| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.8 This is a multi-part question requiring understanding of conditional probability, the relationship P(S∩{X=x}) = P(S|{X=x})·P(X=x), and solving simultaneous equations with the constraint that probabilities sum to 1. Part (a) requires algebraic manipulation to show c=8b/5, part (b) requires solving for all four probabilities using two constraints, and part (c) tests conceptual understanding of model validity. While systematic, it requires careful reasoning across multiple probability concepts beyond routine textbook exercises. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \(x\) | 20 | 50 | 80 | 100 |
| \(\mathrm { P } ( X = x )\) | \(a\) | \(b\) | \(c\) | \(d\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(S \cap \{X=50\}) = P(S \cap \{X=80\})\ [= \text{a constant},\ V] \Rightarrow b \times \dfrac{k}{50} = c \times \dfrac{k}{80}\) | M1 | For use of \(P(S \mid X=x) \times P(X=x)\) for \(x=50\) and \(x=80\). Any expression or equation must be based on the probability statements in the question. NB: Use of values e.g. \(b = \frac{50}{20+50+80+100}\) to prove (a) is M0A0 but scores 2nd M1A1 in (b). |
| Answer | Marks | Guidance |
|---|---|---|
| So \(c = \dfrac{8}{5}b\) * | A1cso (2) | A1cso for rearranging to required result, no incorrect work seen, condone poor notation. |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = 2b\) or \(a = \dfrac{2}{5}b\) or \(c = 4a\) or \(d = 5a\) or \(d = \dfrac{5}{4}c\) | M1 | 1st M1 for at least one other relationship (either probability the subject) from the list. Or allow: \(\dfrac{ak}{20} = \dfrac{bk}{50} = \dfrac{ck}{80} = \dfrac{dk}{100}\). |
| A1 | 1st A1 for a second different relationship (either probability the subject) from the list. | |
| \(\dfrac{2}{5}b + b + \dfrac{8}{5}b + 2b = 1\) | M1 | 2nd M1 for using sum of prob's \(= 1\). May be implied by one correct probability. |
| \(\Rightarrow 5b = 1\) so \(b = \dfrac{1}{5}\) (o.e.) | A1 | 2nd A1 for one correct probability e.g. \(b = \frac{1}{5}\) or exact equivalent such as 0.2. |
| \(a = \dfrac{2}{25} \quad b = \dfrac{1}{5} \quad c = \dfrac{8}{25} \quad d = \dfrac{2}{5}\) | A1 (5) | 3rd A1 for all correct probabilities. Allow exact equivalents e.g. \(c = 0.32\). Sight of correct distribution or list of probs with no obvious incorrect working is 5/5. |
| Answer | Marks | Guidance |
|---|---|---|
| So model won't work since \(P(S \mid X=20) = \dfrac{30}{20}\) or \(\dfrac{0.12}{0.08}\) and so would be greater than 1 | B1 (1) | For deducing \(k = 30\) and giving a suitable example to show model breaks down. |
# Question 5:
## Part (a)
$P(S \cap \{X=50\}) = P(S \cap \{X=80\})\ [= \text{a constant},\ V] \Rightarrow b \times \dfrac{k}{50} = c \times \dfrac{k}{80}$ | M1 | For use of $P(S \mid X=x) \times P(X=x)$ for $x=50$ **and** $x=80$. Any expression or equation must be based on the probability statements in the question. **NB:** Use of values e.g. $b = \frac{50}{20+50+80+100}$ to prove (a) is M0A0 but scores 2nd M1A1 in (b).
May see: $\dfrac{k}{50} = \dfrac{V}{b}$ and $\dfrac{k}{80} = \dfrac{V}{c}$ (condone any letter for $V$ even $S$)
So $c = \dfrac{8}{5}b$ * | A1cso **(2)** | A1cso for rearranging to required result, no incorrect work seen, condone poor notation.
## Part (b)
$d = 2b$ **or** $a = \dfrac{2}{5}b$ **or** $c = 4a$ **or** $d = 5a$ **or** $d = \dfrac{5}{4}c$ | M1 | 1st M1 for at least one other relationship (either probability the subject) from the list. Or allow: $\dfrac{ak}{20} = \dfrac{bk}{50} = \dfrac{ck}{80} = \dfrac{dk}{100}$.
| A1 | 1st A1 for a second different relationship (either probability the subject) from the list.
$\dfrac{2}{5}b + b + \dfrac{8}{5}b + 2b = 1$ | M1 | 2nd M1 for using sum of prob's $= 1$. May be implied by one correct probability.
$\Rightarrow 5b = 1$ so $b = \dfrac{1}{5}$ (o.e.) | A1 | 2nd A1 for one correct probability e.g. $b = \frac{1}{5}$ or exact equivalent such as 0.2.
$a = \dfrac{2}{25} \quad b = \dfrac{1}{5} \quad c = \dfrac{8}{25} \quad d = \dfrac{2}{5}$ | A1 **(5)** | 3rd A1 for all correct probabilities. Allow exact equivalents e.g. $c = 0.32$. **Sight of correct distribution or list of probs with no obvious incorrect working is 5/5.**
## Part (c)
[Experiment suggests for Nav] $P(S \mid \{X=100\}) = 0.3 \Rightarrow k = 30$
or $0.3 = \dfrac{V}{0.4} \Rightarrow V = 0.12$
So model won't work since $P(S \mid X=20) = \dfrac{30}{20}$ or $\dfrac{0.12}{0.08}$ and so would be greater than 1 | B1 **(1)** | For deducing $k = 30$ **and** giving a suitable example to show model breaks down.\begin{enumerate}
\item Tisam is playing a game.
\end{enumerate}
She uses a ball, a cup and a spinner.\\
The random variable $X$ represents the number the spinner lands on when it is spun. The probability distribution of $X$ is given in the following table
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 20 & 50 & 80 & 100 \\
\hline
$\mathrm { P } ( X = x )$ & $a$ & $b$ & $c$ & $d$ \\
\hline
\end{tabular}
\end{center}
where $a , b , c$ and $d$ are probabilities.\\
To play the game
\begin{itemize}
\item the spinner is spun to obtain a value of $x$
\item Tisam then stands $x \mathrm {~cm}$ from the cup and tries to throw the ball into the cup
\end{itemize}
The event $S$ represents the event that Tisam successfully throws the ball into the cup.\\
To model this game Tisam assumes that
\begin{itemize}
\item $\mathrm { P } ( S \mid \{ X = x \} ) = \frac { k } { x }$ where $k$ is a constant
\item $\mathrm { P } ( S \cap \{ X = x \} )$ should be the same whatever value of $x$ is obtained from the spinner
\end{itemize}
Using Tisam's model,\\
(a) show that $c = \frac { 8 } { 5 } b$\\
(b) find the probability distribution of $X$
Nav tries, a large number of times, to throw the ball into the cup from a distance of 100 cm .\\
He successfully gets the ball in the cup $30 \%$ of the time.\\
(c) State, giving a reason, why Tisam's model of this game is not suitable to describe Nav playing the game for all values of $X$
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q5 [8]}}