
Originally Posted by
Soroban
Hello, mathsnerd101!
There are 48 books: .$\displaystyle \begin{Bmatrix}\text{15 novels} \\ \text{33 others} \end{Bmatrix}$
There are: .$\displaystyle _{48}C_2 \:=\:1128$ possible outcomes.
The opposite of "at least one novel" is "no novels" (all Others).
No novels: .$\displaystyle _{33}C_2 \:=\:528$ ways.
Hence: .$\displaystyle 1128 - 528 \:=\:600$ ways to get at least one novel.
Therefore: .$\displaystyle P(\text{at least 1 novel}) \;=\;\frac{600}{1128} \;=\;\frac{25}{47}$
There are: .$\displaystyle 6^2 \:=\:36$ possible outcomes.
There are 6 ways to get a "7": .$\displaystyle P(7) \:=\:\tfrac{6}{36} \:=\:\tfrac{1}{6}$
. . She will win $50 one-sixth of the time: .$\displaystyle P(+\$50) \:=\:\tfrac{1}{6}$
She pays $10 for every game: .$\displaystyle P(\text{-}\$10) \:=\:\tfrac{36}{36} \:=\:1$
Her expected value is: .$\displaystyle E \;=\;(50)\left(\tfrac{1}{6}\right) + (\text{-}10)(1) \:=\:-\frac{5}{3}$
She can expect to lose an average of $1.67 per game.
He pays $10 for every game: .$\displaystyle P(-\$10) \:=\:1$
GAME 1 - Half the outcomes are even: .$\displaystyle P(+\$20) \:=\:\tfrac{1}{2}$
Therefore: .$\displaystyle E \;=\;(20)\left(\tfrac{1}{2}\right) + (-10)(1) \;=\;0$
The game is fair . . . Jack can expect to break even.
GAME 2 - There are 5 ways to get a "6" and 5 ways to get an "8"
. . Hence: .$\displaystyle P(\text{6 or 8}) \:=\:P(+\$30) \:=\:\tfrac{10}{36} \:=\:\tfrac{5}{18}$
Therefore: .$\displaystyle E \;=\;\left(\tfrac{5}{18}\right)(+30) + (1)(-10) \;=\;-\frac{5}{3}$
Jack can expect to lose an average of $1.67 per game.
GAME 3 - There are 15 outcomes with a sum less than or equal to 5.
. . Hence: .$\displaystyle P(\leq 5) \:=\:P(+\$40) \:=\:\tfrac{15}{36} \:=\:\tfrac{5}{12}$
Therefore: .$\displaystyle E \;=\;(+40)\left(\frac{5}{12}\right) + (1)(-10) \;=\;\frac{20}{3}$
Jack can expect to win an average of $6.67 per game.
It's a no-brainer . . .