| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.8 This is a straightforward application of Bayes' theorem with clearly defined probabilities and a simple algebraic equation. Part (a) requires basic tree diagram probability calculations with no conceptual challenges, and part (b) involves setting up and solving a linear equation from given probability information. Both parts are routine S1-level exercises requiring only standard techniques. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(W_2) = P(W_1W_2) + P(L_1W_2)\) | B1 | 0.3 × 0.6 alone as num or denom of a fraction |
| \(= 0.3 \times 0.6 + 0.7 \times 0.15 = 0.285\) | M1 | Attempt at P(W₂) as sum of two 2-factor options seen anywhere |
| \(P(W_1 | W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)} = \frac{0.18}{0.285}\) | A1 |
| \(= 0.632, \frac{12}{19}\) | A1 [4] | Correct answer |
| (b) \(x + 4\) oe seen | B1 | Seen anywhere |
| \(\frac{10}{15} \times \frac{7}{x+4} = \frac{7}{18}\) | M1 | Mult two probabilities, one containing \(x\) and equating to \(\frac{7}{18}\) |
| \(x = 8\) | A1 | Correct unsimplified equation |
| A1 [4] | Correct answer |
**(a)** $P(W_2) = P(W_1W_2) + P(L_1W_2)$ | B1 | 0.3 × 0.6 alone as num or denom of a fraction
$= 0.3 \times 0.6 + 0.7 \times 0.15 = 0.285$ | M1 | Attempt at P(W₂) as sum of two 2-factor options seen anywhere
$P(W_1|W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)} = \frac{0.18}{0.285}$ | A1 | Correct unsimplified P(W₂) as num or denom of a fraction
$= 0.632, \frac{12}{19}$ | A1 [4] | Correct answer
**(b)** $x + 4$ oe seen | B1 | Seen anywhere
$\frac{10}{15} \times \frac{7}{x+4} = \frac{7}{18}$ | M1 | Mult two probabilities, one containing $x$ and equating to $\frac{7}{18}$
$x = 8$ | A1 | Correct unsimplified equation
| A1 [4] | Correct answer5
\begin{enumerate}[label=(\alph*)]
\item John plays two games of squash. The probability that he wins his first game is 0.3 . If he wins his first game, the probability that he wins his second game is 0.6 . If he loses his first game, the probability that he wins his second game is 0.15 . Given that he wins his second game, find the probability that he won his first game.
\item Jack has a pack of 15 cards. 10 cards have a picture of a robot on them and 5 cards have a picture of an aeroplane on them. Emma has a pack of cards. 7 cards have a picture of a robot on them and $x - 3$ cards have a picture of an aeroplane on them. One card is taken at random from Jack's pack and one card is taken at random from Emma's pack. The probability that both cards have pictures of robots on them is $\frac { 7 } { 18 }$. Write down an equation in terms of $x$ and hence find the value of $x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2013 Q5 [8]}}