Edexcel S1 2011 January — Question 7 17 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2011
SessionJanuary
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTree Diagrams
TypeCalculate combined outcome probability
DifficultyStandard +0.3 This is a standard S1 tree diagram question with sequential sampling without replacement. While it has multiple parts and requires careful probability calculations including conditional probability, all techniques are routine for this module. The structure is clearly guided (complete diagram, show that...) and requires no novel insight—just systematic application of probability rules.
Spec2.03a 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
  1. The bag \(P\) contains 6 balls of which 3 are red and 3 are yellow.
The bag \(Q\) contains 7 balls of which 4 are red and 3 are yellow.
A ball is drawn at random from bag \(P\) and placed in bag \(Q\). A second ball is drawn at random from bag \(P\) and placed in bag \(Q\).
A third ball is then drawn at random from the 9 balls in bag \(Q\). The event \(A\) occurs when the 2 balls drawn from bag \(P\) are of the same colour. The event \(B\) occurs when the ball drawn from bag \(Q\) is red.
  1. Complete the tree diagram shown below.
    (4) \includegraphics[max width=\textwidth, alt={}, center]{c78ec7b6-dd06-4de1-94c2-052a5577dd10-12_1201_1390_753_269}
  2. Find \(\mathrm { P } ( A )\)
  3. Show that \(\mathrm { P } ( B ) = \frac { 5 } { 9 }\)
  4. Show that \(\mathrm { P } ( A \cap B ) = \frac { 2 } { 9 }\)
  5. Hence find \(\mathrm { P } ( A \cup B )\)
  6. Given that all three balls drawn are the same colour, find the probability that they are all red.
    (3)
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Both \(\frac{2}{3}, \frac{1}{3}\) on treeB1
\(\frac{4}{9}\) on treeB1
Both \(\frac{3}{5}, \frac{2}{5}\) on treeB1
All three of \(\frac{4}{9}, \frac{4}{9}, \frac{5}{9}\) on treeB1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(A) = P(RR) + P(YY) = \frac{1}{2}\times\frac{2}{5} + \frac{1}{2}\times\frac{2}{5} = \frac{2}{5}\)B1 M1 A1 B1 for \(\frac{1}{2}\times\frac{2}{5}\) (oe) seen at least once. M1 and A1 for correct final answer
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(B) = P(RRR) + P(RYR) + P(YRR) + P(YYR)\)M1 M1 for at least 1 case of 3 balls identified (implied by 2nd M1)
\(\left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right) + \left(\frac{1}{2}\times\frac{3}{5}\times\frac{5}{9}\right) + \left(\frac{1}{2}\times\frac{3}{5}\times\frac{5}{9}\right) + \left(\frac{1}{2}\times\frac{2}{5}\times\frac{4}{9}\right) = \frac{5}{9}\)M1, A1cso
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(A\cap B) = P(RRR) + P(YYR)\)M1 M1 for identifying both cases and adding probabilities (may be implied by correct expressions)
\(= \left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right) + \left(\frac{1}{2}\times\frac{2}{5}\times\frac{4}{9}\right) = \frac{2}{9}\)A1cso
Part (e):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(A\cup B) = P(A) + P(B) - P(A\cap B)\)M1 Must have some attempt to use formula
\(= \frac{2}{5} + \frac{5}{9} - \frac{2}{9} = \frac{11}{15}\)A1cao
Question 7 (part f):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{P(RRR)}{P(RRR)+P(YYY)} = \frac{\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}}{\left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right)+\left(\frac{1}{2}\times\frac{2}{5}\times\frac{5}{9}\right)} = \frac{6}{11}\)M1 For identifying the correct probabilities and forming appropriate fraction of probs. Probabilities must come from the product of 3 probs. from their tree diagram.
Correct expression using probabilities from treeA1ft For a correct expression using probabilities from their tree
\(= \frac{6}{11}\)A1 cao Accept exact decimal equivalents. Correct answer only is full marks except in (c) and (d)
(3) 
# Question 7:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Both $\frac{2}{3}, \frac{1}{3}$ on tree | B1 | |
| $\frac{4}{9}$ on tree | B1 | |
| Both $\frac{3}{5}, \frac{2}{5}$ on tree | B1 | |
| All three of $\frac{4}{9}, \frac{4}{9}, \frac{5}{9}$ on tree | B1 | |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A) = P(RR) + P(YY) = \frac{1}{2}\times\frac{2}{5} + \frac{1}{2}\times\frac{2}{5} = \frac{2}{5}$ | B1 M1 A1 | B1 for $\frac{1}{2}\times\frac{2}{5}$ (oe) seen at least once. M1 and A1 for correct final answer |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B) = P(RRR) + P(RYR) + P(YRR) + P(YYR)$ | M1 | M1 for at least 1 case of 3 balls identified (implied by 2nd M1) |
| $\left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right) + \left(\frac{1}{2}\times\frac{3}{5}\times\frac{5}{9}\right) + \left(\frac{1}{2}\times\frac{3}{5}\times\frac{5}{9}\right) + \left(\frac{1}{2}\times\frac{2}{5}\times\frac{4}{9}\right) = \frac{5}{9}$ | M1, A1cso | |

## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A\cap B) = P(RRR) + P(YYR)$ | M1 | M1 for identifying both cases and adding probabilities (may be implied by correct expressions) |
| $= \left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right) + \left(\frac{1}{2}\times\frac{2}{5}\times\frac{4}{9}\right) = \frac{2}{9}$ | A1cso | |

## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ | M1 | Must have some attempt to use formula |
| $= \frac{2}{5} + \frac{5}{9} - \frac{2}{9} = \frac{11}{15}$ | A1cao | |

## Question 7 (part f):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{P(RRR)}{P(RRR)+P(YYY)} = \frac{\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}}{\left(\frac{1}{2}\times\frac{2}{5}\times\frac{2}{3}\right)+\left(\frac{1}{2}\times\frac{2}{5}\times\frac{5}{9}\right)} = \frac{6}{11}$ | M1 | For identifying the correct probabilities and forming appropriate fraction of probs. Probabilities must come from the product of 3 probs. from their tree diagram. |
| Correct expression using probabilities from tree | A1ft | For a correct expression using probabilities from their tree |
| $= \frac{6}{11}$ | A1 cao | Accept exact decimal equivalents. Correct answer only is full marks except in (c) and (d) |
| | **(3)** | |

---
\begin{enumerate}
  \item The bag $P$ contains 6 balls of which 3 are red and 3 are yellow.
\end{enumerate}

The bag $Q$ contains 7 balls of which 4 are red and 3 are yellow.\\
A ball is drawn at random from bag $P$ and placed in bag $Q$. A second ball is drawn at random from bag $P$ and placed in bag $Q$.\\
A third ball is then drawn at random from the 9 balls in bag $Q$.

The event $A$ occurs when the 2 balls drawn from bag $P$ are of the same colour. The event $B$ occurs when the ball drawn from bag $Q$ is red.\\
(a) Complete the tree diagram shown below.\\
(4)\\
\includegraphics[max width=\textwidth, alt={}, center]{c78ec7b6-dd06-4de1-94c2-052a5577dd10-12_1201_1390_753_269}\\
(b) Find $\mathrm { P } ( A )$\\
(c) Show that $\mathrm { P } ( B ) = \frac { 5 } { 9 }$\\
(d) Show that $\mathrm { P } ( A \cap B ) = \frac { 2 } { 9 }$\\
(e) Hence find $\mathrm { P } ( A \cup B )$\\
(f) Given that all three balls drawn are the same colour, find the probability that they are all red.\\
(3)

\hfill \mbox{\textit{Edexcel S1 2011 Q7 [17]}}
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