OCR S1 2010 June — Question 5 12 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeSum or difference of two spinners/dice
DifficultyModerate -0.8 This is a straightforward S1 probability distribution question involving basic counting of outcomes (choosing 2 from 4 cards with one repeated value), calculating probabilities, and applying standard formulas for expectation and variance. The enumeration is simple with only 6 possible pairs, and all required techniques are routine applications of definitions with no problem-solving insight needed.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

Each of four cards has a number printed on it as shown.
1233
Two of the cards are chosen at random, without replacement. The random variable \(X\) denotes the sum of the numbers on these two cards.
  1. Show that P\((X = 6) = \frac{1}{6}\) and P\((X = 4) = \frac{1}{3}\). [3]
  2. Write down all the possible values of \(X\) and find the probability distribution of \(X\). [4]
  3. Find E\((X)\) and Var\((X)\). [5]

Each of four cards has a number printed on it as shown.

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & 2 & 3 & 3 \\
\hline
\end{tabular}
\end{center}

Two of the cards are chosen at random, without replacement. The random variable $X$ denotes the sum of the numbers on these two cards.

\begin{enumerate}[label=(\roman*)]
\item Show that P$(X = 6) = \frac{1}{6}$ and P$(X = 4) = \frac{1}{3}$. [3]
\item Write down all the possible values of $X$ and find the probability distribution of $X$. [4]
\item Find E$(X)$ and Var$(X)$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR S1 2010 Q5 [12]}}