A community for students.
Here's the question you clicked on:
 0 viewing
KABRIC
 one year 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?
KABRIC
 one year 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?

This Question is Closed

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

perl
 one year ago
Best ResponseYou've already chosen the best response.2so 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)

perl
 one year ago
Best 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+...

perl
 one year ago
Best ResponseYou've already chosen the best response.215 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

perl
 one year ago
Best ResponseYou've already chosen the best response.2P( 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

perl
 one year ago
Best ResponseYou've already chosen the best response.2unless i read the problem incorrectly

Hoa
 one year ago
Best ResponseYou've already chosen the best response.0my question is why you just have P(4) ?

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

perl
 one year ago
Best ResponseYou've already chosen the best response.2you only stop playing if you get 5 wins, so you keep playing if you get less than 5 wins

perl
 one year ago
Best ResponseYou've already chosen the best response.2no worries. you answered tough linear algebra :)

neusa88
 one year ago
Best ResponseYou've already chosen the best response.0thank you i didnt found it :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.