| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Challenging +1.2 This question requires students to use the constraint that probabilities sum to 1, combined with logarithm properties to solve log₃₆(a) + log₃₆(b) + log₃₆(c) = 1, leading to abc = 36. Finding distinct positive integers satisfying this with ordering constraints requires systematic problem-solving beyond routine exercises, but the logarithm manipulation and factorization are standard A-level techniques. Part (b) is straightforward application of independence once part (a) is solved. |
| Spec | 2.04a Discrete probability distributions |
| \(x\) | \(a\) | \(b\) | \(c\) |
| \(\mathrm { P } ( X = x )\) | \(\log _ { 36 } a\) | \(\log _ { 36 } b\) | \(\log _ { 36 } c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \([\text{Sum of probs} = 1 \text{ implies}]\) \(\log_{36} a + \log_{36} b + \log_{36} c = 1\) | M1 |
| \(\Rightarrow \log_{36}(abc) = 1\) so \(abc = 36\) | A1 | |
| All probabilities greater than 0 implies each of \(a, b\) and \(c > 1\) | B1 | |
| \(36 = 2^2 \times 3^2\) (or 3 numbers that multiply to give 36 e.g. 2, 2, 9 etc ) | dM1 | |
| Since \(a, b\) and \(c\) are distinct must be \(\mathbf{2, 3, 6}\) \((\mathbf{a = 2, b = 3, c = 6})\) | A1 | |
| (b) | \((\log_{36} a)^2 + (\log_{36} b)^2 + (\log_{36} c)^2\) | M1 |
| \([= 0.0374137...+ 0.0939873...+0.251]\) | A1 | A1 for awrt 0.381 |
| \(= 0.38140...\) awrt \(\mathbf{0.381}\) |
| Answer | Marks |
|---|---|
| Item | Note |
| (a) | 1st M1 for a start to the problem using sum of probabilities leading to eq'n in \(a, b\) and \(c\). 1st A1 for reducing to the equation \(abc = 36\) [Must follow from their equation.] Can go straight from \(abc = 36\) to the answer for full marks for part (a). B1 for deducing that each value \(> 1\) (may be implied by 3 integers all \(> 1\) in the next line). 2nd dM1 (dep on M1A1) for writing 36 as a product of prime factors or 3 values with product = 36 and none = 1. 2nd A1 for \(2, 3\) and \(6\) as a list or \(a = 2, b = 3\) and \(c = 6\) |
| SC: M0M0 If no method marks scored but a correct answer given score: M0A0B1M0A1 (2/5). This gets the SC score of 2/5 [Question says show your working clearly] | |
| (b) | M1 for a correct expression in terms of \(a, b\) and \(c\) or values; ft their integers \(a, b\) and \(c\). Condone invisible brackets if the answer implies they are used. A1 for awrt 0.381 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $[\text{Sum of probs} = 1 \text{ implies}]$ $\log_{36} a + \log_{36} b + \log_{36} c = 1$ | M1 | 1st M1 for a start to the problem using sum of probabilities leading to eq'n in $a, b$ and $c$. 1st A1 for reducing to the equation $abc = 36$ [Must follow from their equation.] Can go straight from $abc = 36$ to the answer for full marks for part (a). B1 for deducing that each value $> 1$ (may be implied by 3 integers all $> 1$ in the next line). dM1 (dep on M1A1) for writing 36 as a product of prime factors or 3 values with product = 36 and none = 1. 2nd A1 for $2, 3$ and $6$ as a list or $a = 2, b = 3$ and $c = 6$. |
| | $\Rightarrow \log_{36}(abc) = 1$ so $abc = 36$ | A1 | |
| | All probabilities greater than 0 implies each of $a, b$ and $c > 1$ | B1 | |
| | $36 = 2^2 \times 3^2$ (or 3 numbers that multiply to give 36 e.g. 2, 2, 9 etc ) | dM1 | |
| | Since $a, b$ and $c$ are distinct must be $\mathbf{2, 3, 6}$ $(\mathbf{a = 2, b = 3, c = 6})$ | A1 | |
| (b) | $(\log_{36} a)^2 + (\log_{36} b)^2 + (\log_{36} c)^2$ | M1 | M1 for a correct expression in terms of $a, b$ and $c$ or values; ft their integers $a, b$ and $c$. Condone invisible brackets if the answer implies they are used. |
| | $[= 0.0374137...+ 0.0939873...+0.251]$ | A1 | A1 for awrt 0.381 |
| | $= 0.38140...$ awrt $\mathbf{0.381}$ | | |
**Notes:**
| Item | Note |
|------|------|
| (a) | 1st M1 for a start to the problem using sum of probabilities leading to eq'n in $a, b$ and $c$. 1st A1 for reducing to the equation $abc = 36$ [Must follow from their equation.] Can go straight from $abc = 36$ to the answer for full marks for part (a). B1 for deducing that each value $> 1$ (may be implied by 3 integers all $> 1$ in the next line). 2nd dM1 (dep on M1A1) for writing 36 as a product of prime factors or 3 values with product = 36 and none = 1. 2nd A1 for $2, 3$ and $6$ as a list or $a = 2, b = 3$ and $c = 6$ |
| | **SC:** M0M0 If no method marks scored but a correct answer given score: **M0A0B1M0A1** (2/5). **This gets the SC score of 2/5** [Question says show your working clearly] |
| (b) | M1 for a correct expression in terms of $a, b$ and $c$ or values; ft their integers $a, b$ and $c$. Condone invisible brackets if the answer implies they are used. A1 for awrt 0.381 |\begin{enumerate}
\item The discrete random variable $X$ has the following probability distribution
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & $a$ & $b$ & $c$ \\
\hline
$\mathrm { P } ( X = x )$ & $\log _ { 36 } a$ & $\log _ { 36 } b$ & $\log _ { 36 } c$ \\
\hline
\end{tabular}
\end{center}
where
\begin{itemize}
\item $\quad a , b$ and $c$ are distinct integers $( a < b < c )$
\item all the probabilities are greater than zero\\
(a) Find\\
(i) the value of a\\
(ii) the value of $b$\\
(iii) the value of $c$
\end{itemize}
Show your working clearly.
The independent random variables $X _ { 1 }$ and $X _ { 2 }$ each have the same distribution as $X$\\
(b) Find $\mathrm { P } \left( X _ { 1 } = X _ { 2 } \right)$
\section*{Question 6 continued.}
\section*{Question 6 continued.}
\hfill \mbox{\textit{Edexcel Paper 3 2021 Q6 [7]}}