| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Conditional probability from tree |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram probability question requiring basic probability rules: (i) simple multiplication along branches, (ii) addition of multiple branch probabilities, (iii) conditional probability using P(A|B) = P(A∩B)/P(B). All techniques are standard S1 content with no conceptual challenges or novel problem-solving required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks |
|---|---|
| (i) | P(Wet and bus) = 0.4 0.7 |
| = 0.28 | M1 for multiplying |
| Answer | Marks | Guidance |
|---|---|---|
| A1 CAO | 2 | Fractional answer = 7/25 (Allow 28/100) |
| (ii) | P(Walk or bike) = |
| Answer | Marks |
|---|---|
| = 0.66 | M1 for any two correct |
| Answer | Marks | Guidance |
|---|---|---|
| A1 CAO | 3 | Or = 0.6 0.9 + 0.4 0.3 gets M1 for either term |
| Answer | Marks |
|---|---|
| (iii) | P(Dry and walk or bike) |
| Answer | Marks |
|---|---|
| 0.66 0.66 11 | M1 for numerator |
| Answer | Marks | Guidance |
|---|---|---|
| A1 FT | 3 | Allow 0.82, not 0.819 More accurate answer |
| Answer | Marks |
|---|---|
| TOTAL | 8 |
Question 7:
--- 7
(i) ---
7
(i) | P(Wet and bus) = 0.4 0.7
= 0.28 | M1 for multiplying
probabilities
A1 CAO | 2 | Fractional answer = 7/25 (Allow 28/100)
(ii) | P(Walk or bike) =
0.6 0.5 + 0.6 0.4 + 0.4 0.2 + 0.4 0.1 or
0.3+0.24+0.08+0.04
= 0.66 | M1 for any two correct
pairs
M1 for sum of all four
correct terms
With no extra terms for
second M1
A1 CAO | 3 | Or = 0.6 0.9 + 0.4 0.3 gets M1 for either term
= 0.54 + 0.12 gets M1 for sum of both
A1 CAO
Or = 1 – 0.6×0.1 – 0.4×0.7 = 0.66. M1 for 1 – one
correct term, M1 for complete correct expression and
A1 for correct evaluation.
(iii) | P(Dry and walk or bike)
P(Dry given walk or bike)
P(Walk or bike)
0.60.9 0.54 9
= = = = 0.818
0.66 0.66 11 | M1 for numerator
leading to 0.54
M1 for denominator
Ft their P(Walk or bike)
from (ii) provided
between 0 and 1
A1 FT | 3 | Allow 0.82, not 0.819 More accurate answer
=0.81818 Fractional answer = 54/66 = 27/33 = 9/11
Condone answer of 0.8181
Do not give final A1 if ans ≥ 1
TOTAL | 8Andy can walk to work, travel by bike or travel by bus. The tree diagram shows the probabilities of any day being dry or wet and the corresponding probabilities for each of Andy's methods of travel.
\includegraphics{figure_7}
A day is selected at random. Find the probability that
\begin{enumerate}[label=(\roman*)]
\item the weather is wet and Andy travels by bus, [2]
\item Andy walks or travels by bike, [3]
\item the weather is dry given that Andy walks or travels by bike. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q7 [8]}}