| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring basic probability calculations from independent trials and standard expectation/variance formulas. Part (i) involves listing simple cases (0+2, 1+1, 2+0) and multiplying probabilities, while part (ii) is direct application of memorized formulas with arithmetic. No problem-solving insight needed, just routine procedure. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Number | Probability |
| 0 | 0.7 |
| 1 | 0.2 |
| 2 | 0.1 |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | 0.49 | 0.28 | 0.18 | 0.04 | 0.01 |
1 Each time a certain triangular spinner is spun, it lands on one of the numbers 0,1 and 2 with probabilities as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Number & Probability \\
\hline
0 & 0.7 \\
\hline
1 & 0.2 \\
\hline
2 & 0.1 \\
\hline
\end{tabular}
\end{center}
The spinner is spun twice. The total of the two numbers on which it lands is denoted by $X$.\\
(i) Show that $\mathrm { P } ( X = 2 ) = 0.18$.
The probability distribution of $X$ is given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & 0.49 & 0.28 & 0.18 & 0.04 & 0.01 \\
\hline
\end{tabular}
\end{center}
(ii) Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S1 2009 Q1 [8]}}