Solution: 2022 Winter Final - 14

Author: Michiel Smid

Question

Let $D_1,\ldots,D_{2n}$ be the result of rolling a $6$-sided die $2n$ times, and consider the length-$n$ sequence \[ S=\langle (D_1+D_2), (D_3+D_4), (D_5+D_6),\ldots, (D_{2n-1}+D_n)\rangle \] whose entries are all in the set $\{2,3,\ldots,12\}$. Let $k$ be an integer in $\{0,\ldots,n\}$. What is the probability that $S$ contains exactly $k$ many $5$'s?

(a)

$\binom{n}{k}\cdot(\tfrac{1}{36})^k\cdot (\tfrac{35}{36})^{n-k}$

(b)

$\binom{n}{k}\cdot(\tfrac{1}{12})^k\cdot (\tfrac{11}{12})^{n-k}$

(c)

$\binom{n}{k}\cdot(\tfrac{1}{5})^k\cdot (\tfrac{4}{5})^{n-k}$

(d)

$\binom{n}{k}\cdot(\tfrac{1}{9})^k\cdot (\tfrac{8}{9})^{n-k}$

(e)

$\binom{n}{k}\cdot(\tfrac{1}{11})^k\cdot (\tfrac{10}{11})^{n-k}$

Solution

First, we choose which $k$ of the $n$ pairs will contain a 5: $ \binom{n}{k} $

As can be seen $ \Pr(Di+D{i+1}=5) = \frac{4}{36} = \frac{1}{9} $

The remaining sum roles can be anything but 5: $ {(\frac{8}{9})}^{n-k} $

Thus, the final answer is $ \binom{n}{k} \cdot {(\frac{1}{9})}^{k} \cdot {(\frac{8}{9})}^{n-k} $