| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Topic | Geometric Distribution |
| Type | Non-geometric distribution identification |
| Difficulty | Standard +0.3 This is a straightforward probability distributions question covering basic calculations (expectation, variance) and standard geometric distribution applications. All parts require direct formula application with minimal problem-solving—slightly easier than average A-level due to the routine nature of all components. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm{P}(X = x)\) | 0.30 | 0.25 | 0.20 | 0.16 | 0.06 | 0.03 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(E(X) = 1.52\) | B1 | |
| \(\text{Var}(X) = \sum x^2 p(x) - [E(X)]^2\) used, seen or implied | M1 | |
| \(= 1.8896\) or rounding to 1.89 | A1 | [3] |
| (ii) 0.25 | B1 | [1] |
| (iii) \((1 - 0.3)^2\) or \(0.7^2\) seen | M1 | |
| \(= 0.49\) | A1 | [2] |
| (b) (i) The paper is bought repeatedly until a success is obtained. | B1 | |
| \(F(y) = (0.002)(0.998)^{y-1}\) | B1 | [2] |
| (ii) \(F(20) = (p)(1-p)^{19}\) | M1 | |
| \(= 0.00193\) or better | A1 | [2] |
| (iii) Attempt \(P(Y \leq 3) = P(Y=1) + P(Y=2) + P(Y=3)\) | M1 | |
| \(= 0.00599\) or better | A1 | [2] |
(a) (i) $E(X) = 1.52$ | B1 |
$\text{Var}(X) = \sum x^2 p(x) - [E(X)]^2$ used, seen or implied | M1 |
$= 1.8896$ or rounding to 1.89 | A1 | [3]
(ii) 0.25 | B1 | [1]
(iii) $(1 - 0.3)^2$ or $0.7^2$ seen | M1 |
$= 0.49$ | A1 | [2]
(b) (i) The paper is bought repeatedly until a success is obtained. | B1 |
$F(y) = (0.002)(0.998)^{y-1}$ | B1 | [2]
(ii) $F(20) = (p)(1-p)^{19}$ | M1 |
$= 0.00193$ or better | A1 | [2]
(iii) Attempt $P(Y \leq 3) = P(Y=1) + P(Y=2) + P(Y=3)$ | M1 |
$= 0.00599$ or better | A1 | [2]\begin{enumerate}[label=(\alph*)]
\item In a game show contestants are asked up to five questions in succession to qualify for the next round. An incorrect answer eliminates a contestant from the game show.
Let $X$ denote the number of questions correctly answered by a contestant. The probability distribution of $X$ is given below.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm{P}(X = x)$ & 0.30 & 0.25 & 0.20 & 0.16 & 0.06 & 0.03 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Find the expected number of correctly answered questions and the variance of the distribution. [3]
\item Find the probability that a randomly selected contestant will correctly answer 3 or more questions. [1]
\item Each show had two contestants. Find the probability that both the contestants will correctly answer at least one question. [2]
\end{enumerate}
\item In a promotion, a newspaper included a token in every copy of the newspaper. A proportion, 0.002, are winning tokens and occur randomly. A reader keeps buying copies of the newspaper until he buys one with a winning token and then stops.
Let $Y$ denote the number of copies bought.
\begin{enumerate}[label=(\roman*)]
\item Explain briefly why this situation may be modelled by a geometric distribution and write down a formula for $\mathrm{P}(Y = y)$. [2]
\item Find the probability that the reader gets a winning token with the twentieth copy bought. [2]
\item Find the probability that the reader will not have to buy more than three copies in order to get a winning token. [2]
\end{enumerate]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2010 Q14 [12]}}