A CASINO

Algebra Level 2

Suppose you go to a casino where you see a game. There is 1 bag kept in which there are 3 green balls and 2 red balls. Without looking, you have to take out 1 ball. If the ball turns out to be green, then you have to put it on the table. Once again, you have to put your hand in the bag (without looking) and take out 1 more ball. If this ball also turns out to be green(i.e., both the balls to be green), then you have won the game and you will get 1$. The entry-fee for the game is 35 cents (0.35$). If you play the game then you will be in-

Equal profit equal loss No profit no loss Profit Loss

Omkar Pande by Omkar Pande , 22, India

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1 solution

Benjamin Wong
Feb 14, 2014

P(Win)=(3/5)(2/4)=3/10=three wins out of ten bets

So 3.00 win for 3.50 bet, loss

5 Helpful 0 Interesting 0 Brilliant 0 Confused

where did the 2/4 come?

Edward Jeremy Lo - 7 years, 3 months ago

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If the 1st ball turns out to be green, then- New no. of favorable outcomes = 2
New no. of possible outcomes = 5-1 = 4 Hence 2/4.

Omkar Pande - 7 years, 3 months ago

good one!!!

TIRTHANKAR GHOSH - 7 years, 1 month ago

nice solution.......i realized that though the probability of getting a green ball is marginally higher the first time, it becomes equal to the red ball the second time..so, if you don't get a green ball the second time( 50% chance you won't)...then you are at a loss

Krishna Ramesh - 7 years, 1 month ago

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