CAIE FP2 2015 November — Question 6 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeFind minimum n for P(X ≤ n) > threshold
DifficultyStandard +0.3 This is a straightforward geometric distribution question requiring standard formula application. Part (i) establishes p=2/3 using mean=1/p and variance=(1-p)/p², then parts (ii-iv) involve direct substitution into geometric distribution formulas. The algebra is simple and all steps are routine for Further Maths students, making it slightly easier than average.
Spec5.02f 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

A biased coin is tossed repeatedly until a head is obtained. The random variable \(X\) denotes the number of tosses required for a head to be obtained. The mean of \(X\) is equal to twice the variance of \(X\). Show that the probability that a head is obtained when the coin is tossed once is \(\frac{2}{3}\). [2] Find
  1. P(\(X = 4\)), [1]
  2. P(\(X > 4\)), [2]
  3. the least integer \(N\) such that P(\(X \leq N\)) \(> 0.999\). [3]

A biased coin is tossed repeatedly until a head is obtained. The random variable $X$ denotes the number of tosses required for a head to be obtained. The mean of $X$ is equal to twice the variance of $X$. Show that the probability that a head is obtained when the coin is tossed once is $\frac{2}{3}$. [2]

Find
\begin{enumerate}[label=(\roman*)]
\item P($X = 4$), [1]
\item P($X > 4$), [2]
\item the least integer $N$ such that P($X \leq N$) $> 0.999$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2015 Q6 [8]}}