Question

2. (Binomial model) Consider a roulette wheel with 38 slots, of which 18 are red, 18 are black, and 2 are green (0 and 00). You spin the wheel 6 times.

(a) What is the probability that 2 of those times the ball ends up in a green slot? (

b) What is the probability that 4 of those times the ball ends up in a red slot?

3. (Normal approximation to binomial model) Take the roulette wheel from question 2. Assume that the wheel is spun 100 times, and you are interested in whether the ball ends up in a red slot.

(a) Verify that you can use the normal model here.

(b) Find the probability that the ball ends up in a red slot at least 60 times.

Answer 1

2. (Binomial model) Consider a roulette wheel with 38 slots, of which 18 are red, 18 are black, and 2 are green (0 and 00). You spin the wheel 6 times.

(a) What is the probability that 2 of those times the ball ends up in a green slot?

-> p = probability of green = 2/(18+18+2) = 1/19

n = 6, r =2

Formula = ⁿC_r*p^r*(1-p)^n-r

=   ⁶C₂*(2/38)²*(1- 2/38)⁴

= 0.0335

b) What is the probability that 4 of those times the ball ends up in a red slot?

-> p = probability of red = = 18/(18+18+2) = 18/38,

n = 6, r =4

Formula = ⁿC_r*p^r*(1-p)^n-r

=   ⁶C₄*(18/38)⁴*(1- 18/38)²

= 0.2092

3. (Normal approximation to binomial model) Take the roulette wheel from question 2. Assume that the wheel is spun 100 times, and you are interested in whether the ball ends up in a red slot.

(a) Verify that you can use the normal model here.

-> sample size is 100 which is large and P(red) = 18/38 which is close to 1/2. Thus, we can use normal approximation to binomial model

(b) Find the probability that the ball ends up in a red slot at least 60 times.

-> Taking u = np = 100*18/38 = 47.37

and = sqrt(np(1-p)) = sqrt(100*18/38*20/38) = 5

P(the ball ends up in a red slot at least 60 times)

= P( (x-47.37)/5 > (60-47.37)/5)

= P(z > 2.53)

= 0.0057