CAIE S1 2019 June — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndependent Events
TypeTest independence using definition
DifficultyModerate -0.3 This is a straightforward application of the independence definition P(S∩T) = P(S)P(T) using systematic enumeration of outcomes from two dice. While it requires careful counting and organization, it's a standard textbook exercise with no conceptual subtlety—slightly easier than average due to the mechanical nature of the task.
Spec2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles

1 Two ordinary fair dice are thrown and the numbers obtained are noted. Event \(S\) is 'The sum of the numbers is even'. Event \(T\) is 'The sum of the numbers is either less than 6 or a multiple of 4 or both'. Showing your working, determine whether the events \(S\) and \(T\) are independent.

Question 1:
AnswerMarks Guidance
AnswerMark Guidance
\(P(S) = \frac{1}{2}\)B1
\(P(T) = \frac{16}{36} \left(\frac{4}{9}\right)\)B1
\(P(S \cap T) = \frac{10}{36} \left(\frac{5}{18}\right)\)M1 \(P(S \cap T)\) found by multiplication scores M0; M1 awarded if *their* value is identifiable in their sample space diagram or Venn diagram or list of terms or probability distribution table (oe)
\(P(S) \cdot P(T) \neq P(S \cap T)\) so not independentA1 \(\frac{8}{36}\), \(\frac{10}{36}\): \(P(S) \times P(T)\) and \(P(S \cap T)\) seen in workings and correct conclusion stated, www
Alternative method for Question 1:
AnswerMarks Guidance
AnswerMark Guidance
\(P(S) = \frac{1}{2}\)B1
\(P(T) = \frac{16}{36} \left(\frac{4}{9}\right)\)B1
\(P(S \cap T) = \frac{10}{36} \left(\frac{5}{18}\right)\)M1 \(P(S \cap T)\) found by multiplication scores M0; M1 awarded if *their* value is identifiable in their sample space diagram or Venn diagram or list of terms or probability distribution table (oe)
\(P(S\T) = \frac{10}{16}\) or \(P(T\ S) = \frac{10}{18}\); \(P(S\
Total4
## Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(S) = \frac{1}{2}$ | B1 | |
| $P(T) = \frac{16}{36} \left(\frac{4}{9}\right)$ | B1 | |
| $P(S \cap T) = \frac{10}{36} \left(\frac{5}{18}\right)$ | M1 | $P(S \cap T)$ found by multiplication scores M0; M1 awarded if *their* value is identifiable in their sample space diagram **or** Venn diagram **or** list of terms **or** probability distribution table (oe) |
| $P(S) \cdot P(T) \neq P(S \cap T)$ so not independent | A1 | $\frac{8}{36}$, $\frac{10}{36}$: $P(S) \times P(T)$ and $P(S \cap T)$ seen in workings and correct conclusion stated, www |

**Alternative method for Question 1:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(S) = \frac{1}{2}$ | B1 | |
| $P(T) = \frac{16}{36} \left(\frac{4}{9}\right)$ | B1 | |
| $P(S \cap T) = \frac{10}{36} \left(\frac{5}{18}\right)$ | M1 | $P(S \cap T)$ found by multiplication scores M0; M1 awarded if *their* value is identifiable in their sample space diagram **or** Venn diagram **or** list of terms **or** probability distribution table (oe) |
| $P(S\|T) = \frac{10}{16}$ or $P(T\|S) = \frac{10}{18}$; $P(S\|T) \neq P(S)$ or $P(T\|S) \neq P(T)$ so not independent | A1 | **Either** $\frac{18}{36}$, $\frac{10}{16}$, $P(S)$ and $P(S\|T)$ seen in workings and correct conclusion stated, www **Or** $\frac{16}{36}$, $\frac{10}{18}$, $P(T)$ and $P(T\|S)$ seen in workings and correct conclusion stated, www |
| **Total** | **4** | |
1 Two ordinary fair dice are thrown and the numbers obtained are noted. Event $S$ is 'The sum of the numbers is even'. Event $T$ is 'The sum of the numbers is either less than 6 or a multiple of 4 or both'. Showing your working, determine whether the events $S$ and $T$ are independent.\\

\hfill \mbox{\textit{CAIE S1 2019 Q1 [4]}}