A store sells two types of tables: plain and deluxe. When a customer buys a table, there is an 80% chance that it will be a plain table. Assume that whether or not the type of table that one customer buys is independent of the types of the tables that any other customer buys.
a. Suppose that on each of the days Monday, Tuesday, and Wednesday five tables are sold. What is the probability that all the tables sold on Monday and Tuesday are plain and that on Wednesday at least one deluxe table is sold?
b. Suppose the number of tables sold each day is a Poisson random variable with parameter x = 5. What is the probability that exactly five tables will be sold on a given day?
Prob(5plain tables) on Monday = .8^5
prob(5 plain) on Tuesday = .8^5
prob(5 plain on Monday AND 5 plain on Tuesday) = (.8^5)(.8^5) = .8^10 = .107
Prob(at least one deluxe) = 1 - Prob(all 5 plain)
= 1 - .8^5 = .672
b. don't know about Poisson random variables.
A poisson distribution follows the formula: (e^(-x) * x^k)/k!
where x is the expected mean and k is the number of "true" events. You are given that the mean number of sales of tables is 5 (x=5) and you are asked for the probability that exactly 5 tables are sold (k=5)
Plug in the formula:
e^(-5) = .066738
5^5 = 3125
5! = 120
poisson=.1755 = 17.55%
See: http://www.answers.com/topic/poisson-distribution
To solve these probability questions, we will break them down step by step and apply the relevant formulas and concepts.
a. Probability that all tables sold on Monday and Tuesday are plain, and at least one deluxe table is sold on Wednesday:
Step 1: Calculate the probability that all tables sold on Monday and Tuesday are plain.
Using the given information, we know that there is an 80% chance of a table being plain, and the probability of a table being deluxe is therefore 20%. Since the table purchases on Monday and Tuesday are independent events, we can use the multiplication rule for independent events to calculate the probability of all tables being plain on these two days.
P(Monday and Tuesday are plain) = P(Monday is plain) * P(Tuesday is plain)
= 0.8 * 0.8
= 0.64
Step 2: Calculate the probability that at least one deluxe table is sold on Wednesday.
Since the probabilities of selling plain and deluxe tables are complementary (i.e., P(deluxe) = 1 - P(plain)), we can find the probability of at least one deluxe table being sold by subtracting the probability that all tables sold on Wednesday are plain from 1.
P(at least one deluxe table on Wednesday) = 1 - P(Wednesday is plain)
= 1 - 0.8 * 0.8 * 0.8
= 1 - 0.512
= 0.488
Step 3: Calculate the probability that all tables sold on Monday and Tuesday are plain and at least one deluxe table is sold on Wednesday.
To calculate the probability of these two events happening together, we multiply their individual probabilities.
P(all plain on Mon and Tue, at least one deluxe on Wed) = P(Mon and Tue plain) * P(at least one deluxe on Wed)
= 0.64 * 0.488
= 0.31232
Therefore, the probability that all the tables sold on Monday and Tuesday are plain and that at least one deluxe table is sold on Wednesday is approximately 0.31232 or 31.23%.
b. Probability of exactly five tables being sold on a given day, given that the number of tables sold each day is a Poisson random variable with a parameter of x = 5.
To solve this problem, we will use the Poisson probability formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- X is the random variable representing the number of tables sold on a given day
- k is the number of tables sold
- λ is the parameter of the Poisson distribution, which represents the average number of tables sold per day
In this case, λ = 5 (as given in the problem).
P(X = 5) = (e^(-5) * 5^5) / 5!
= (e^(-5) * 3125) / (5*4*3*2*1)
≈ 0.1755 or 17.55%
Therefore, the probability that exactly five tables will be sold on a given day is approximately 0.1755 or 17.55%.