1. ## probability question

A large bag contains 1200 red marbles and 2001 blue marbles. Joe and Lisa will play a game in which they draw a single marble from the bag with replacement. For every game they will bet $35. Lisa wins the money if the marble is red and Jow wins if it's blue. If they play this game for 7 straigth days at an average rate of one game per every 10 seconds, who should win money and exactly how much should he/she expect to win? I have a data management test tomorrow(( I dunno if that's going to help but it's binomial distribution and we're suposed to use this formula (n)*p^k*q^n-k (k) I tried to solve it and the best solution I came up with is sorry it's not 2001 blue marbles, it's 1201. sorry about the typo (2001/2401)*604800*35=10592820 I'm not sure if it's right thou 2. Originally Posted by anna12345 A large bag contains 1200 red marbles and 2001 blue marbles. Joe and Lisa will play a game in which they draw a single marble from the bag with replacement. For every game they will bet$35. Lisa wins the money if the marble is red and Jow wins if it's blue. If they play this game for 7 straigth days at an average rate of one game per every 10 seconds, who should win money and exactly how much should he/she expect to win?

I have a data management test tomorrow((

I dunno if that's going to help but it's binomial distribution and we're suposed to use this formula

(n)*p^k*q^n-k
(k)

I tried to solve it and the best solution I came up with is

(2001/2401)*604800*35=10592820

I'm not sure if it's right thou
n = (6)(60)(24)(7) = ....

p = Pr(Red in a single trial) = 1200/(1200 + 2001).

Now consider the expected number of wins (and losses) for red.

3. Hello, anna12345!

A large bag contains 1200 red marbles and 2001 blue marbles.
Joe and Lisa will play a game in which they draw a single marble from the bag with replacement.
For every game they will bet $35. Lisa wins the money if the marble is red, and Jow wine if it's blue. If they play this game for 7 days at a rate of one game every 10 seconds, who should win money and exactly how much should he/she expect to win? Just "eyeballing" the problem, we see that Joe has the advantage. . . (There are more blue marbles than red marbles.) There is a total of 3201 marbles: 1200 red and 2001 blue. $P(\text{red}) \:=\:\frac{1200}{3201} \:\approx\:\frac{3}{8}$ $P(\text{blue}) \:=\:\frac{2001}{3201} \:\approx\:\frac{5}{8}$ Joe wins $\tfrac{5}{8}$ of the time and loses $\tfrac{3}{8}$ of the time. He comes out ahead $\tfrac{5}{8} - \tfrac{3}{8} \:=\:\tfrac{2}{8} \:=\:\tfrac{1}{4}$ of the time. In seven days, they will play 60,480 games. Joe comes out ahead in: $\tfrac{1}{4} \times 60,\!480 \:=\:15,\!120$ games. At$35 per game, Joe expects to win: . $15,\!120 \times \35 \:=\:\529,\!200$