| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Find population parameter from probability |
| Difficulty | Standard +0.3 This is a straightforward hypergeometric distribution problem requiring students to set up P(X=3) using combinations, solve a cubic equation (which factors nicely), and calculate probabilities. The algebra is routine and the question provides significant scaffolding by showing what to prove and verify, making it easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.01a Permutations and combinations: evaluate probabilities |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm {~F} ( x )\) | \(a\) | \(b\) | \(\frac { 37 } { 38 }\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | \(P(X = 3) = F(3) - F(2) = \frac{1}{38}\) | M1 |
| \(P(X = 3) = \frac{7}{n} \times \frac{6}{n-1} \times \frac{5}{n-2}\) | M1 | 2nd M1 product of 3 probabilities where the denominators are \(n\), (\(n - 1\)) and (\(n - 2\)) and the numerators are decreasing \(k\), (\(k - 1\)) and (\(k - 2\)) This may be seen as a single term in a longer expression |
| \(\frac{7 \times 6 \times 5}{n \times n-1 \times n-2} = \frac{1}{38} \rightarrow n(n-1)(n-2) = 7980\) | M1 A1cso (4) | 3rd M1 setting up equation for \(P(X = 3) = \) product of correct 3 probabilities without replacement. A1cso fully correct solution with no errors seen |
| (b) | \(21 \times 20 \times 19 = 7980\) | B1cso (1) |
| (c) | \(a = F(0) = P(X = 0) = \frac{14}{21} \times \frac{13}{20} \times \frac{12}{19}\) | M1 |
| 1st A1 \(a = \frac{26}{95}\) o.e must be clear this is the value for \(a\) | ||
| \(P(X = 1) = 3 \times \frac{14}{21} \times \frac{13}{20} \times \frac{7}{19} = \left[\frac{91}{190}\right]\) or \(P(X = 2) = 3 \times \frac{7}{21} \times \frac{6}{20} \times \frac{14}{19} = \left[\frac{21}{95}\right]\) | M1 M1 | 2nd M1 product of 3 probabilities for \(P(X = 1)\) or \(P(X = 2)\) or \(\frac{91}{190}\) or \(\frac{91}{570}\) or \(\frac{21}{95}\) or \(\frac{7}{95}\) o.e seen. Condone incorrect labelling. The three probabilities can be in any arrangement |
| \(b = F(1) = P(X = 0) + P(X = 1) = "n \frac{26}{95}" + \frac{91}{190}\) or \(b = \frac{37}{38} = "\frac{21}{95}" \) or \(= "b = \frac{143}{190}\) | dM1 | 3rd M1 × 3 or adding the 3 sets of the 3 fractions or \(\frac{91}{190}\) or \(\frac{91}{570}\) or \(\frac{21}{95}\) or \(\frac{7}{95}\) o.e seen. Condone incorrect labelling. 4th dM1 their \(P(X = 0) + \) their \(P(X = 1)\) or \(F(2) - P(X = 2)\) (dep on 2nd M1 being scored) |
| \(b = \frac{143}{190}\) | A1 (6) | 2nd A1 \(b = \frac{143}{190}\) o.e must be clear this is the value for \(b\) |
| [11] | NB if \(a = 0.273\ldots\) and \(b = 0.7526\) implies the method marks. |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | $P(X = 3) = F(3) - F(2) = \frac{1}{38}$ | M1 | 1st M1 for use of $F(3) - F(2)$ Accept $\frac{1}{38}$ |
| | $P(X = 3) = \frac{7}{n} \times \frac{6}{n-1} \times \frac{5}{n-2}$ | M1 | 2nd M1 product of 3 probabilities where the denominators are $n$, ($n - 1$) and ($n - 2$) and the numerators are decreasing $k$, ($k - 1$) and ($k - 2$) This may be seen as a single term in a longer expression |
| | $\frac{7 \times 6 \times 5}{n \times n-1 \times n-2} = \frac{1}{38} \rightarrow n(n-1)(n-2) = 7980$ | M1 A1cso (4) | 3rd M1 setting up equation for $P(X = 3) = $ product of correct 3 probabilities without replacement. A1cso fully correct solution with no errors seen |
| (b) | $21 \times 20 \times 19 = 7980$ | B1cso (1) | B1cso correctly evaluated product. Allow $21(21-1)(21-2) = 7980$ |
| (c) | $a = F(0) = P(X = 0) = \frac{14}{21} \times \frac{13}{20} \times \frac{12}{19}$ | M1 | 1st M1 product of 3 probabilities for $P(X = 0)$ The three probabilities can be in any arrangement May be implied by $\frac{26}{95}$ |
| | | | 1st A1 $a = \frac{26}{95}$ o.e must be clear this is the value for $a$ |
| | $P(X = 1) = 3 \times \frac{14}{21} \times \frac{13}{20} \times \frac{7}{19} = \left[\frac{91}{190}\right]$ or $P(X = 2) = 3 \times \frac{7}{21} \times \frac{6}{20} \times \frac{14}{19} = \left[\frac{21}{95}\right]$ | M1 M1 | 2nd M1 product of 3 probabilities for $P(X = 1)$ or $P(X = 2)$ or $\frac{91}{190}$ or $\frac{91}{570}$ or $\frac{21}{95}$ or $\frac{7}{95}$ o.e seen. Condone incorrect labelling. The three probabilities can be in any arrangement |
| | $b = F(1) = P(X = 0) + P(X = 1) = "n \frac{26}{95}" + \frac{91}{190}$ or $b = \frac{37}{38} = "\frac{21}{95}" $ or $= "b = \frac{143}{190}$ | dM1 | 3rd M1 × 3 or adding the 3 sets of the 3 fractions or $\frac{91}{190}$ or $\frac{91}{570}$ or $\frac{21}{95}$ or $\frac{7}{95}$ o.e seen. Condone incorrect labelling. 4th dM1 their $P(X = 0) + $ their $P(X = 1)$ or $F(2) - P(X = 2)$ (dep on 2nd M1 being scored) |
| | $b = \frac{143}{190}$ | A1 (6) | 2nd A1 $b = \frac{143}{190}$ o.e must be clear this is the value for $b$ |
| | | [11] | **NB** if $a = 0.273\ldots$ and $b = 0.7526$ implies the method marks. |7. A bag contains $n$ marbles of which 7 are green.
From the bag, 3 marbles are selected at random.\\
The random variable $X$ represents the number of green marbles selected.\\
The cumulative distribution function of $X$ is given by
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm {~F} ( x )$ & $a$ & $b$ & $\frac { 37 } { 38 }$ & 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $n ( n - 1 ) ( n - 2 ) = 7980$
\item Verify that $n = 21$ satisfies the equation in part (a).
Given that $n = 21$
\item find the exact value of $a$ and the exact value of $b$
\begin{center}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fa1cb8a2-dab9-4133-b7a1-9108888c37d7-28_2655_1947_114_116}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2022 Q7 [11]}}