Question

TUDIOT (1 point) In the game of roulette, a steel ball is rolled onto a wheel that contains 18 red, 18 black, and 2 green slots. If the ball is rolled 34 times, find the probability of the following events. A. The ball falls into the green slots 4 or more times. Probability = B. The ball does not fall into any green slots. Probability = C. The ball falls into black slots 15 or more times. Probability D. The ball falls into red slots 9 or fewer times. Probability Note: You can earn partial credit on this problem, Preview My Answers Submit Answers You have attempted this problem 0 times You have unlimited attempts remaining, Email Instructor

Answer 1

A) Probability of falling into green slot, p = 2/ (18+18+2) = 1/19

As each roll is independent and number of trials is finite , n= 34

Let X be the number of times the ball falls into green slot

X follow Binomial with n= 34 , p =1/19

To find P( X $\geq$ 4)

Probability mass function of a Binomial distribution  is

$P(X=x)= \binom{n}{x}p^{x}(1-p)^{n-x}$

$P(X\geq 4)=1-P(X\leq 3)$

$=1-\sum_{0}^{3}\binom{34}{x}(1/19)^{x}(1-1/19)^{34-x}$

$=1-(\binom{34}{0}(1/19)^{0}(1-1/19)^{34-0}+....+\binom{34}{3}(1/19)^{3}(1-1/19)^{34-3})$

= 1-0.8983

= 0.1017

Probability that the ball falls into green slot 4 or more times = 0.1017

b)

$P(X=0)=\binom{34}{0}(1/19)^{0}(1-1/19)^{34-0}=0.1591$

Probability that the ball does not fall into any green slot =0.1591

c)

Probability of falling into black slot, p = 18/ (18+18+2) = 9/19

As each roll is independent and number of trials is finite , n= 34

Let Y be the number of times the ball falls into black slot

Y follow Binomial with n= 34 , p =9/19

To find P( Y $\geq$ 15)

$P(Y\geq 15)=1-P(Y\leq 14)$

$=1-\sum_{0}^{14}\binom{34}{y}(9/19)^{y}(1-9/19)^{34-y}$

= 1-0.2918

= 0.7082

Probability that the ball falls into black slot 15 or more times = 0.7082

Note : for simplifying calculation we can use excel formula for P( X $\leq$ 14)" =BINOM.DIST(14,34,9/19,cumulative )"

d) Probability of falling into red slot, p = 18/ (18+18+2) = 9/19

Let Y1 be the number of times the ball falls into black slot

Y1 follow Binomial with n= 34 , p =9/19

$P(Y1\leq 9)$

$=\sum_{0}^{9}\binom{34}{y1}(9/19)^{y1}(1-9/19)^{34-y1}$

=0.0105

Probability that the ball falls into red slot 9 or fewer times = 0.0.0150

Note : using excel formula "=BINOM.DIST(9,34,9/19,cumulative)"

or we can calculate manually

$\binom{34}{0}(9/19)^{0}(1-9/19)^{34-0}+....+\binom{34}{9}(9/19)^{9}(1-9/19)^{34-9}$