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anonymous
 3 years ago
Q.If you bet on “red” at roulette, you have chance 18/38 of winning. (There will be more on roulette later in the course; for now, just treat it as a generic gambling game.) Suppose you make a sequence of independent bets on “red” at roulette, with the decision that you will stop playing once you’ve won 5 times. What is the chance that after 15 bets you are still playing?
anonymous
 3 years ago
Q.If you bet on “red” at roulette, you have chance 18/38 of winning. (There will be more on roulette later in the course; for now, just treat it as a generic gambling game.) Suppose you make a sequence of independent bets on “red” at roulette, with the decision that you will stop playing once you’ve won 5 times. What is the chance that after 15 bets you are still playing?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are still playing if in 15 bets you have won zero times or once or twice or three times or four times. That means:\[P(x <5)=\sum_{k=0}^{4}\left(\begin{matrix}15 \\ k\end{matrix}\right)\left( \frac{ 18 }{ 38 } \right)^k \left(\frac{ 20 }{38 }\right)^{15k}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Substracting P(x=5) from 1 gives you the probability of obtaining less than 5 but also more than 5 in 15 bets in case you went on playing. In any case, you have to substract from one the probability P(x>=5), which leads to the same result that I have proposed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Khlara, that´s correct

perl
 3 years ago
Best ResponseYou've already chosen the best response.0on calculator binomcdf(15, 18/38, 4)
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