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 16 times, find the probability of the following eventsA. 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 11 or more times.

Probability =

D. The ball falls into red slots 12 or fewer times.

Probability =

A) 0.00836

B) 0.4212

C) 0.0717

D) 0.9939

Step-by-step explanation:

Let the probability of the ball falling into the red, black and green slots be P(R), P(B) and P(G) respectively.

Total number of slots = 38

P(R) = 18/38 = 9/19 = 0.474

P(B) = 18/38 = 9/19 = 0.474

P(G) = 2/38 = 1/19 = 0.0526

a) In 16 ball rolls, probability of the ball falling into the green slots 4 or more times.

For this, we grade the slots as green and others. Then, it becomes a binomial distribution problem

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 16

x = Number of successes required

p = probability of success = 0.0526

q = probability of failure = 1 - 0.0526 = 0.9474

probability of the ball falling into the green slots 4 or more times = 1 - (Probability of the ball falling into the green slot less than 4 times)

Probability of the ball falling into the green slot less than 4 times = P(X < 4) = Σ ⁿCₓ pˣ qⁿ⁻ˣ (summing from x = 0 to x = 3)

P(X ≥ 4) = 1 - P(X < 4) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3]

P(X ≥ 4) = 1 - P(X < 4) = 1 - (0.99164) = 0.00836

b) The probability that the ball does not fall into any green slots = P(X=0)

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 16

x = Number of successes required = 0

p = probability of success = 0.0526

q = probability of failure = 1 - 0.0526 = 0.9474

P(X = 0) = ¹⁶C₀ (0.0526)⁰ (0.9474)¹⁶⁻⁰

P(X = 0) = ¹⁶C₀ (0.0526)⁰ (0.9474)¹⁶ = 0.4212

C) Probability that the ball falls into black slots 11 or more times.

For this, we grade the slots as black and others. Then, it becomes a binomial distribution problem

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 16

x = Number of successes required

p = probability of success = probability of ball entering a black slot = 0.474

q = probability of failure = probability of ball not entering a black slot = 1 - 0.474 = 0.526

P(X ≥ 11) = Σ ⁿCₓ pˣ qⁿ⁻ˣ (summing from x = 11 to x = 16)

P(X ≥ 11) = [P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15) + P(X=16)]

P(X ≥ 11) = 0.0717

D) Probability that the ball falls into red slots 12 or fewer times = P(X ≤ 12)

P(X ≤ 12) = 1 - P(X > 12)

Again, we grade the slots as red and others. Then, it becomes a binomial distribution problem

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 16

x = Number of successes required

p = probability of success = probability of ball entering a red slot = 0.474

q = probability of failure = probability of ball not entering a red slot = 1 - 0.474 = 0.526

P(X ≤ 12) = 1 - P(X > 12)

P(X > 12) = Σ ⁿCₓ pˣ qⁿ⁻ˣ (summing from x = 13 to x = 16)

P(X > 12) = [P(X=13) + P(X=14) + P(X=15) + P(X=16)]

P(X > 12) = 0.00605

P(X ≤ 12) = 1 - P(X > 12) = 1 - 0.00605 = 0.9939

please take note, there are binomial distribution calculator. These calculators make it very easy to obtain whichever binomial probability is required. A screenshot of a binomial distribution calculator is attached to this solution.

Supplying the data on the first 3 rows gives whichever answer is required.

this was already solved by me with a different person. if you look it up in the search bar at the top of the page, you will get the answer.

step-by-step explanation:

all you have to do is multiply them both to see how many out comes you can put:

3 x 2 = 6

you can make 6 diferent combinations of crusts and toppings.

~batmans wife dun dun