| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Standard +0.8 This question requires understanding of conditional probability rules, systematic enumeration of outcomes, and rigorous application of the independence definition P(A∩B)=P(A)P(B) for multiple pairs. Part (i) demands careful case analysis with conditional scoring rules, while part (ii) requires calculating six different probabilities and testing three pairs—going well beyond routine exercises to test deep understanding of independence. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) \(P(\text{final score is } 12) = P(6,6) = 1/36\) | B1 [1] | Correct answer |
| (b) \(P[(1,5) + (1,4) + (2,3) + (3,2) + (4,1)]\) | M1 | Considering \(P(1, 5)\) |
| M1 | Considering \(P[(1,4) + (2,3) + (3,2) + (4,1)]\) | |
| \(= 5/36\) | A1 [3] | Correct answer |
| (ii) \(P(A) = 1/6\) | B1 | Any two of \(P(A)\), \(P(B)\) and \(P(C)\) correct |
| \(P(B) = P[(1,5) + (2,4) + (3,3) + (4,2) + (5,1)] = 5/36\) | B1 | Third probability correct |
| \(P(C) = 1 - P(O, O) = 3/4\) | ||
| \(P(A \text{ and } B) = P(1 \text{ and } 5) = 1/36 \neq P(A) \times P(B)\) | M1 | Numerical attempt to compare \(P(X \text{ and } Y)\) with \(P(X) \times P(Y)\), must be three positive probs |
| \(P(A \text{ and } C) = P[(2,5) + (4,5) + (6,5)] = 3/36 \neq P(A) \times P(C)\) | ||
| \(P(B \text{ and } C) = P[(2,4) + (4,2)] = 2/36 \neq P(B) \times P(C)\) | ||
| None are independent. | A1\(\vee\) | One correct comparison and conclusion, fl their probabilities |
| A1 [5] | Correct conclusion(s) following legitimate working |
**(i)** **(a)** $P(\text{final score is } 12) = P(6,6) = 1/36$ | B1 [1] | Correct answer
**(b)** $P[(1,5) + (1,4) + (2,3) + (3,2) + (4,1)]$ | M1 | Considering $P(1, 5)$
| M1 | Considering $P[(1,4) + (2,3) + (3,2) + (4,1)]$
$= 5/36$ | A1 [3] | Correct answer
**(ii)** $P(A) = 1/6$ | B1 | Any two of $P(A)$, $P(B)$ and $P(C)$ correct
$P(B) = P[(1,5) + (2,4) + (3,3) + (4,2) + (5,1)] = 5/36$ | B1 | Third probability correct
$P(C) = 1 - P(O, O) = 3/4$ | |
$P(A \text{ and } B) = P(1 \text{ and } 5) = 1/36 \neq P(A) \times P(B)$ | M1 | Numerical attempt to compare $P(X \text{ and } Y)$ with $P(X) \times P(Y)$, must be three positive probs
$P(A \text{ and } C) = P[(2,5) + (4,5) + (6,5)] = 3/36 \neq P(A) \times P(C)$ | |
$P(B \text{ and } C) = P[(2,4) + (4,2)] = 2/36 \neq P(B) \times P(C)$ | |
None are independent. | A1$\vee$ | One correct comparison and conclusion, fl their probabilities
| A1 [5] | Correct conclusion(s) following legitimate working
---4 Tim throws a fair die twice and notes the number on each throw.\\
(i) Tim calculates his final score as follows. If the number on the second throw is a 5 he multiplies the two numbers together, and if the number on the second throw is not a 5 he adds the two numbers together. Find the probability that his final score is
\begin{enumerate}[label=(\alph*)]
\item 12,
\item 5 .\\
(ii) Events $A , B , C$ are defined as follows.\\
$A$ : the number on the second throw is 5\\
$B$ : the sum of the numbers is 6\\
$C$ : the product of the numbers is even\\
By calculation find which pairs, if any, of the events $A , B$ and $C$ are independent.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2011 Q4 [9]}}