Nested Probability Distributions

What are Nested Probability Distributions?

These questions involve two probability distributions working in sequence. You first compute a probability from one distribution, then use that probability as the parameter for a second distribution. Recognising the structure is the key skill.

General 3-Step Method

1

Identify both distributions — determine which distribution appears first (inner) and which is applied to the results (outer).

2

Find the linking probability P — use the inner distribution to calculate the probability of a "success" event.

3

Apply the outer distribution — use P from Step 2 as the success probability in the outer distribution to answer the final question.

All Five Combinations — click any card to explore

Each combination has a reminder, a worked example, and a full solution.

Normal → Binomial

Find a probability from a normal distribution, then count successes in repeated independent trials.

Rice packets · acceptable mass · shipment of 12

Uniform → Binomial

Use the rectangle-area formula on a uniform distribution, then apply binomially over many trials.

Bus arrivals · short wait · 7 mornings

Geometric → Binomial

Compute a tail probability from the geometric distribution, then count how many players satisfy it.

Quiz machine · 8 independent players

Binomial → Geometric

Find a "box-level" probability using the binomial, then model how many boxes until the first success.

Defective chips · boxes inspected

Uniform → Geometric

Compute a probability from a uniform distribution, then find how many trials until first success.

Electronic sensor lifetime · first long-lasting

Quick Reference — Distribution Formulas

Distribution	Notation	Key Formula / Rule
Normal	\(X \sim N(\mu, \sigma^2)\)	Standardise: \(Z = \dfrac{X - \mu}{\sigma}\), use tables or calculator
Uniform	\(X \sim U(a, b)\)	\(P(c < X < d) = \dfrac{d - c}{b - a}\)
Geometric	\(X \sim \text{Geom}(p)\)	\(P(X = r) = (1-p)^{r-1}p\) · \(P(X > k) = (1-p)^k\)
Binomial	\(X \sim B(n, p)\)	\(P(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}\)

Normal Distribution → Binomial Distribution

Inner: Normal | Outer: Binomial | Link: P(value in range)

Technique

1

Identify: A quantity is normally distributed. A number of independent items are then sampled, and you need to count how many satisfy some condition.

2

Find P: Use the Normal distribution to compute \(P = P(c < X < d)\) — the probability that a single item satisfies the condition. Standardise if needed: \(Z = \dfrac{X - \mu}{\sigma}\).

3

Apply Binomial: Let \(Y \sim B(n, P)\) where \(n\) = number of items. Use the binomial formula to answer the count question.

Normal Distribution Reminder

For \(X \sim N(\mu, \sigma^2)\), standardise to \(Z \sim N(0,1)\):

\[ Z = \frac{X - \mu}{\sigma} \]

Read \(\Phi(z) = P(Z \leq z)\) from tables or compute directly on a calculator. Use symmetry: \(\Phi(-z) = 1 - \Phi(z)\).

Worked Example

The masses of packets of rice produced by a factory are normally distributed with mean 500 g and standard deviation 8 g.

A packet is classified as acceptable if its mass is between 495 g and 510 g.

(a)

Find the probability that a randomly selected packet is acceptable.

Solution (a)

\(X \sim N(500,\, 8^2)\)

\[ P(495 < X < 510) \]

Standardise both bounds:

\[ z_1 = \frac{495 - 500}{8} = -0.625 \qquad z_2 = \frac{510 - 500}{8} = 1.25 \]

Using tables / calculator:

\[ \Phi(1.25) - \Phi(-0.625) = \Phi(1.25) - (1 - \Phi(0.625)) \]

\[ \approx 0.8944 - (1 - 0.7340) = 0.8944 - 0.2660 \]

\(P(\text{acceptable}) \approx 0.6284\)

(b)

A supermarket receives a shipment of 12 packets. Find the probability that exactly 9 packets are acceptable.

Solution (b)

Let \(Y\) = number of acceptable packets in 12. Using \(P \approx 0.6284\) from (a):

\[ Y \sim B(12,\; 0.6284) \]

\[ P(Y = 9) = \binom{12}{9}(0.6284)^9(0.3716)^3 \]

\[ = 220 \times (0.6284)^9 \times (0.3716)^3 \]

\(P(Y = 9) \approx 0.189\)

Key insight: The probability \(P\) from the Normal calculation is carried forward exactly (or to at least 4 s.f.) into the Binomial. Rounding too early loses accuracy in the final answer.

