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.
