# Statistics and Probability [closed]

A casino game has three possible winnings, $\$15$,$\$10$, and $\$5$, each being twice as likely as the one before it. Player's don't win anything$65\%$of the time. i) If it costs$\$3$ to play this game, how much does a player expect to lose on average? (find the expected profit, which will be negative. This is the player's loss).

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Please show any work you have done for this problem. (We have just experienced a recent episode where someone decided not to do so, and it ended up turning out badly.) –  Clarinetist Jun 15 at 1:08
So you're most likely to get more money rather than less? I like this game already. –  Ataraxia Jun 15 at 1:24
Math.SE is not a homework service. We are here to answer concept questions, not do your homework for you. –  Nicholas R. Peterson Jun 15 at 1:27
downvoted because you didn't show any attempted work –  user67258 Jun 15 at 1:38
@Clarinetist I got .05 for the first part. –  Sara Jun 15 at 1:45
Observe that when you win, $\frac{1}{7}$ of the time you win \$15,$\frac{2}{7}$you win \$10 and $\frac{4}{7}$ you win \$5 and you lose with a probability 0.65. So your expected winnings are $$\left(\frac{1}{7}15+\frac{2}{7}10+\frac{4}{7}5\right)0.35=\2.75$$ If it costs \$3.00 to play you expect to lose \$0.25 - add comment Hint: $$a+b+c=35$$ $$b=2a$$ $$c=2b$$ You have 3 equations and 3 unknowns. Solve for$a$,$b$, and$c\$.