Uniform Distribution → Binomial Distribution

Inner: Uniform | Outer: Binomial | Link: P(value in sub-interval)

Uniform Distribution Reminder

For \(X \sim U(a,b)\) the probability density is constant at \(\dfrac{1}{b-a}\) between \(a\) and \(b\):

\[ P(c < X < d) = \frac{d - c}{b - a} \qquad (a \leq c < d \leq b) \]

Equivalently: probability = (width of target interval) ÷ (total width). No standardising needed.

Technique

1

Identify: A continuous variable is uniformly distributed over \([a, b]\). Repeated independent observations are made, and you count how many fall in a sub-interval.

2

Find P: Calculate \(P = \dfrac{d - c}{b - a}\) — the ratio of the target sub-interval length to the full interval length. This is the probability of "success" on one trial.

3

Apply Binomial: Let \(Y \sim B(n, P)\). Compute the required binomial probability.

Worked Example

A bus arrives at a stop at a random time uniformly distributed between 0 and 20 minutes after 8:00 am. The waiting time is called short if the bus arrives within 6 minutes.

(a)

Find the probability that the waiting time is short.

Solution (a)

\(T \sim U(0, 20)\). A short wait means \(T < 6\).

\[ P(T < 6) = \frac{6 - 0}{20 - 0} = \frac{6}{20} \]

\(P = 0.3\)

(b)

The passenger travels on 7 different mornings. Find the probability that on at least 5 mornings the waiting time is short.

Solution (b)

Let \(Y \sim B(7,\; 0.3)\). We need \(P(Y \geq 5) = P(Y=5) + P(Y=6) + P(Y=7)\).

\[ P(Y=5) = \binom{7}{5}(0.3)^5(0.7)^2 = 21 \times 0.00243 \times 0.49 \approx 0.02501 \]

\[ P(Y=6) = \binom{7}{6}(0.3)^6(0.7)^1 = 7 \times 0.000729 \times 0.7 \approx 0.003571 \]

\[ P(Y=7) = (0.3)^7 \approx 0.000219 \]

\[ P(Y \geq 5) \approx 0.02501 + 0.003571 + 0.000219 \]

\(P(Y \geq 5) \approx 0.0288\)

Key insight: The uniform distribution requires only a simple ratio — no tables or standardising. Sketch the rectangle to confirm which sub-interval you want.

Geometric Distribution → Binomial Distribution

Inner: Geometric | Outer: Binomial | Link: P(X > k) = tail probability

Geometric Distribution Reminder

For \(X \sim \text{Geom}(p)\) (trials until first success, success probability \(p\)):

\[ P(X = r) = (1-p)^{r-1}p \qquad r = 1, 2, 3, \ldots \]

Tail probability (very useful in nested questions):

\[ P(X > k) = (1-p)^k \]

Reason: \(X > k\) means the first \(k\) trials are all failures, each with probability \((1-p)\).

Technique

1

Identify: One player/process follows a geometric model. The question then asks about several independent players, so a binomial count is needed.

2

Find P: Compute the required geometric probability for one player — often a tail probability \(P(X > k) = (1-p)^k\). This becomes the success probability for the outer binomial.

3

Apply Binomial: Let \(Y \sim B(n, P)\). Calculate the required probability for the group of \(n\) players.

Worked Example

A quiz machine gives a prize with probability 0.18 on each play. A player continues playing until winning their first prize. Let \(X\) = number of plays needed.

(a)

State the distribution of \(X\).

Solution (a)

\(X \sim \text{Geom}(0.18)\)

(b)

Find the probability that the player needs more than 4 plays to obtain the first prize.

Solution (b)

"More than 4 plays" means the first 4 plays are all failures.

\[ P(X > 4) = (1 - 0.18)^4 = (0.82)^4 \]

\(P(X > 4) \approx 0.4521\)

(c)

Eight different players each play independently until winning their first prize. Find the probability that exactly 3 players need more than 4 plays.

Solution (c)

For each player, \(P(\text{needs} > 4 \text{ plays}) \approx 0.4521\) from (b). Let \(Y\) = number of such players:

\[ Y \sim B(8,\; 0.4521) \]

\[ P(Y = 3) = \binom{8}{3}(0.4521)^3(0.5479)^5 \]

\[ = 56 \times 0.09233 \times 0.05117 \approx 0.2646 \]

\(P(Y = 3) \approx 0.265\)

Key insight: The tail probability \(P(X > k) = (1-p)^k\) is the most common link between the geometric and binomial layers. Memorise this result — deriving it from scratch in an exam is slow.

