| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Combined hypergeometric and binomial |
| Difficulty | Standard +0.3 Part (a) is a straightforward hypergeometric distribution with small numbers (3 letters from 8, counting A's), requiring basic probability calculations. Part (b) is a standard binomial probability calculation using complement rule. Both parts are routine applications of standard distributions with no conceptual challenges beyond recognizing the setup. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(X = 0, 1, 2\) only | B1 | Identifying that \(X\) can only take values 0, 1 and 2 (if other values stated, must be associated with probability of 0) |
| \([P(X=0) =]\, \frac{6}{8} \times \frac{5}{7} \times \frac{4}{6}\) | M1 | Correct expression for \(P(X=0)\) |
| \([P(X=1) =]\, 3 \times \frac{2}{8} \times \frac{6}{7} \times \frac{5}{6}\) or \([P(X=2) =]\, 3 \times \frac{2}{8} \times \frac{1}{7} \times \frac{6}{6}\) | M1 | Correct expression for either \(P(X=1)\) or \(P(X=2)\). Note: \(\frac{6}{8} \times \frac{5}{7} = \frac{15}{28}\) is incorrect attempt at \(P(X=0)\) and scores M0A0 |
| \(x\) | \(0\) | \(1\) |
| \(P(X=x)\) | \(\dfrac{5}{14}\) | \(\dfrac{15}{28}\) |
| A1 | One correct probability | |
| A1 | Complete probability distribution; need not be in table but each value of \(x\) must be associated with its probability. Allow awrt 0.357, awrt 0.536, awrt 0.107 |
| Answer | Marks | Guidance |
|---|---|---|
| \(J \sim B\!\left(10,\ \tfrac{1}{9}\right)\) | M1 | Identifying Binomial distribution with \(n=10\) is appropriate; if distribution not stated may be implied by use of \((10Cr)p^r(1-p)^{10-r}\) or \(0.981(57\ldots)\) |
| \(P(J \geq 4) = 1 - P(J \leq 3)\) or \(P(J=4)+P(J=5)+\cdots+P(J=10)\) or \(1 - 0.981(57\ldots)\) | M1 | Writing or using a correct probability statement |
| \(= \text{awrt } 0.0184\) | A1 | Correct answer scores 3 out of 3 |
## Question 5:
**Part (a):**
$X = 0, 1, 2$ only | B1 | Identifying that $X$ can only take values 0, 1 and 2 (if other values stated, must be associated with probability of 0)
$[P(X=0) =]\, \frac{6}{8} \times \frac{5}{7} \times \frac{4}{6}$ | M1 | Correct expression for $P(X=0)$
$[P(X=1) =]\, 3 \times \frac{2}{8} \times \frac{6}{7} \times \frac{5}{6}$ or $[P(X=2) =]\, 3 \times \frac{2}{8} \times \frac{1}{7} \times \frac{6}{6}$ | M1 | Correct expression for either $P(X=1)$ or $P(X=2)$. Note: $\frac{6}{8} \times \frac{5}{7} = \frac{15}{28}$ is incorrect attempt at $P(X=0)$ and scores M0A0
| $x$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $P(X=x)$ | $\dfrac{5}{14}$ | $\dfrac{15}{28}$ | $\dfrac{3}{28}$ |
| A1 | One correct probability
| A1 | Complete probability distribution; need not be in table but each value of $x$ must be associated with its probability. Allow awrt 0.357, awrt 0.536, awrt 0.107
**Part (b):**
$J \sim B\!\left(10,\ \tfrac{1}{9}\right)$ | M1 | Identifying Binomial distribution with $n=10$ is appropriate; if distribution not stated may be implied by use of $(10Cr)p^r(1-p)^{10-r}$ or $0.981(57\ldots)$
$P(J \geq 4) = 1 - P(J \leq 3)$ or $P(J=4)+P(J=5)+\cdots+P(J=10)$ or $1 - 0.981(57\ldots)$ | M1 | Writing or using a correct probability statement
$= \text{awrt } 0.0184$ | A1 | Correct answer scores 3 out of 3\begin{enumerate}
\item Julia selects 3 letters at random, one at a time without replacement, from the word
\end{enumerate}
\section*{VARIANCE}
The discrete random variable $X$ represents the number of times she selects a letter A.\\
(a) Find the complete probability distribution of $X$.
Yuki selects 10 letters at random, one at a time with replacement, from the word
\section*{DEVIATION}
(b) Find the probability that he selects the letter E at least 4 times.
\hfill \mbox{\textit{Edexcel AS Paper 2 2023 Q5 [8]}}