| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Repeated trials until stopping condition |
| Difficulty | Moderate -0.3 This is a straightforward tree diagram problem with conditional probability. Part (a) is routine construction, part (b) requires careful enumeration but is guided by 'show that', parts (c) and (d) are standard applications of independence and conditional probability formulas. The stopping condition adds mild complexity but the calculations are methodical rather than insightful. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| First throw fully correct with all probabilities \(\frac{1}{6}\) and outcomes \((1,2,3,4,5,6)\) on branches | B1 | 1st throw fully correct with probabilities and outcomes identified |
| Second throw fully correct with all probabilities \(\frac{1}{6}\) and outcomes \((1,2,3,4,5,6)\) on branches | B1 | 2nd throw fully correct with probabilities and outcomes identified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 5 comes from 1+4 or 5: \(P(5) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} = \frac{7}{36}\) | B1 | \(P(5)\) or \(P(7)\) identified and correct unsimplified, accept if supported by correct scenarios shown or from tree diagram |
| 6 comes from 1+5: \(P(6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) | ||
| 7 comes from 1+6 or 6+1: \(P(7) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}\) | ||
| 8 comes from 6+2: \(P(8) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) | ||
| 9 comes from 6+3: \(P(9) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) | ||
| \(P(A) = \frac{7}{36} + \frac{1}{36} + \frac{2}{36} + \frac{1}{36} + \frac{1}{36}\) | M1 | Adding only the values from 5 correct scenarios |
| \(= \frac{12}{36} = \frac{1}{3}\) | A1 | Scenarios identified (may be on tree diagram in 5(a)), all probabilities seen, WWW AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(B) = \frac{1}{3}\), \(P(A \cap B) = \frac{6}{36}\) | M1 | Both identified and evaluated, consistent with *their* tree diagram or correct |
| \(P(A)P(B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\) | A1 | \(P(A) \times P(B)\) seen and evaluated, all notation present and correct. Correct conclusion WWW |
| \(\frac{6}{36} \neq \frac{1}{9}\), so not independent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(B \mid A') = \frac{P(B \cap A')}{P(A')} = \frac{\text{their } \frac{6}{36}}{\frac{2}{3}}\) | B1 | \(\frac{6}{36}\) oe as numerator of a fraction |
| M1 | \(\dfrac{\text{their } \frac{6}{36}}{\text{their } 1 - \frac{1}{3}}\) seen, consistent with *their* tree diagram | |
| \(\frac{1}{4}\), 0.25 | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| First throw fully correct with all probabilities $\frac{1}{6}$ and outcomes $(1,2,3,4,5,6)$ on branches | B1 | 1st throw fully correct with probabilities and outcomes identified |
| Second throw fully correct with all probabilities $\frac{1}{6}$ and outcomes $(1,2,3,4,5,6)$ on branches | B1 | 2nd throw fully correct with probabilities and outcomes identified |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| 5 comes from 1+4 or 5: $P(5) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} = \frac{7}{36}$ | B1 | $P(5)$ or $P(7)$ identified and correct unsimplified, accept if supported by correct scenarios shown or from tree diagram |
| 6 comes from 1+5: $P(6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ | | |
| 7 comes from 1+6 or 6+1: $P(7) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$ | | |
| 8 comes from 6+2: $P(8) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ | | |
| 9 comes from 6+3: $P(9) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ | | |
| $P(A) = \frac{7}{36} + \frac{1}{36} + \frac{2}{36} + \frac{1}{36} + \frac{1}{36}$ | M1 | Adding only the values from 5 correct scenarios |
| $= \frac{12}{36} = \frac{1}{3}$ | A1 | Scenarios identified (may be on tree diagram in 5(a)), all probabilities seen, WWW AG |
**Total: 3 marks**
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(B) = \frac{1}{3}$, $P(A \cap B) = \frac{6}{36}$ | M1 | Both identified and evaluated, consistent with *their* tree diagram or correct |
| $P(A)P(B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ | A1 | $P(A) \times P(B)$ seen and evaluated, all notation present and correct. Correct conclusion WWW |
| $\frac{6}{36} \neq \frac{1}{9}$, so not independent | | |
**Total: 2 marks**
---
## Question 5(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(B \mid A') = \frac{P(B \cap A')}{P(A')} = \frac{\text{their } \frac{6}{36}}{\frac{2}{3}}$ | B1 | $\frac{6}{36}$ oe as numerator of a fraction |
| | M1 | $\dfrac{\text{their } \frac{6}{36}}{\text{their } 1 - \frac{1}{3}}$ seen, consistent with *their* tree diagram |
| $\frac{1}{4}$, 0.25 | A1 | |
**Total: 3 marks**
---5 A game is played with an ordinary fair 6-sided die. A player throws the die once. If the result is $2,3,4$ or 5 , that result is the player's score and the player does not throw the die again. If the result is 1 or 6 , the player throws the die a second time and the player's score is the sum of the two numbers from the two throws.
\begin{enumerate}[label=(\alph*)]
\item Draw a fully labelled tree diagram to represent this information.
Events $A$ and $B$ are defined as follows.\\
$A$ : the player's score is $5,6,7,8$ or 9\\
$B$ : the player has two throws
\item Show that $\mathrm { P } ( A ) = \frac { 1 } { 3 }$.
\item Determine whether or not events $A$ and $B$ are independent.
\item Find $\mathrm { P } \left( B \mid A ^ { \prime } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q5 [10]}}