Online Course Discussion Forum

Math Challenge I-A + I-B Counting and Probability

 
 
LiangNeo的头像
Math Challenge I-A + I-B Counting and Probability
LiangNeo - 2019年08月5日 Monday 10:34
 

Lecture 9, Problem 9.8 b

for problem 9.8b, i got $\frac {1}{7^5}$ because it is at least 3 7's, so the answer should be $\frac {1}{7}*\frac {1}{7}*\frac {1}{7}$    The answer says that it is $\frac {5C3*6^2+5C4*6+1}{7^5}$

 
ReynosoDavid的头像
Re: Math Challenge I-A + I-B Counting and Probability
ReynosoDavid - 2019年08月7日 Wednesday 17:51
 

For the median of the $5$ books to be $7$ we need that $3$, $4$ or $5$ of them choose book $7$ as their favorite.

There are $\displaystyle \binom{5}{3} \times 6^2$ was to have $3$ of them choose book $7$ as their favorite, since we need to choose which $3$ of the $5$ chose it, and we also need to choose some of the other $6$ books for each of the other $2$ people.

Similarly, there are $\displaystyle \binom{5}{4} \times 6$ and $\displaystyle \binom{5}{5} \times 6^0$ ways for $4$ of them or $5$ of them, respectively, to choose book $7$ as their favorite. 

Thus, there are $\displaystyle \binom{5}{3} \times 6^2 + \binom{5}{4} \times 6^1 + \binom{5}{5} \times 6^0$ ways to have the median of the books to be $7$.