Binomial Distribution → Geometric Distribution

Inner: Binomial | Outer: Geometric | Link: P(batch satisfies condition)

Technique

1

Identify: Items within a batch follow a Binomial model. Batches are then inspected one at a time until the first batch meeting some condition — this is the Geometric layer.

2

Find P: Use the Binomial distribution to compute \(P = P(X = k)\) or \(P(X \leq k)\) for a single batch — the probability of that batch being a "success".

3

Apply Geometric: Let \(Y \sim \text{Geom}(P)\) where \(P\) is from Step 2. Compute geometric probabilities for the batch-inspection count.

Worked Example

A production line manufactures electronic chips. Each chip is defective with probability 0.07 independently. A box contains 5 chips. Let \(X\) = number of defective chips in a box.

(a)

State the distribution of \(X\).

Solution (a)

\(X \sim B(5,\; 0.07)\)

(b)

Find the probability that a box contains no defective chips.

Solution (b)

\[ P(X = 0) = \binom{5}{0}(0.07)^0(0.93)^5 = (0.93)^5 \]

\(P(X = 0) \approx 0.6957\)

(c)

Boxes are inspected one at a time until the first box containing no defective chips. Let \(Y\) = number of boxes inspected. State the distribution of \(Y\).

Solution (c)

A "success" is a box with no defective chips; probability = 0.6957 from (b).

\(Y \sim \text{Geom}(0.6957)\)

(d)

Find the probability that more than 3 boxes must be inspected.

Solution (d)

"More than 3 boxes" means the first 3 are all failures (boxes with defective chips).

Failure probability for one box: \(1 - 0.6957 = 0.3043\)

\[ P(Y > 3) = (0.3043)^3 \]

\(P(Y > 3) \approx 0.0282\)

Key insight: The Binomial operates at the item level within a batch; the Geometric operates at the batch level. Make sure you clearly identify what constitutes a "trial" and a "success" at each layer.

Uniform Distribution → Geometric Distribution

Inner: Uniform | Outer: Geometric | Link: P(value in sub-interval)

Technique

1

Identify: A continuous quantity is uniformly distributed. Items are tested one at a time until the first item meeting a condition — this sequential stopping rule signals the Geometric distribution.

2

Find P: Use \(P = \dfrac{d - c}{b - a}\) to find the probability a single item satisfies the condition (its uniform value falls in the target sub-interval).

3

Apply Geometric: Let \(Y \sim \text{Geom}(P)\). Use \(P(Y = r) = (1-P)^{r-1}P\) or the tail formula \(P(Y > k) = (1-P)^k\).

Worked Example

The lifetime, in hours, of a small electronic sensor is uniformly distributed between 40 and 100 hours. A sensor is called long-lasting if it survives more than 85 hours. Sensors are tested one at a time until the first long-lasting sensor is found.

(a)

Find the probability that a randomly selected sensor is long-lasting.

Solution (a)

\(L \sim U(40, 100)\). Long-lasting means \(L > 85\).

\[ P(L > 85) = \frac{100 - 85}{100 - 40} = \frac{15}{60} = \frac{1}{4} \]

\(p = 0.25\)

(b)

State the distribution of the number of sensors tested.

Solution (b)

Let \(Y\) = number of sensors tested until the first long-lasting one:

\(Y \sim \text{Geom}(0.25)\)

(c)

Find the probability that the first long-lasting sensor is found on the 6th test.

Solution (c)

First 5 tests fail, 6th succeeds. Failure probability per test = 0.75.

\[ P(Y = 6) = (0.75)^5 \times 0.25 \]

\[ = 0.2373 \times 0.25 \]

\(P(Y = 6) \approx 0.0593\)

Key insight: The phrase "tested/inspected/selected one at a time until the first …" is the tell-tale sign of a Geometric distribution in the outer layer. The inner distribution (here, Uniform) supplies the value of \(p\).

