At a carnival, the probability that a person will win a prize at the ring-toss game has been found to be zo. 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?

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!!!

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.