| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Standard +0.3 This is a straightforward discrete probability question requiring basic probability axioms (part a: sum to 1), simple probability calculations with independence (part b: geometric-type probability), constructing a probability distribution (part c), and careful interpretation of events (part d). All techniques are standard AS-level material with no novel insight required, though part (d) requires careful thinking about what the condition means. Slightly easier than average due to the routine nature of most parts. |
| Spec | 2.04a Discrete probability distributions5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| \(x\) | 6 | 7 | 8 | 10 |
| \(\mathrm { P } ( X = x )\) | 0.5 | 0.2 | \(q\) | \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([2q = 0.3] \quad \left[q = \frac{1-(0.5+0.2)}{2}\right] \quad [q=] \quad \mathbf{0.15}\) | B1 | For \(q = 0.15\) o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Realising require sequence: \(\bar{7}, \bar{7}, 7\); may see \(0.8 \times 0.8 \times 0.2\) o.e. \(= \mathbf{0.128}\) | M1 | For evidence that a correct sequence has been applied. Allow a clear list of all 9 possibilities |
| A1* | For 0.128 from a correct expression with no incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Possible values for \(S\) are: 1, 2, 3 or 4 only | B1 | For a correct sample space for \(S\). If any other values stated they must be attached to probability of 0 |
| \([P(S=1)] = 0.2\) and \([P(S=2)] = 0.8 \times 0.2 = 0.16\) | M1 | For using the given model to find both values of \(P(S=1)\) and \(P(S=2)\) |
| \(P(S=4) = 0.8^3 \times 0.2 + 0.8^4\ [= 0.512]\) or \(1-[P(S=1)'+P(S=2)'+0.128]\) | M1 | For a correct method to find \(P(S=4)\) |
| Full correct table: \(P(S=1)=0.2=\frac{1}{5}\), \(P(S=2)=0.16=\frac{4}{25}\), \(P(S=3)=0.128=\frac{16}{125}\), \(P(S=4)=0.512=\frac{64}{125}\) | A1 | For a fully correct probability distribution in table or listed separately. Must be in terms of \(S\) for this mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([= 1 - P(S=1=N)] = 1 - 0.2 = \mathbf{0.8}\) | B1 | For 0.8 |
## Question 5:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[2q = 0.3] \quad \left[q = \frac{1-(0.5+0.2)}{2}\right] \quad [q=] \quad \mathbf{0.15}$ | B1 | For $q = 0.15$ o.e. |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Realising require sequence: $\bar{7}, \bar{7}, 7$; may see $0.8 \times 0.8 \times 0.2$ o.e. $= \mathbf{0.128}$ | M1 | For evidence that a correct sequence has been applied. Allow a clear list of all 9 possibilities |
| | A1* | For 0.128 from a correct expression with no incorrect working seen |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Possible values for $S$ are: 1, 2, 3 or 4 only | B1 | For a correct sample space for $S$. If any other values stated they must be attached to probability of 0 |
| $[P(S=1)] = 0.2$ **and** $[P(S=2)] = 0.8 \times 0.2 = 0.16$ | M1 | For using the given model to find both values of $P(S=1)$ **and** $P(S=2)$ |
| $P(S=4) = 0.8^3 \times 0.2 + 0.8^4\ [= 0.512]$ **or** $1-[P(S=1)'+P(S=2)'+0.128]$ | M1 | For a correct method to find $P(S=4)$ |
| Full correct table: $P(S=1)=0.2=\frac{1}{5}$, $P(S=2)=0.16=\frac{4}{25}$, $P(S=3)=0.128=\frac{16}{125}$, $P(S=4)=0.512=\frac{64}{125}$ | A1 | For a fully correct probability distribution in table or listed separately. Must be in terms of $S$ for this mark |
**Part (d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $[= 1 - P(S=1=N)] = 1 - 0.2 = \mathbf{0.8}$ | B1 | For 0.8 |\begin{enumerate}
\item A biased 4 -sided spinner has the numbers $6,7,8$ and 10 on it.
\end{enumerate}
The discrete random variable $X$ represents the score when the spinner is spun once and has the following probability distribution,
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
$x$ & 6 & 7 & 8 & 10 \\
\hline
$\mathrm { P } ( X = x )$ & 0.5 & 0.2 & $q$ & $q$ \\
\hline
\end{tabular}
\end{center}
where $q$ is a probability.\\
(a) Find the value of $q$
Karen spins the spinner repeatedly until she either gets a 7 or she has taken 4 spins.\\
(b) Show that the probability that Karen stops after taking her 3rd spin is 0.128
The random variable $S$ represents the number of spins Karen takes.\\
(c) Find the probability distribution for $S$
The random variable $N$ represents the number of times Karen gets a 7\\
(d) Find $\mathrm { P } ( S > N )$
\hfill \mbox{\textit{Edexcel AS Paper 2 2024 Q5 [8]}}