OCR H240/02 — Question 14 8 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Marks8
PaperDownload PDF ↗
TopicDiscrete Probability Distributions
TypeConditional probability with random variables
DifficultyStandard +0.8 This question requires understanding of discrete probability distributions and conditional probability. Part (a) is straightforward calculation, but part (b) requires careful application of conditional probability P(A|B) = P(A∩B)/P(B), involving summing specific terms from the distribution where x is both even AND less than 10, then dividing by the sum over all even values. The algebraic manipulation of sums and the two-step conditional probability reasoning elevate this above routine exercises.
Spec2.04a Discrete probability distributions

14 A random variable \(X\) has probability distribution given by \(\mathrm { P } ( X = x ) = \frac { 1 } { 860 } ( 1 + x )\) for \(x = 1,2,3 , \ldots , 40\).
  1. Find \(\mathrm { P } ( X > 39 )\).
  2. Given that \(x\) is even, determine \(\mathrm { P } ( X < 10 )\). \section*{END OF QUESTION PAPER}

14 A random variable $X$ has probability distribution given by $\mathrm { P } ( X = x ) = \frac { 1 } { 860 } ( 1 + x )$ for $x = 1,2,3 , \ldots , 40$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 39 )$.
\item Given that $x$ is even, determine $\mathrm { P } ( X < 10 )$.

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR H240/02  Q14 [8]}}