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KABRIC
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?
since you are still plaing after 15 bets, the first 15 bets you can win 0,1,2,3,or 4 times.
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)
= P(X=0) + P(X=1) + ... P(X=4) = 15 C0 * (18/38)^0 *(20/38)^15 + 15C1 *(18/38)^1 * (20/38)^14+...
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
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
unless i read the problem incorrectly
my question is why you just have P(4) ?
P(4) is one possibility, you can also have P(3) and P(2), etc
you only stop playing if you get 5 wins, so you keep playing if you get less than 5 wins
no worries. you answered tough linear algebra :)
thank you i didnt found it :)