The question goes as follows. My attempts are below it.
A motel has ten rooms, all located on the same side of a single corridor and numbered 1 to 10 in numerical order. The motel always randomly allocates rooms to its guests. There are no other guests besides those mentioned.
a) Friends Molly and Polly have been allocated two separate rooms at the motel. What is the likely number of rooms between their rooms?
b) Molly believes there is a greater than 1/3 chance that at most one room will separate them, but Polly disagrees. Who is right? Explain why.
c) On another occasion, Molly, Polly and a third friend, Ollie, were allocated three separate rooms. Molly believes there is a better than 1/3 chance that they are all within a block of five consecutive rooms. Ollie believes that there is exactly 1/3 chance and Polly believes there is less than 1/3 chance. Who is right? Explain why.
d) Ollie arrived after rooms were allocated to Molly and Polly. There was then a 50% chance he would be in a room adjacent to Molly or Polly or both. In how many ways could a pair of rooms have been allocated to Molly and Polly?
Here I am bit sure of how to work through (c). Do I use permutations or combinations – could you please give me a step by step solution as I am still a bit wobbly on combinatorics.
Also could someone please check my answer for (d) (30)?
Thanks.
Best Answer
For C:
The number of ways to have three rooms within 5 consecutive rooms is $3!$ multiplied with the number of non negative solutions of:
$$ \begin{aligned} A+B+C+D&=7\\ A+D&=7-i\\ C+D&=i \end{aligned} $$
Where $i=0,1,2$ which is $3!\times 40=240$. So the probability is $\frac{240}{10\times 9\times 8}=\frac{1}{3}$
For D:
The number of ways we can allocate rooms to molly and polly equals to $2$ multiplied with the number of non negative solutions of:
$$ A+B+C=4 $$
Which is $2 \times 15 = 30$ so You are right