| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Sampling without replacement from bags/boxes |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question with standard sampling without replacement calculations. Part (a) is basic complementary probability, (b) requires setting up and simplifying a probability equation (routine algebra), (c) is solving a quadratic, and (d) applies the conditional probability formula P(A|B) = P(A∩B)/P(B). All techniques are standard S1 material with clear scaffolding through the parts. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(G_1) + P(R_1 \cap G_2) + P(Y_1 \cap G_2)\) or \(P(GY)+P(GR)+P(RG)+P(YG)\) | M1 | For at least 2 correct cases. May be in symbols or probs. May be in tree diagram. Use of \(r=16\) or \(y=47\) can score maximum of 1st M1 then A0M0A0 |
| \(= \frac{1}{64} + \frac{r}{64}\times\frac{1}{63} + \frac{y}{64}\times\frac{1}{63} = \frac{1}{64} + \frac{r+y}{64\times63}\) or \(2\times\frac{r+y}{64\times63}\) | A1 | For all cases and their associated probs added |
| \(= \frac{1}{64} + \frac{63}{64\times63}\) or \(\frac{2\times63}{64\times63}\) or \(\frac{1}{64}+\frac{1}{64}\) | M1 | For combining probabilities and using \(r+y=63\) |
| \(= \frac{1}{32}\) or \(0.03125\) | A1 | For \(\frac{1}{32}\) or an exact equivalent (correct answer only 4/4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(R_1 \cap R_2) = \frac{r}{64}\times\frac{r-1}{63} = \frac{5}{84}\) | M1A1 | For \(\frac{r}{64}\times g(r) = ...\) where \(g(r)\) is any linear function of \(r\). For any correct equation in \(r\) |
| \(r(r-1) = 5\times64\times63\div84 = 240\) hence \(r^2 - r - 240 = 0\) or \(r^2 - r = 240\) | A1cso | For correctly simplifying to the given equation with no incorrect working seen. There should be at least 1 intermediate step seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r^2 - r - 240 = (r-16)(r+15)\{=0\}\) or \(16^2-16-240=256-256\) | M1 | For correct factors or completing square or use of formula or substitution |
| so \(r=16\) and rejecting \(-15\) | A1cso | For concluding \(r=16\) and rejecting \(-15\) (e.g. crossing out etc) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\geqslant 1\text{ red}) = P(RG)+P(GR)+P(RY)+P(YR)+P(RR)\) or \(\frac{2}{252}+\frac{2y}{252}+\frac{15}{252}\) | M1 | For a correct expression for at least one red. May be in symbols or probs. or in a tree |
| or \(P(R_1)+P(R_1'\cap R_2)\) or \(\frac{16}{64}+\frac{48}{64}\times\frac{16}{63}\) or \(1-\frac{48}{64}\times\frac{47}{63} = \frac{37}{84}\) | A1 | For \(\frac{37}{84}\) (o.e.) as a single fraction or awrt 0.440 [May be implied by correct answer] |
| \(\frac{P(R_1\cap R_2)}{P(\text{at least one red})} = \frac{\frac{5}{84}}{\text{``}\frac{37}{84}\text{''}} = \frac{5}{37}\) or \(0.1\dot{3}\dot{5}\) | M1, A1 | For a ratio of probabilities (denom may be in symbols) with numerator of \(\frac{5}{84}\) (o.e.). For \(\frac{5}{37}\) or an exact equivalent |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(G_1) + P(R_1 \cap G_2) + P(Y_1 \cap G_2)$ or $P(GY)+P(GR)+P(RG)+P(YG)$ | M1 | For at least 2 correct cases. May be in symbols or probs. May be in tree diagram. Use of $r=16$ or $y=47$ can score maximum of 1st M1 then A0M0A0 |
| $= \frac{1}{64} + \frac{r}{64}\times\frac{1}{63} + \frac{y}{64}\times\frac{1}{63} = \frac{1}{64} + \frac{r+y}{64\times63}$ or $2\times\frac{r+y}{64\times63}$ | A1 | For all cases and their associated probs added |
| $= \frac{1}{64} + \frac{63}{64\times63}$ or $\frac{2\times63}{64\times63}$ or $\frac{1}{64}+\frac{1}{64}$ | M1 | For combining probabilities and using $r+y=63$ |
| $= \frac{1}{32}$ or $0.03125$ | A1 | For $\frac{1}{32}$ or an exact equivalent (correct answer only 4/4) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R_1 \cap R_2) = \frac{r}{64}\times\frac{r-1}{63} = \frac{5}{84}$ | M1A1 | For $\frac{r}{64}\times g(r) = ...$ where $g(r)$ is any linear function of $r$. For any correct equation in $r$ |
| $r(r-1) = 5\times64\times63\div84 = 240$ hence $r^2 - r - 240 = 0$ or $r^2 - r = 240$ | A1cso | For correctly simplifying to the given equation with no incorrect working seen. There should be at least 1 intermediate step seen |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r^2 - r - 240 = (r-16)(r+15)\{=0\}$ or $16^2-16-240=256-256$ | M1 | For correct factors or completing square or use of formula or substitution |
| so $r=16$ and rejecting $-15$ | A1cso | For concluding $r=16$ and rejecting $-15$ (e.g. crossing out etc) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\geqslant 1\text{ red}) = P(RG)+P(GR)+P(RY)+P(YR)+P(RR)$ or $\frac{2}{252}+\frac{2y}{252}+\frac{15}{252}$ | M1 | For a correct expression for at least one red. May be in symbols or probs. or in a tree |
| or $P(R_1)+P(R_1'\cap R_2)$ or $\frac{16}{64}+\frac{48}{64}\times\frac{16}{63}$ or $1-\frac{48}{64}\times\frac{47}{63} = \frac{37}{84}$ | A1 | For $\frac{37}{84}$ (o.e.) as a single fraction or awrt 0.440 [May be implied by correct answer] |
| $\frac{P(R_1\cap R_2)}{P(\text{at least one red})} = \frac{\frac{5}{84}}{\text{``}\frac{37}{84}\text{''}} = \frac{5}{37}$ or $0.1\dot{3}\dot{5}$ | M1, A1 | For a ratio of probabilities (denom may be in symbols) with numerator of $\frac{5}{84}$ (o.e.). For $\frac{5}{37}$ or an exact equivalent |
---4.A bag contains 64 coloured beads.There are $r$ red beads,$y$ yellow beads and 1 green bead and $r + y + 1 = 64$
Two beads are selected at random,one at a time without replacement.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the green bead is one of the beads selected.
The probability that both of the beads are red is $\frac { 5 } { 84 }$
\item Show that $r$ satisfies the equation $r ^ { 2 } - r - 240 = 0$
\item Hence show that the only possible value of $r$ is 16
\item Given that at least one of the beads is red,find the probability that they are both red.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q4 [13]}}