2012 Paper 1 Q13

Year: 2012
Paper: 1
Question Number: 13

Course: LFM Stats And Pure
Section: Probability Definitions

Difficulty: 1500.0 Banger: 1529.2

probability conditional expected value combinatorics counting digits inclusion-exclusion

Problem

I choose at random an integer in the range 10000 to 99999, all choices being equally likely. Given that my choice does not contain the digits 0, 6, 7, 8 or 9, show that the expected number of different digits in my choice is 3.3616.

Solution

We are choosing any \(5\) digit number from \(\{1,2,3,4,5\}\). There are \(5^5\) such numbers. \begin{align*} && \mathbb{E}(\text{different digits}) &= \frac1{5^5} \left (1 \cdot 5 + 2 \cdot \binom{5}{2}(2^5-2)+3 \cdot \binom{5}{3}(3^5-3 \cdot 2^5+3)+4 \cdot \binom{5}{4}(4^5 - 4 \cdot 3^5+6 \cdot 2^5-4) + 5 \cdot 5! \right) \\ &&&= \frac{2101}{625} = 3.3616 \end{align*}

Examiner's report

— 2012 STEP 1, Question 13

Below Average Described as 'not attempted by many candidates'

This question was not attempted by many candidates. There were some good answers showing a clear thought process to reach the required value, but many of the other solutions offered suffered from a lack of explanation of the method meaning that the ideas being applied were difficult to follow.

Source: Cambridge STEP 2012 Examiner's Report · 2012-full.pdf

Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1529.2

Banger Comparisons: 2

Show LaTeX source

Problem source

I choose at random an integer in the 
range 10000 to 99999, all choices being
equally likely. Given that my choice
does not contain the digits 0, 6, 7, 8 or 9,
show that the expected number of different digits
in my choice is 3.3616.

Solution source

We are choosing any $5$ digit number from $\{1,2,3,4,5\}$.

There are $5^5$ such numbers.

\begin{align*}
&& \mathbb{E}(\text{different digits}) &= \frac1{5^5} \left (1 \cdot 5 + 2 \cdot \binom{5}{2}(2^5-2)+3 \cdot \binom{5}{3}(3^5-3 \cdot 2^5+3)+4 \cdot \binom{5}{4}(4^5 - 4 \cdot 3^5+6 \cdot 2^5-4) + 5 \cdot 5! \right) \\
&&&= \frac{2101}{625} = 3.3616
\end{align*}

Back to List