| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This question requires constructing a new probability distribution from independent trials and calculating expectation. While it involves multiple steps (listing outcomes, computing probabilities using independence, organizing into a table), the concepts are straightforward S1 material with no novel insight required. The systematic enumeration of cases and basic probability calculations make this slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 2 | 4 | 6 |
| \(\mathrm { P } ( X = x )\) | 0.5 | 0.4 | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| y | 0 | 2 |
| P(Y = y) | 0.42 | 0.48 |
| B1 | 0, 2, 4 only seen for Y no probs needed. Accept other vals if P(value) = 0 seen in table, allow 0002244 with probs | |
| M1 | Summing two or more 2-factor probs (can be implied) | |
| A1 | Once correct prob | |
| A1 [4] | Correct table or list | |
| (ii) \(0.96 + 0.4 = 1.36\) | B1ft [1] | Ft their table for Y or X \(\Sigma p = 1\) |
(i)
| y | 0 | 2 | 4 |
|P(Y = y)| 0.42 | 0.48 | 0.1 |
| B1 | 0, 2, 4 only seen for Y no probs needed. Accept other vals if P(value) = 0 seen in table, allow 0002244 with probs
| M1 | Summing two or more 2-factor probs (can be implied)
| A1 | Once correct prob
| A1 [4] | Correct table or list
(ii) $0.96 + 0.4 = 1.36$ | B1ft [1] | Ft their table for Y or X $\Sigma p = 1$2 The random variable $X$ has the probability distribution shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 2 & 4 & 6 \\
\hline
$\mathrm { P } ( X = x )$ & 0.5 & 0.4 & 0.1 \\
\hline
\end{tabular}
\end{center}
Two independent values of $X$ are chosen at random. The random variable $Y$ takes the value 0 if the two values of $X$ are the same. Otherwise the value of $Y$ is the larger value of $X$ minus the smaller value of $X$.\\
(i) Draw up the probability distribution table for $Y$.\\
(ii) Find the expected value of $Y$.
\hfill \mbox{\textit{CAIE S1 2012 Q2 [5]}}