Year: 2007
Paper: 1
Question Number: 13
Course: LFM Stats And Pure
Section: Conditional Probability
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1469.5
Banger Comparisons: 2
A bag contains eleven small discs, which are identical except that six of the discs are blank and five of the discs are numbered, using the numbers 1, 2, 3, 4 and 5.
The bag is shaken, and four discs are taken one at a time without replacement.
Calculate the probability that:
\begin{questionparts}
\item all four discs taken are numbered;
\item all four discs taken are numbered,
given that the disc numbered ``3'' is taken first;
\item exactly two numbered discs are taken,
given that the disc numbered ``3'' is taken first;
\item exactly two numbered discs are taken,
given that the disc numbered ``3'' is taken;
\item
exactly two numbered discs are taken,
given that a numbered disc is taken first;
\item exactly two numbered discs are taken,
given that a numbered disc is taken.
\end{questionparts}There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.
\begin{questionparts}
\item
There are we are choose $4$ objects in order from $5$ (ie ${^5\P_4}$) to obtain valid draws, this is out of a total of picking $4$ objects from $11$ (${^{11}\P_4}$). Ie the probability is: $\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}$
Alternatively, there are $\binom{5}{4}$ ways to choose four numbered discs, out of $\binom{11}{4}$ ways to choose four discs. ie $\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}$
\item
\begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\
&= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\
&= \frac{4! \cdot 7! \cdot 11}{11!} \\
&= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\
&= \frac{1}{30}
\end{align*}
Alternatively,
\begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\
&= \frac{1}{30}
\end{align*}
Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in $\binom{4}{3}$ ways, and there are $11 \cdot \binom{10}{3}$ was overall.
Another alternative using Bayes rule:
\begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\
&= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\
&= \frac1{30}
\end{align*}
\item \begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\
&= \frac12
\end{align*}
Alternatively,
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\
&= \frac12
\end{align*}
\item
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\
&= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\
&= \frac{\frac{2}{11}}{\frac4{11}} \\
&= \frac12
\end{align*}
Using Bayes rule:
$\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}$
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\
&= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\
&= \frac{\frac4{10}{5 / 11}}{4/11} \\
&= \frac{1}{2}
\end{align*}
\item
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\
&= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\
&= \frac{1}{2}
\end{align*}
\item
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\
&= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\
&= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\
&= \frac{\frac{5}{11}}{\frac{21}{22}} \\
&= \frac{10}{21} \neq \frac12
\end{align*}
\end{questionparts}
A lot of attempts at this question were seen, but conceptual errors undermined many solutions. In particular, a lot of candidates seemed not to realise that they were being asked to calculate conditional probabilities in parts (ii) to (vi).