| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Moderate -0.3 This is a straightforward probability question requiring systematic case enumeration and basic probability rules. Part (a) requires finding P(X=0 and second roll=2) + P(X=2 and second roll=0) + P(X=1 and second roll=1). Parts (b) and (c) follow similar logic with conditional probability in (c). The calculations are routine with no conceptual challenges beyond organizing cases, making it slightly easier than average for S1. |
| Spec | 2.04a Discrete probability distributions |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T=2) = 3 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{12}\) oe | M1 A1 | M1: correct expression; A1: allow exact equivalent (\(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\) is M0A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T=3) = [P(0,3) + P(1,2) + P(2,1)] + P(3)\) | ||
| \(= \left(\frac{1}{6}\times\frac{1}{2}\right)+\left(\frac{1}{6}\times\frac{1}{6}\right)+\left(\frac{1}{6}\times\frac{1}{6}\right)+\frac{1}{2}\) | M1 M1 | First M1: for \(\frac{1}{2}\) + at least one correct product; Second M1: fully correct expression |
| \(= \frac{23}{36}\) oe | A1 | Allow exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T=3 \mid \text{rolled twice}) = \frac{P((T=3) \cap \text{die rolled twice})}{P(\text{die rolled twice})}\) | M1 | For correct conditional probability ratio; using numerator \(P(T=3) \times P(\text{rolled twice})\) is M0M0A0 |
| \(= \frac{\frac{5}{36}}{\frac{1}{2}}\) | M1 | For correct numerical ratio; allow ft of (their (b) \(- \frac{1}{2}\)) as numerator |
| \(= \frac{5}{18}\) oe | A1 | Allow exact equivalent |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T=2) = 3 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{12}$ oe | M1 A1 | M1: correct expression; A1: allow exact equivalent ($\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$ is M0A0) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T=3) = [P(0,3) + P(1,2) + P(2,1)] + P(3)$ | | |
| $= \left(\frac{1}{6}\times\frac{1}{2}\right)+\left(\frac{1}{6}\times\frac{1}{6}\right)+\left(\frac{1}{6}\times\frac{1}{6}\right)+\frac{1}{2}$ | M1 M1 | First M1: for $\frac{1}{2}$ + at least one correct product; Second M1: fully correct expression |
| $= \frac{23}{36}$ oe | A1 | Allow exact equivalent |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T=3 \mid \text{rolled twice}) = \frac{P((T=3) \cap \text{die rolled twice})}{P(\text{die rolled twice})}$ | M1 | For correct conditional probability ratio; using numerator $P(T=3) \times P(\text{rolled twice})$ is M0M0A0 |
| $= \frac{\frac{5}{36}}{\frac{1}{2}}$ | M1 | For correct numerical ratio; allow ft of (their (b) $- \frac{1}{2}$) as numerator |
| $= \frac{5}{18}$ oe | A1 | Allow exact equivalent |
---\begin{enumerate}
\item A biased tetrahedral die has faces numbered $0,1,2$ and 3 . The die is rolled and the number face down on the die, $X$, is recorded. The probability distribution of $X$ is
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 2 }$ \\
\hline
\end{tabular}
\end{center}
If $X = 3$ then the final score is 3\\
If $X \neq 3$ then the die is rolled again and the final score is the sum of the two numbers.\\
The random variable $T$ is the final score.\\
(a) Find $\mathrm { P } ( T = 2 )$\\
(b) Find $\mathrm { P } ( T = 3 )$\\
(c) Given that the die is rolled twice, find the probability that the final score is 3\\
$\_\_\_\_$ VAYV SIHI NI JIIIM ION OC\\
VAYV SIHIL NI JIIIMM ION OC\\
VJYV SIHI NI JLIYM ION OC
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel S1 2018 Q5 [8]}}