| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Standard +0.3 This is a straightforward S1 geometric distribution question with standard calculations (probability of success on specific trial, probability distribution, expectation, variance). Part (a) is shown for students, and all parts follow textbook methods with no novel problem-solving required. Slightly above average difficulty only due to the multi-part nature and part (e) requiring interpretation of F(1+E(A)), but all techniques are routine for S1 level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{correct at third attempt}) = 0.4 \times 0.4 \times 0.6 = 0.096\) | M1 A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 1, 2, 3, 4\) | B1 | |
| \(P(A=a)\): \(0.6,\ 0.24,\ 0.096,\ 0.064\) — all correct | B1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{correct number}) = 1 - (0.4)^4 = 0.9744\) | M1 A1 (2) | Accept awrt 0.974 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(A) = \sum a\,P(A=a) = (1 \times 0.6) + \ldots + (4 \times 0.064) = 1.624\) | M1 A1 | Accept awrt 1.62 |
| \(E(A^2) = \sum a^2\,P(A=a) = (1^2 \times 0.6) + \ldots + (4^2 \times 0.064) = 3.448\) | M1 A1 | |
| \(\therefore \text{Var}(A) = 3.448 - (1.624)^2 = 0.810624\) | M1 A1 (6) | Accept awrt 0.811 |
| \(F(1 + E(A)) = P(A \leq 1 + E(A)) = P(A \leq 2.624) = 0.84\) | M1 A1 (2) | (14) |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{correct at third attempt}) = 0.4 \times 0.4 \times 0.6 = 0.096$ | M1 A1 (2) | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 1, 2, 3, 4$ | B1 | |
| $P(A=a)$: $0.6,\ 0.24,\ 0.096,\ 0.064$ — all correct | B1 (2) | |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{correct number}) = 1 - (0.4)^4 = 0.9744$ | M1 A1 (2) | Accept awrt 0.974 |
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(A) = \sum a\,P(A=a) = (1 \times 0.6) + \ldots + (4 \times 0.064) = 1.624$ | M1 A1 | Accept awrt 1.62 |
| $E(A^2) = \sum a^2\,P(A=a) = (1^2 \times 0.6) + \ldots + (4^2 \times 0.064) = 3.448$ | M1 A1 | |
| $\therefore \text{Var}(A) = 3.448 - (1.624)^2 = 0.810624$ | M1 A1 (6) | Accept awrt 0.811 |
| $F(1 + E(A)) = P(A \leq 1 + E(A)) = P(A \leq 2.624) = 0.84$ | M1 A1 (2) | **(14)** |
---4. A customer wishes to withdraw money from a cash machine. To do this it is necessary to type a PIN number into the machine. The customer is unsure of this number. If the wrong number is typed in, the customer can try again up to a maximum of four attempts in total. Attempts to type in the correct number are independent and the probability of success at each attempt is 0.6 .
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that the customer types in the correct number at the third attempt is 0.096 .\\
(2 marks)\\
The random variable $A$ represents the number of attempts made to type in the correct PIN number, regardless of whether or not the attempt is successful.
\item Find the probability distribution of $A$.
\item Calculate the probability that the customer types in the correct number in four or fewer attempts.
\item Calculate $\mathrm { E } ( A )$ and $\operatorname { Var } ( A )$.
\item Find $\mathrm { F } ( 1 + \mathrm { E } ( A ) )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q4 [14]}}