| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.8 This AS-level question requires setting up and solving a system of three equations (probability sum, conditional probability relationship, and using an unstated constraint like variance or expectation) to find three unknowns, then calculating a compound probability across two independent trials. The multi-step algebraic manipulation and careful enumeration of ways to score 6 from two games makes this moderately challenging for AS level, though the underlying concepts are standard. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general |
| \(s\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( S = s )\) | \(a\) | \(b\) | \(c\) | 0.1 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Overall method | M1 | Fully correct method with all required steps. Gaining 2 correct equations with at least one correct (allow unsimplified). Attempting to solve to find value of \(c\) followed by correct method to find the probability |
| \(a+b=2c+0.5\) oe or \(a+b=2(1-a-b)\) | B1 | Forming a correct equation from information given |
| \(a+b+c=0.75\) oe | B1 | Correct equation using sum of probabilities equals 1 |
| \(3c=0.25 \quad \left[c=0.0833\ldots \text{ or } \frac{1}{12}\right]\) | M1 | Correct method for solving 2 equations to find \(c\), implied by \(c=\frac{1}{12}\) |
| \(P(\text{scoring 2,4 or 4,2 or 3,3}) = 2\times\text{"}\frac{1}{12}\text{"}\times0.15+0.1^2\) | M1 | Recognising the ways to get a total of 6. Condone missing arrangements or repeats. Do not ignore extras unless ignored in calculation. May be implied by \(m\times\text{"}\frac{1}{12}\text{"}\times0.15+n\times0.1^2\) where \(m,n\) are positive integers |
| \(= 0.035\) oe | A1cso | cao \(0.035\), \(\frac{7}{200}\) oe |
| (6) | 6 marks |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall method | M1 | Fully correct method with all required steps. Gaining 2 correct equations with at least one correct (allow unsimplified). Attempting to solve to find value of $c$ followed by correct method to find the probability |
| $a+b=2c+0.5$ oe **or** $a+b=2(1-a-b)$ | B1 | Forming a correct equation from information given |
| $a+b+c=0.75$ oe | B1 | Correct equation using sum of probabilities equals 1 |
| $3c=0.25 \quad \left[c=0.0833\ldots \text{ or } \frac{1}{12}\right]$ | M1 | Correct method for solving 2 equations to find $c$, implied by $c=\frac{1}{12}$ |
| $P(\text{scoring 2,4 or 4,2 or 3,3}) = 2\times\text{"}\frac{1}{12}\text{"}\times0.15+0.1^2$ | M1 | Recognising the ways to get a total of 6. Condone missing arrangements or repeats. Do not ignore extras unless ignored in calculation. May be implied by $m\times\text{"}\frac{1}{12}\text{"}\times0.15+n\times0.1^2$ where $m,n$ are positive integers |
| $= 0.035$ oe | A1cso | cao $0.035$, $\frac{7}{200}$ oe |
| | (6) | **6 marks** |\begin{enumerate}
\item In a game, a player can score $0,1,2,3$ or 4 points each time the game is played.
\end{enumerate}
The random variable $S$, representing the player's score, has the following probability distribution where $a , b$ and $c$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$s$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( S = s )$ & $a$ & $b$ & $c$ & 0.1 & 0.15 \\
\hline
\end{tabular}
\end{center}
The probability of scoring less than 2 points is twice the probability of scoring at least 2 points.
Each game played is independent of previous games played.\\
John plays the game twice and adds the two scores together to get a total.\\
Calculate the probability that the total is 6 points.
\hfill \mbox{\textit{Edexcel AS Paper 2 2020 Q3 [6]}}