Answer:
a) 0.20363
b) 0.00113
c) 0.77378
d) 0.18868
Step-by-step explanation:
Complete Question
At a carnival, the probability that a person will win a prize at the ring-toss game has been found to be (1/20). What is the probability that a prize will be won by exactly
a. 1 of the next 5 players?
b. 3 of the next 5 players?
c. 0 of the next 5 players?
d. 2 of the next 20 players?
Solution
This is a binomial distribution problem
A binomial experiment is one in which the probability of success doesn't change with every run or number of trials.
It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure.
The outcome of each trial/run of a binomial experiment is independent of one another.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 5, 20
x = Number of successes required = 1, 3, 0, 2
p = probability of success = probability of winning a prize in the ring-toss = (1/20) = 0.05
q = probability of failure = probability of NOT winning a prize in the ring-toss = 1 - (1/20) = (19/20) = 0.95
a. 1 of the next 5 players?
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 5
x = Number of successes required = 1
p = probability of success = 0.05
q = probability of failure = 0.95
P(X = 1) = ⁵C₁ (0.05)¹ (0.95)⁵⁻¹ = 0.2036265625 = 0.20363
b) 3 of the next 5 players?
n = total number of sample spaces = 5
x = Number of successes required = 3
p = probability of success = 0.05
q = probability of failure = 0.95
P(X = 3) = ⁵C₃ (0.05)³ (0.95)⁵⁻³ = 0.001128125 = 0.00113
c. 0 of the next 5 players?
n = total number of sample spaces = 5
x = Number of successes required = 0
p = probability of success = 0.05
q = probability of failure = 0.95
P(X = 0) = ⁵C₀ (0.05)⁰ (0.95)⁵⁻⁰ = 0.7737809375 = 0.77378
d. 2 of the next 20 players?
n = total number of sample spaces = 20
x = Number of successes required = 2
p = probability of success = 0.05
q = probability of failure = 0.95
P(X = 2) = ²⁰C₂ (0.05)² (0.95)²⁰⁻² = 0.18867680127 = 0.18868
Hope this Helps!!!
