I Just got given homework on things i don't get.

Here are a few questions i'm stuck on:

Brock has 48 books in his bookcase and 15 of the books are novels. Calculate the probability that of the first 2 books chosen at random for reading (assuming the same book is not chosen twice) that at least one of them is a novel.

Probability can be used to determine if a gambling game is fair (ie. A fair game is when the cost is equal to the expected winnings)

EX(winnings)=P(winning)x$amount won Expected winnings = Probability of winning x value of prize a) If chloe pays$10 to play a game in which she rolls two dice and the sum of the dice is her scroe, determine if this is a fair game if she wins $50 whenever she rolls a 7 b) Jack was offered the opportunity to play any number of games. In each case he pays$10 to play and roll a pair of dice.

GAME 1 - if the sum of the numbers is even jack receives $20 GAME 2 - if the sum of the numbers is 6 or 8 jack receives$30
GAME 3 - if the sum of the numbers is 5 or less Jak receives $40 Which Game/s should Jack choose to play? fully justify your answer by comparing all 4 games. Thanks to anyone that helps, also i would rather an explanation on how to do it then just an answer. Although how to do it and an answer would be great! 2. first pick probability P= 15/48 second pick probability P= 15/47 add that to get the probability of one or the other being a novel. did you not learn the or 'or', 'and' rules? 3. Hello, mathsnerd101! Brock has 48 books in his bookcase and 15 of the books are novels. .Calculate the probability that of the first 2 books chosen at random at least one of them is a novel. There are 48 books: .$\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}$a) If Chloe pays$10 to play a game in which she rolls two dice,
determine if this is a fair game if she wins $50 whenever she rolls a 7 There are: .$\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.

b) Jack was offered the opportunity to play any number of games.
In each case he pays $10 to play and roll a pair of dice. GAME 1 - if the sum of the numbers is even, Jack receives$20.

GAME 2 - if the sum of the numbers is 6 or 8, Jack receives $30. GAME 3 - if the sum of the numbers is 5 or less, Jack receives$40.

Which Game/s should Jack choose to play?
Fully justify your answer by comparing all 4 games. . Four games?
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 $6.67 per game. It's a no-brainer . . . 4. Originally Posted by Soroban Hello, mathsnerd101! There are 48 books: .$\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 . . .
Thank you so much!