Extended Question — Normal Distribution (A-Level Style)

Finding σ from a percentile | Normal approx. to Binomial | Hypothesis test for mean

Skills Covered in This Question

Using a given percentile to find the standard deviation of a normal distribution
Finding an unknown upper bound \(k\) using inverse normal
Applying the normal approximation to a binomial (with continuity correction)
Conducting a one-tailed \(z\)-test for a population mean (known variance)

Question (Total: 13 marks)

A machine fills bottles of shampoo. The amount of shampoo put into each bottle, \(S\) ml, follows a normal distribution with mean 200 ml.

It is known that 20% of bottles contain less than 198.32 ml.

(a) 5 marks

Find, to 2 decimal places, the value of \(k\) such that \(\,P(198.32 < S < k) = 0.35\).

Solution (a)

Step 1 — Find \(\sigma\).

Given \(P(S < 198.32) = 0.20\), standardise:

\[ \frac{198.32 - 200}{\sigma} = \Phi^{-1}(0.20) = -0.8416 \]

\[ \sigma = \frac{-1.68}{-0.8416} = 1.9961\ldots \approx 2.0 \]

So \(S \sim N(200,\; 2^2)\).

Step 2 — Find \(k\).

We need \(P(S < k)\):

\[ P(S < k) = P(S < 198.32) + 0.35 = 0.20 + 0.35 = 0.55 \]

Standardise and use inverse normal:

\[ \frac{k - 200}{2} = \Phi^{-1}(0.55) = 0.1257 \]

\[ k = 200 + 2 \times 0.1257 = 200.2514\ldots \]

\(k = 200.25 \text{ ml}\quad\text{(2 d.p.)}\)

(b) 3 marks

A random sample of 100 bottles is taken.

Using a normal approximation, find the probability that more than 40 of these bottles contain between 198.32 ml and \(k\) ml.

Solution (b)

From part (a), \(P(198.32 < S < k) = 0.35\). Let \(Y\) = number of bottles in the range:

\[ Y \sim B(100,\; 0.35) \]

Normal approximation parameters:

\[ \mu = np = 35, \qquad \sigma^2 = np(1-p) = 100 \times 0.35 \times 0.65 = 22.75 \]

\[ Y \approx N(35,\; 22.75) \]

With continuity correction, \(P(Y > 40) = P(Y \geq 41) \approx P\!\left(Z > \dfrac{40.5 - 35}{\sqrt{22.75}}\right)\):

\[ z = \frac{40.5 - 35}{4.770} = \frac{5.5}{4.770} = 1.153 \]

\[ P(Z > 1.153) = 1 - \Phi(1.153) \approx 1 - 0.8756 \]

\(P(Y > 40) \approx 0.124\)

(c) 5 marks

The machine is adjusted so that the standard deviation of the shampoo volume is now 0.4 ml.

Following the adjustment, it is believed that the mean amount of shampoo per bottle has increased from 200 ml.

A random sample of 25 bottles is taken and the sample mean is found to be 200.18 ml.

Test this belief at the 5% level of significance. State your hypotheses clearly.

Solution (c)

Hypotheses:

\(H_0: \mu = 200\) (the mean has not changed)
\(H_1: \mu > 200\) (one-tailed; the mean has increased)

Distribution of sample mean under \(H_0\):

\[ \bar{S} \sim N\!\left(200,\; \frac{0.4^2}{25}\right) = N\!\left(200,\; \frac{0.16}{25}\right) = N(200,\; 0.0064) \]

Test statistic:

\[ z = \frac{\bar{s} - \mu_0}{\sigma/\sqrt{n}} = \frac{200.18 - 200}{0.4/\sqrt{25}} = \frac{0.18}{0.4/5} = \frac{0.18}{0.08} = 2.25 \]

Critical value (one-tailed, 5%):

\[ z_{\text{crit}} = 1.6449 \]

Comparison: \(z = 2.25 > 1.6449\)

Conclusion: Since \(2.25 > 1.6449\), the test statistic lies in the critical region. We reject \(H_0\). There is sufficient evidence at the 5% level to conclude that the mean amount of shampoo per bottle has increased above 200 ml.

Examiner tips: In part (a), always find \(\sigma\) first before attempting to locate \(k\) — two separate inverse-normal lookups are needed and both contribute marks. In part (b), state \(B(n,p)\) explicitly before switching to the normal approximation, and always apply the continuity correction. In part (c), write the hypotheses in terms of \(\mu\) (not \(\bar{x}\)), state the critical value, and write a conclusion in context.