| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Expected profit or cost problem |
| Difficulty | Moderate -0.8 This is a straightforward expected value problem with independent events. Part (i) is simple probability calculation (0.8 × 0.8), part (ii) requires basic tree diagram thinking, and part (iii) is direct application of E(X) = Σxp(x). All steps are routine S1 techniques with no problem-solving insight required, making it easier than average. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Amy's gain (\$) | |||
| Probability | 0.64 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{loses \\)1}) = P(\text{F and F}) = 0.8 \times 0.8\( | M1 | \)0.8 \times 0.8\( or \)(1-0.2)(1-0.2)\( or \)P(F) \times P(F)\( or \)P(F)+P(F)$ seen or implied |
| \(= 0.64\) AG | A1 | Must see probabilities multiplied together with final answer and a clear probability statement or implied by labelled tree diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Amount gained (\\(): \)-1\(, \)0.50\(, \)2\( | B1 | \)-1$ linked with 0.64 in table |
| Prob: \([0.64]\), \(0.16\), \(0.2\) | B1 | 0.5 seen in table |
| B1 | 0.16 seen in table linked to their 0.5 | |
| B1 | FT \(P(2.00 \text{ gained}) = 0.36 - P(0.50 \text{ gained})\) or correct, and all amount gained linked correctly in table |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(\text{winnings}) = -1 \times 0.64 + 0.5 \times 0.16 + 2 \times 0.2 = -(\\))0.16\(, \)-16\( cents | B1 | FT Accept (\\))0.16 or 16 cents loss. FT unsimplified \(E(\text{winnings})\) from their table provided \(\Sigma p = 1\) |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{loses \$1}) = P(\text{F and F}) = 0.8 \times 0.8$ | M1 | $0.8 \times 0.8$ or $(1-0.2)(1-0.2)$ or $P(F) \times P(F)$ or $P(F)+P(F)$ seen or implied |
| $= 0.64$ AG | A1 | Must see probabilities multiplied together with final answer and a clear probability statement or implied by labelled tree diagram |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Amount gained (\$): $-1$, $0.50$, $2$ | B1 | $-1$ linked with 0.64 in table |
| Prob: $[0.64]$, $0.16$, $0.2$ | B1 | 0.5 seen in table |
| | B1 | 0.16 seen in table linked to their 0.5 |
| | B1 | **FT** $P(2.00 \text{ gained}) = 0.36 - P(0.50 \text{ gained})$ or correct, and all amount gained linked correctly in table |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(\text{winnings}) = -1 \times 0.64 + 0.5 \times 0.16 + 2 \times 0.2 = -(\$)0.16$, $-16$ cents | B1 | **FT** Accept (\$)0.16 or 16 cents **loss**. FT unsimplified $E(\text{winnings})$ from their table provided $\Sigma p = 1$ |
---6 At a funfair, Amy pays $\$ 1$ for two attempts to make a bell ring by shooting at it with a water pistol.
\begin{itemize}
\item If she makes the bell ring on her first attempt, she receives $\$ 3$ and stops playing. This means that overall she has gained $\$ 2$.
\item If she makes the bell ring on her second attempt, she receives $\$ 1.50$ and stops playing. This means that overall she has gained $\$ 0.50$.
\item If she does not make the bell ring in the two attempts, she has lost her original $\$ 1$.
\end{itemize}
The probability that Amy makes the bell ring on any attempt is 0.2 , independently of other attempts.\\
(i) Show that the probability that Amy loses her original $\$ 1$ is 0.64 .\\
(ii) Complete the probability distribution table for the amount that Amy gains.
\begin{center}
\begin{tabular}{ | l | c | l | l | }
\hline
Amy's gain (\$) & & & \\
\hline
Probability & 0.64 & & \\
\hline
\end{tabular}
\end{center}
(iii) Calculate Amy's expected gain.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [7]}}