OCR MEI AS Paper 2 Specimen — Question 4 5 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Definitions
TypeTwo-item selection from population
DifficultyModerate -0.8 This is a straightforward probability question requiring basic understanding of complementary probability (part a) and systematic enumeration of cases (part b). The calculations are routine with no conceptual challenges—students simply apply P(at least one O) = 1 - P(neither O) and sum probabilities for different blood group pairs. This is easier than average A-level work as it requires only basic probability rules without any problem-solving insight.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space

4 There are four human blood groups; these are called \(\mathrm { O } , \mathrm { A } , \mathrm { B }\) and AB . Each person has one of these blood groups. The table below shows the distribution of blood groups in a large country.
Blood group
Proportion of
population
O\(49 \%\)
A\(38 \%\)
B\(10 \%\)
AB\(3 \%\)
Two people are selected at random from this country.
  1. Find the probability that at least one of these two people has blood group O .
  2. Find the probability that each of these two people has a different blood group.

Question 4(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(1 - 0.51^2\)M1 (3.1b)
\(= 0.7399\)A1 (1.1) Accept 0.74 or 0.740
[2]
Question 4(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(1 - 0.49^2 - 0.38^2 - 0.1^2 - 0.03^2\)M1 (3.1b) For squaring probabilities OR products of pairs
M1 (1.1)For complementary event OR doubling products of pairs; \(2\times(0.1862 + 0.049 + 0.0147 + 0.038 + 0.0114 + 0.003)\)
\(= 0.6046\)A1 (1.1)
[3]
## Question 4(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.51^2$ | M1 (3.1b) | |
| $= 0.7399$ | A1 (1.1) | Accept 0.74 or 0.740 |
| **[2]** | | |

---

## Question 4(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.49^2 - 0.38^2 - 0.1^2 - 0.03^2$ | M1 (3.1b) | For squaring probabilities OR products of pairs |
| | M1 (1.1) | For complementary event OR doubling products of pairs; $2\times(0.1862 + 0.049 + 0.0147 + 0.038 + 0.0114 + 0.003)$ |
| $= 0.6046$ | A1 (1.1) | |
| **[3]** | | |
4 There are four human blood groups; these are called $\mathrm { O } , \mathrm { A } , \mathrm { B }$ and AB . Each person has one of these blood groups. The table below shows the distribution of blood groups in a large country.

\begin{center}
\begin{tabular}{ | c | c | }
\hline
Blood group & \begin{tabular}{ c }
Proportion of \\
population \\
\end{tabular} \\
\hline
O & $49 \%$ \\
\hline
A & $38 \%$ \\
\hline
B & $10 \%$ \\
\hline
AB & $3 \%$ \\
\hline
\end{tabular}
\end{center}

Two people are selected at random from this country.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that at least one of these two people has blood group O .
\item Find the probability that each of these two people has a different blood group.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI AS Paper 2  Q4 [5]}}