| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Standard +0.3 This is a straightforward tree diagram problem requiring basic probability calculations: (i) simple addition of probabilities along branches, (ii) multiplication along paths and addition across cases, (iii) conditional probability using Bayes' theorem. The multi-stage structure and stopping condition add slight complexity, but the calculations are routine and clearly guided by the question structure. Slightly easier than average due to clear setup and standard techniques. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{1st correct}) = 0.7 + 0.2 \times 0.95 = 0.89\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Considering any 2 of CC, CHA, HAC or HAHP [where \(C\) = Peter correct, \(H\) = ask for help, \(A\) = audience correct, \(P\) = phone correct] or tree diagram with 'top half' labels and probs shown | M1 | |
| Considering other 2 | M1 | |
| Summing 4 probabilities | M1 | |
| \(P(CC) = 0.7 \times 0.7 (= 0.49)\) and \(P(CHA) = 0.7 \times 0.2 \times 0.95 (= 0.133)\) and \(P(HAC) = 0.2 \times 0.95 \times 0.7 (= 0.133)\) and \(P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 (= 0.0247)\) | B1, B1 | Two correct probabilities; Three correct probabilities |
| \(P(\text{both correctly answered}) = 0.781\) | A1 | Correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{audience} \mid \text{both correct}) = \frac{P(CHA) + P(HAC) + P(HAHP)}{ans\text{ (ii)}}\) | M1* | Summing two or three 3-factor terms in numerator of a fraction |
| \(\frac{0.7 \times 0.2 \times 0.95 + 0.2 \times 0.95 \times 0.7 + 0.2 \times 0.95 \times 0.2 \times 0.65}{0.7807}\) | M1 dep | Dividing by their (ii) |
| \(= 0.2907/0.7807 = 0.372\) | A1 | Correct answer |
**(i) Probability:**
$P(\text{1st correct}) = 0.7 + 0.2 \times 0.95 = 0.89$ | B1 | | **[1]**
**(ii) Probability tree:**
Considering any 2 of CC, CHA, HAC or HAHP [where $C$ = Peter correct, $H$ = ask for help, $A$ = audience correct, $P$ = phone correct] or tree diagram with 'top half' labels and probs shown | M1 |
Considering other 2 | M1 |
Summing 4 probabilities | M1 |
$P(CC) = 0.7 \times 0.7 (= 0.49)$ and $P(CHA) = 0.7 \times 0.2 \times 0.95 (= 0.133)$ and $P(HAC) = 0.2 \times 0.95 \times 0.7 (= 0.133)$ and $P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 (= 0.0247)$ | B1, B1 | Two correct probabilities; Three correct probabilities |
$P(\text{both correctly answered}) = 0.781$ | A1 | Correct | **[6]**
**(iii) Conditional probability:**
$P(\text{audience} \mid \text{both correct}) = \frac{P(CHA) + P(HAC) + P(HAHP)}{ans\text{ (ii)}}$ | M1* | Summing two or three 3-factor terms in numerator of a fraction |
$\frac{0.7 \times 0.2 \times 0.95 + 0.2 \times 0.95 \times 0.7 + 0.2 \times 0.95 \times 0.2 \times 0.65}{0.7807}$ | M1 dep | Dividing by their (ii) |
$= 0.2907/0.7807 = 0.372$ | A1 | Correct answer | **[3]**7 In a television quiz show Peter answers questions one after another, stopping as soon as a question is answered wrongly.
\begin{itemize}
\item The probability that Peter gives the correct answer himself to any question is 0.7 .
\item The probability that Peter gives a wrong answer himself to any question is 0.1 .
\item The probability that Peter decides to ask for help for any question is 0.2 .
\end{itemize}
On the first occasion that Peter decides to ask for help he asks the audience. The probability that the audience gives the correct answer to any question is 0.95 . This information is shown in the tree diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{e7e0fcbe-ab96-4292-b3ad-c57b74f15301-3_394_649_1779_386}\\
\includegraphics[max width=\textwidth, alt={}, center]{e7e0fcbe-ab96-4292-b3ad-c57b74f15301-3_270_743_2010_1023}\\
(i) Show that the probability that the first question is answered correctly is 0.89 .
On the second occasion that Peter decides to ask for help he phones a friend. The probability that his friend gives the correct answer to any question is 0.65 .\\
(ii) Find the probability that the first two questions are both answered correctly.\\
(iii) Given that the first two questions were both answered correctly, find the probability that Peter asked the audience.
\hfill \mbox{\textit{CAIE S1 2010 Q7 [10]}}