix $t$ , count $N$ ) gives the same formula, $\sigma_{R}/R = 1/\sqrt{N}$ , but you only learn $N$ — and hence the precision — after the count.

8 Quick revision

(1) Binomial law

$P(r)=\binom{n}{r}p^{r}q^{n-r},\;q=1-p$

(2) Binomial mean

$\bar{r}=np$

(3) Binomial variance

$\sigma^{2}=npq=np(1-p)$

(4) Poisson law

$P(r)=\mu^{r}e^{-\mu}/r!$

(5) Poisson mean = variance

$\bar{r}=\mu,\;\sigma^{2}=\mu$

(6) Counting error

$\sigma_{N}=\sqrt{N},\;\sigma_{\rm rel}=1/\sqrt{N}$

(7) Net-rate variance

$\sigma_{R_{n}}^{2}=R_{s}/t_{s}+R_{b}/t_{b}$

(8) Optimal time split

$t_{s}/t_{b}=\sqrt{R_{s}/R_{b}}$

Symbol	Meaning	Symbol	Meaning
$P(r)$	probability of exactly $r$ successes/counts	$\mu$	Poisson mean (= $np$ in the limit)
$n$	number of independent trials	$N$	total counts observed
$r$	number of successes observed	$R$	count rate $N/t$
$p$	success probability per trial	$t$	counting time
$q=1-p$	failure probability per trial	$R_{s}, R_{b}$	sample (gross) and background rates
$\bar{r}$	mean of the distribution	$\chi^{2}$	$\sum(x_{i}-\bar{x})^{2}/\bar{x}$ , ≈ $n-1$ if statistical
$\sigma^{2}, \sigma$	variance, standard deviation	$\lambda$	decay constant = decay probability per unit time

1 Chapter overview

2 Core concepts

2.1 · Radioactivity is a statistical phenomenon

2.2 · The probability rules you need

2.3 · The three distributions of this chapter

2.4 · Where the syllabus topics sit

3 Key derivations

3.1 · The binomial distribution law

3.2 · Mean of the binomial distribution: rˉ=np\bar{r} = nprˉ=np

3.3 · Variance of the binomial distribution: σ2=npq\sigma^{2} = npqσ2=npq

3.4 · The Poisson distribution as the limit of the binomial

3.5 · Mean and variance of the Poisson distribution

3.6 · Standard deviation of counting data: σ=N\sigma = \sqrt{N}σ=N​

3.7 · Background subtraction and error propagation

3.8 · Optimisation: how to split the counting time

4 Worked examples

Example 1 — Rate and its error from one count

Example 2 — How many counts for 1% precision?

Example 3 — χ² check: is the counter behaving statistically?

Example 4 — Background subtraction with error propagation

Example 5 — Optimal split of counting time

5 Figures

6 PYQ bank

7 Exam Q&A

8 Quick revision

3.2 · Mean of the binomial distribution: $\bar{r} = np$

3.3 · Variance of the binomial distribution: $\sigma^{2} = npq$

3.6 · Standard deviation of counting data: $\sigma = \sqrt{N}$