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.

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

answer: 40.95

step-by-step explanation: