Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
KABRIC
Group Title
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?
 one year ago
 one year ago
KABRIC Group Title
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?
 one year ago
 one year ago

This Question is Closed

perl Group TitleBest 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.
 one year ago

perl Group TitleBest 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)
 one year ago

perl Group TitleBest 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+...
 one year ago

perl Group TitleBest 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
 one year ago

Hoa Group TitleBest ResponseYou've already chosen the best response.0
ok, I follow you
 one year ago

perl Group TitleBest 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
 one year ago

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

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

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

perl Group TitleBest 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
 one year ago

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

neusa88 Group TitleBest ResponseYou've already chosen the best response.0
thank you i didnt found it :)
 one year ago
See more questions >>>
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.