Mathematics, 30.07.2019 17:20 BrainlyAvenger
Aroulette wheel has 38 numbers. eighteen of the numbers are black, eighteen are red, and two are green. when the wheel is spun, the ball is equally likely to land on any of the 38 numbers. each spin of the wheel is independent of all other spins of the wheel. one roulette bet is a bet on black—that the ball will stop on one of the black numbers. the payoff for winning a bet on black is $2 for every $1 one bets; that is, if you win, you get the dollar ante back and an additional dollar, for a net gain of $1, while if you lose, you gets nothing back, for a net loss of $1. each $1 bet thus results in the gain or loss of $1. suppose one repeatedly places $1 bets on black, and plays until either winning $7 more than he has lost, or loses $7 more than he has won. equivalently, one plays until the first time that | net winnings | = | $won − $bet | = |$2×(#bets won) − $1×#bets | = $7, where |x| is the absolute value of x. if one quits according to that rule, the distribution of the number of times one plays has a (q8) the chance that one places exactly 6 bets before stopping is (q9) 0 . the chance that one places exactly 7 bets before stopping is (q10) 0.01653813221 . the chance that one places exactly 8 bets before stopping is (q11) 0 . the chance that one places exactly 9 bets before stopping is (q12) 0.02056958826 .
Answers: 2
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