| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Three or more independent values |
| Difficulty | Challenging +1.2 Part (a) is trivial substitution (1 mark). Part (b) requires conditional probability with enumeration of cases where three values sum to 9, which is moderately challenging but systematic. Part (c) is a standard negative binomial application. The question tests understanding of discrete distributions and conditional probability but follows recognizable patterns without requiring deep insight. |
| Spec | 2.03d Calculate conditional probability: from first principles2.04a Discrete probability distributions |
In this question you must show detailed reasoning.
The random variable $X$ has probability distribution defined as follows.
$$P(X = x) = \begin{cases}
\frac{15}{64} \times \frac{2^x}{x!} & x = 2, 3, 4, 5, \\
0 & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $P(X = 2) = \frac{15}{32}$. [1]
\end{enumerate}
The values of three independent observations of $X$ are denoted by $X_1$, $X_2$ and $X_3$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that $X_1 + X_2 + X_3 = 9$, determine the probability that at least one of these three values is equal to 2. [6]
\end{enumerate}
Freda chooses values of $X$ at random until she has obtained $X = 2$ exactly three times. She then stops.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Determine the probability that she chooses exactly 10 values of $X$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2020 Q15 [10]}}