| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes and their probabilities. Students must list all possible combinations (1p or 2p from A; two coins from B), calculate totals and medians, then aggregate probabilities. While it requires careful organization and basic probability calculations with ratios (P(1p from A) = 1/4, P(2p from A) = 3/4, etc.), it's a standard S2 exercise with no conceptual surprises—slightly easier than average due to its mechanical nature. |
| Spec | 5.02a Discrete probability distributions: general |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Box A: \(P(1) = \frac{1}{4}\) \(P(2) = \frac{3}{4}\) | B1 |
| Box B: \(P(2) = \frac{1}{5}\) \(P(5) = \frac{4}{5}\) | B1 | |
| Totals (T) 5, 6, 8, 9, 11, 12 | B1 | |
| (1, 2, 2) (1, 2, 5) (1, 5, 5) (2, 2, 2) (2, 2, 5) (2, 5, 5) [(1, 5, 2)] [(2, 5, 2)] | M1 | |
| \([P(T=5) =] \frac{1}{4} \times \frac{1}{5} \times \frac{1}{5} = \left[\frac{1}{100}\right]\) | M1 | |
| \([P(T=6) =] \frac{3}{4} \times \frac{1}{5} \times \frac{1}{5} = \left[\frac{3}{100}\right]\) | M1 | |
| \([P(T=8) =] 2 \times \frac{1}{4} \times \frac{1}{5} \times \frac{4}{5} = \left[\frac{8}{100}\right]\) | M1 | |
| \([P(T=9) =] 2 \times \frac{3}{4} \times \frac{1}{5} \times \frac{4}{5} = \left[\frac{24}{100}\right]\) | M1 | |
| \([P(T=11) =] \frac{1}{4} \times \frac{4}{5} \times \frac{4}{5} = \left[\frac{16}{100}\right]\) | M1 | |
| \([P(T=12) =] \frac{3}{4} \times \frac{4}{5} \times \frac{4}{5} = \left[\frac{48}{100}\right]\) | M1 | |
| (7) | ||
| t | 5 | 6 |
| \(P(T=t)\) | \(\frac{1}{100}\) | \(\frac{3}{100}\) |
| (b) | \(m = 2\) \(m = 5\) | B1 |
| \([P(M=2) =]\) \(\frac{1}{100} + \frac{8}{100} + \frac{3}{100} + \frac{24}{100} = \left[\frac{36}{100}\right]\) | M1 | |
| \([P(M=5) =]\) \(\frac{16}{100} + \frac{48}{100} = \left[\frac{64}{100}\right]\) or \(P(M=5) = 1 - 'P(M=2)'\) | M1 | |
| m | 2 | 5 |
| \(P(M=m)\) | \(\frac{36}{100}\) | \(\frac{64}{100}\) |
| Total | 11 |
| (a) | Box A: $P(1) = \frac{1}{4}$ $P(2) = \frac{3}{4}$ | B1 | |
| | Box B: $P(2) = \frac{1}{5}$ $P(5) = \frac{4}{5}$ | B1 | |
| | Totals (T) 5, 6, 8, 9, 11, 12 | B1 | |
| | (1, 2, 2) (1, 2, 5) (1, 5, 5) (2, 2, 2) (2, 2, 5) (2, 5, 5) [(1, 5, 2)] [(2, 5, 2)] | M1 | |
| | $[P(T=5) =] \frac{1}{4} \times \frac{1}{5} \times \frac{1}{5} = \left[\frac{1}{100}\right]$ | M1 | |
| | $[P(T=6) =] \frac{3}{4} \times \frac{1}{5} \times \frac{1}{5} = \left[\frac{3}{100}\right]$ | M1 | |
| | $[P(T=8) =] 2 \times \frac{1}{4} \times \frac{1}{5} \times \frac{4}{5} = \left[\frac{8}{100}\right]$ | M1 | |
| | $[P(T=9) =] 2 \times \frac{3}{4} \times \frac{1}{5} \times \frac{4}{5} = \left[\frac{24}{100}\right]$ | M1 | |
| | $[P(T=11) =] \frac{1}{4} \times \frac{4}{5} \times \frac{4}{5} = \left[\frac{16}{100}\right]$ | M1 | |
| | $[P(T=12) =] \frac{3}{4} \times \frac{4}{5} \times \frac{4}{5} = \left[\frac{48}{100}\right]$ | M1 | |
| | | | (7) |
| | | | |
| | | t | 5 | 6 | 8 | 9 | 11 | 12 |
| | | $P(T=t)$ | $\frac{1}{100}$ | $\frac{3}{100}$ | $\frac{8}{100}$ $\frac{2}{25}$ | $\frac{24}{100}$ $\frac{6}{25}$ | $\frac{16}{100}$ $\frac{4}{25}$ | $\frac{48}{100}$ $\frac{12}{25}$ | A1 | (7) |
| (b) | $m = 2$ $m = 5$ | B1 | |
| | $[P(M=2) =]$ $\frac{1}{100} + \frac{8}{100} + \frac{3}{100} + \frac{24}{100} = \left[\frac{36}{100}\right]$ | M1 | |
| | $[P(M=5) =]$ $\frac{16}{100} + \frac{48}{100} = \left[\frac{64}{100}\right]$ or $P(M=5) = 1 - 'P(M=2)'$ | M1 | |
| | | m | 2 | 5 |
| | | $P(M=m)$ | $\frac{36}{100}$ | $\frac{64}{100}$ | A1 | (4) |
| | **Total** | **11** | |
**Notes for Question 6:**
- **(a) B1:** All 4 correct probabilities – may be seen in an equation
- **(a) B1:** All 6 totals correct with no extras (ignore units if stated) (condone 8 or 9 listed twice)
- **(a) B1:** All 6 basic combinations correct, either seen or used (implied by the 3rd M1 mark)
- Condone any permutation of the 6 basic combinations for this mark
- **(a) M1:** Correct method for **one** probability (ft their probabilities)
- **(a) M1:** Correct method for **five** probabilities (ft their probabilities)
- **(a) M1:** Correct method for **all six** probabilities (ft their probabilities)
- **(a) A1:** cao Need not be in a table but probabilities must be attached to the correct total
- **(b) B1:** For both values of $m$ (no extras) If $m = 1$ is stated it must be stated that its probability is 0
- **(b) M1:** Ft part (a) For a correct method to find $P(M = '2')$ For this mark there must only be 2 probability calculations
- **(b) M1:** Ft part (a) For a correct method to find $P(M = '5')$ For this mark there must only be 2 probability calculations
- **(b) A1:** cao Need not be in a table but probabilities must be attached to the correct total
---\begin{enumerate}
\item Two boxes, A and B , each contain a large number of coins.
\end{enumerate}
In box A
\begin{itemize}
\item there are only 1 p coins and 2 p coins
\item the ratio of 1 p coins to 2 p coins is $1 : 3$
\end{itemize}
In box B
\begin{itemize}
\item there are only 2 p coins and 5 p coins
\item the ratio of 2 p coins to 5 p coins is $1 : 4$
\end{itemize}
One coin is randomly selected from box A and two coins are randomly selected from box B
The random variable $T$ represents the total of the values of the three coins selected.\\
(a) Find the sampling distribution of $T$
The random variable $M$ represents the median of the values of the three coins selected.\\
(b) Find the sampling distribution of $M$
\hfill \mbox{\textit{Edexcel S2 2024 Q6}}