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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?
 11 months ago
 11 months ago
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?
 11 months ago
 11 months ago

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perlBest ResponseYou've already chosen the best response.2
since you are still plaing after 15 bets, the first 15 bets you can win 0,1,2,3,or 4 times.
 11 months ago

perlBest ResponseYou've already chosen the best response.2
so you have played roullette 15 times, the probability of success (by betting at red) is equal to 18/38. Let X = number of times you win betting on red X is a binomial random variable, X can equal 0,1,2,3,4...15 we are interested in P( X <= 4) = P( X = 0 or X = 1 or X=2 or X=3 or X=4)
 11 months ago

perlBest ResponseYou've already chosen the best response.2
= P(X=0) + P(X=1) + ... P(X=4) = 15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+...
 11 months ago

perlBest ResponseYou've already chosen the best response.2
15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+ 15C2 *(18/38)^2 * (20/38)^13 +15C3 *(18/38)^3 * (20/38)^12 + 15C4 *(18/38)^4 * (20/38)^11 Please check over
 11 months ago

perlBest ResponseYou've already chosen the best response.2
P( X <= 4) = P( X = 0 or X = 1 or X=2 or X=3 or X=4) = P(X=0) + P(X=1) + ... P(X=4) =15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+ 15C2 *(18/38)^2 * (20/38)^13 +15C3 *(18/38)^3 * (20/38)^12 + 15C4 *(18/38)^4 * (20/38)^11 Now on calculator this is fast binomcdf(15, 18/38, 4) = .087399
 11 months ago

perlBest ResponseYou've already chosen the best response.2
unless i read the problem incorrectly
 11 months ago

HoaBest ResponseYou've already chosen the best response.0
my question is why you just have P(4) ?
 11 months ago

perlBest ResponseYou've already chosen the best response.2
P(4) is one possibility, you can also have P(3) and P(2), etc
 11 months ago

perlBest ResponseYou've already chosen the best response.2
you only stop playing if you get 5 wins, so you keep playing if you get less than 5 wins
 11 months ago

perlBest ResponseYou've already chosen the best response.2
no worries. you answered tough linear algebra :)
 11 months ago

neusa88Best ResponseYou've already chosen the best response.0
thank you i didnt found it :)
 11 months ago
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