| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Standard +0.3 This is a straightforward S1 question requiring standard probability calculations. Part (i) asks to verify a given probability using P(X=6) = P(at least one 6) = 1 - P(no 6s) = 1 - (5/6)³, which is a direct application of complement rule. Part (ii) requires routine calculation of E(X) and Var(X) using the given probability distribution table with the standard formulas. Both parts involve only mechanical computation with no problem-solving insight required, making this slightly easier than average. |
| Spec | 2.04a Discrete probability distributions |
| \(r\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 216 }\) | \(\frac { 7 } { 216 }\) | \(\frac { 19 } { 216 }\) | \(\frac { 37 } { 216 }\) | \(\frac { 61 } { 216 }\) | \(\frac { 91 } { 216 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=6) = 1 - P(X<6) = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216}\) | M1 | For \(\left(\frac{5}{6}\right)^3\) |
| M1 | For \(1 - \left(\frac{5}{6}\right)^3\) | |
| \(= \frac{91}{216}\) | A1 | NB ANSWER GIVEN |
| OR: \(= \left(\frac{1}{6}\right)^3 + 3\times\left(\frac{5}{6}\right)\times\left(\frac{1}{6}\right)^2 + 3\times\left(\frac{5}{6}\right)^2\times\left(\frac{1}{6}\right)\) | M1 | For second or third product term — Correct, including \(\times 3\) or probabilities seen on correct tree diagram |
| M1 | For attempt at three terms — With no extras, but allow omission of \(\times 3\); NB Zero for \(1-\)(sum of probs given in part (ii)) | |
| \(= \frac{91}{216}\) | A1 | NB ANSWER GIVEN |
| OR: \(1 + 15 + 75\), \(= \frac{1+15+75}{216}\) | M1, M1 | For 15 or 75 seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \left(1\times\frac{1}{216}\right)+\left(2\times\frac{7}{216}\right)+\left(3\times\frac{19}{216}\right)+\left(4\times\frac{37}{216}\right)+\left(5\times\frac{61}{216}\right)+\left(6\times\frac{91}{216}\right)\) | M1 | For \(\Sigma rp\) (at least 3 terms correct) |
| \(= \frac{1071}{216} = \frac{119}{24} = 4.96\) (exact: 4.9583333) | A1 | CAO — Accept fractional answers; do not allow answer of 5 unless more accurate answer given first |
| \(E(X^2) = \left(1\times\frac{1}{216}\right)+\left(4\times\frac{7}{216}\right)+\left(9\times\frac{19}{216}\right)+\left(16\times\frac{37}{216}\right)+\left(25\times\frac{61}{216}\right)+\left(36\times\frac{91}{216}\right)\) | M1* | For \(\Sigma r^2 p\) (at least 3 terms correct) — Use of \(E(X-\mu)^2\) gets M1 for attempt at \((x-\mu)^2\), should see \((-3.96)^2\), \((-2.96)^2\), \((-1.96)^2\), \((-0.96)^2\), \(0.04^2\), \(1.04^2\) |
| \(= \frac{5593}{216} = 25.89\) | ||
| \(\text{Var}(X) = 25.89\ldots - 4.958\ldots^2\) | M1* dep | For \(- \) their \((E(X))^2\) — Division by 6 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0 |
| \(= 1.31\) Accept answers in range 1.28 to 1.31 with correct working or \(\frac{2261}{1728}\) (Exact: 1.308449…) | A1 | FT their \(E(X)\) provided \(\text{Var}(X) > 0\) — Do not FT \(E(X)=5\) if full marks given for \(E(X)\); deduct at most 1 mark for over-specification of either mean or variance or both |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=6) = 1 - P(X<6) = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216}$ | M1 | For $\left(\frac{5}{6}\right)^3$ |
| | M1 | For $1 - \left(\frac{5}{6}\right)^3$ |
| $= \frac{91}{216}$ | A1 | **NB ANSWER GIVEN** |
| **OR:** $= \left(\frac{1}{6}\right)^3 + 3\times\left(\frac{5}{6}\right)\times\left(\frac{1}{6}\right)^2 + 3\times\left(\frac{5}{6}\right)^2\times\left(\frac{1}{6}\right)$ | M1 | For second or third product term — Correct, including $\times 3$ or probabilities seen on correct tree diagram |
| | M1 | For attempt at three terms — With no extras, but allow omission of $\times 3$; NB Zero for $1-$(sum of probs given in part (ii)) |
| $= \frac{91}{216}$ | A1 | **NB ANSWER GIVEN** |
| **OR:** $1 + 15 + 75$, $= \frac{1+15+75}{216}$ | M1, M1 | For 15 or 75 seen |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \left(1\times\frac{1}{216}\right)+\left(2\times\frac{7}{216}\right)+\left(3\times\frac{19}{216}\right)+\left(4\times\frac{37}{216}\right)+\left(5\times\frac{61}{216}\right)+\left(6\times\frac{91}{216}\right)$ | M1 | For $\Sigma rp$ (at least 3 terms correct) |
| $= \frac{1071}{216} = \frac{119}{24} = 4.96$ (exact: 4.9583333) | A1 | CAO — Accept fractional answers; do not allow answer of 5 unless more accurate answer given first |
| $E(X^2) = \left(1\times\frac{1}{216}\right)+\left(4\times\frac{7}{216}\right)+\left(9\times\frac{19}{216}\right)+\left(16\times\frac{37}{216}\right)+\left(25\times\frac{61}{216}\right)+\left(36\times\frac{91}{216}\right)$ | M1* | For $\Sigma r^2 p$ (at least 3 terms correct) — Use of $E(X-\mu)^2$ gets M1 for attempt at $(x-\mu)^2$, should see $(-3.96)^2$, $(-2.96)^2$, $(-1.96)^2$, $(-0.96)^2$, $0.04^2$, $1.04^2$ |
| $= \frac{5593}{216} = 25.89$ | | |
| $\text{Var}(X) = 25.89\ldots - 4.958\ldots^2$ | M1* dep | For $- $ their $(E(X))^2$ — Division by 6 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0 |
| $= 1.31$ Accept answers in range 1.28 to 1.31 with correct working or $\frac{2261}{1728}$ (Exact: 1.308449…) | A1 | FT their $E(X)$ provided $\text{Var}(X) > 0$ — Do not FT $E(X)=5$ if full marks given for $E(X)$; deduct at most 1 mark for over-specification of either mean or variance or both |
---2 Three fair six-sided dice are thrown. The random variable $X$ represents the highest of the three scores on the dice.\\
(i) Show that $\mathrm { P } ( X = 6 ) = \frac { 91 } { 216 }$.
The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 216 }$ & $\frac { 7 } { 216 }$ & $\frac { 19 } { 216 }$ & $\frac { 37 } { 216 }$ & $\frac { 61 } { 216 }$ & $\frac { 91 } { 216 }$ \\
\hline
\end{tabular}
\end{center}
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 Q2 [8]}}