Part a)
I calculated the Pr by ($10\choose{1}$ * $5\choose{1}$) / $10\choose{5}$
Part b)
I defined X to be the 5 beer subset that Lindsay and Simon can pick. Multiplied each Probability by 2 for both Simon and Lindsay
E(X) = 1 * (($10\choose{1}$ * $5\choose{1}$) / $10\choose{5}$)*2) +
2 * (($10\choose{2}$ * $5\choose{2}$) / $10\choose{5}$)*2) +
3 * (($10\choose{3}$ * $5\choose{3}$) / $10\choose{5}$)*2) +
4 * (($10\choose{4}$ * $5\choose{4}$) / $10\choose{5}$)*2) +
5 * (($10\choose{5}$ * $5\choose{5}$) / $10\choose{5}$)*2)
For Part c)
This was a tricky one, but here is my attempt on it:
I defined $X =1$ to be the number of days that $B_{ai}$ = $B_i$ and $0$ otherwise
I took $n =3$ and since Simon uses the uniform random permutations to select, it becomes:
{$B_1$,$B_2$,$B_3$}
{$B_3$,$B_2$,$B_1$}
{$B_2$,$B_1$,$B_3$}
{$B_1$,$B_3$,$B_2$}
{$B_1$,$B_3$,$B_2$}
{$B_3$,$B_1$,$B_2$}
And for Lindsay, it is simply {$B_1$,$B_2$,$B_3$}
Now for $3$ days, I compared when $B_i's$ are the same for both.
Day $1$: Simon Possibilities: $B_1$, $B_3$, $B_2$, $B_1$, $B_3$
Lindsay Possibilities: $B_1$
These are same only Pr = $\frac{2}{5}$ times
Same thing with Day $2$ and Day $3$ = Pr = $\frac{2}{5}$ times
E(X) = $1$ * $\frac{2}{5}$ + $2$ * $\frac{2}{5}$ + $3$ * $\frac{2}{5}$ = $2.4$
This was for the scenario of taking $n=3$.
I am still learning on how to attempt these question so this was a wild attempt by me. I feel like my logic is still getting there so any help on these would be appreciated.
Best Answer
For part A, you get ${9 \choose 4} / {10\choose 5}$. The numerator is the number of ways that you can select 4 beers out of the remaining 9.
For part B, assume that Linday selects beers 1 through 5 (just relabel the beers if necessary), and let $X_i$ take the value 1 if Simon selects beer $i$ and 0 otherwise. The question asks for $E(\sum_{i=1}^5 X_i)$. We have computed $EX_i$ in part A, so The answer is 5 times the answer for part A.
For part C, let $X_i$ take the value 1 if $a_i=i$, and 0 otherwise. The question asks for $E(\sum_{i=1}^n X_i)$.