probability question

Question

Question 4

In a café, coffee and tea are on average sold at 3 cups and 1 cup respectively every 5 minutes. Let C and T denote the number of cups of coffee and tea sold in an hour interval. (a) State any assumptions required for us to use the Poisson distribution to model the random variables C and T . Comment on how reasonable the assumption is in practice. (4 marks)

(b) Assume that we can model C and T after Poisson distributions. Find the probability that a total of 45 cups of beverages (you may assume the only beverages sold in the café are either coffee or tea) in a randomly chosen one hour interval. State any assumptions you have made in your computations. (4 marks)

(c) It is known that from 15:00 to 16:00, 45 cups of beverages are sold. Find the probability that the number of cups of coffee sold in this time period is between 30 to 33 cups, inclusive. State any assumptions you have made in your computations. (6 marks)

(d) Let W denotes the time (in hours) passed after a cup of beverage is sold before the next is sold. State the probability distribution of W and write down the values of E(W) and Var(W ) (5 marks)

(e) I am in a queue at the café waiting to order my drink. There are two counters, A and B, in front of me that I can turn to. On average, I need to wait for 10 minutes for my turn to arrive to order at counter A and 8 minutes for that at counter B. I will proceed to whichever counter is available first. Find the probability that I will have to wait for at least 7 minutes for my turn to make my order. State any assumptions you have made in your computations. (5 marks)

Answer 1

(a) The assumptions required to use the Poisson distribution to model the random variables C and T are:

1. The number of cups sold in any time interval is independent of the number sold in any other time interval.
2. The probability of selling a cup of coffee or tea is the same for any given time interval.
3. The number of cups sold in a time interval is proportional to the length of the interval.
4. The average number of cups sold in a time interval is constant over time.

Assumption (1) is reasonable in practice as the sales of cups of coffee and tea are likely to be independent of each other, and the sales of cups in any given time interval are unlikely to be affected by sales in previous or future time intervals. Assumptions (2)-(4) may not hold exactly in practice, but they are reasonable approximations for modeling the random variables C and T.

(d) Let W denote the time passed after a cup of beverage is sold before the next is sold. The probability distribution of W is the exponential distribution with parameter λ = 1/mean time between sales.

The mean and variance of W are:

E(W) = 1/λ = 1/((1/10) + (1/8)) = 4.44 minutes Var(W) = 1/λ^2 = 1/((1/10) + (1/8))^2 = 3.21 minutes^2

The assumption made in this computation is that the time between sales of cups of coffee and tea follows an exponential distribution.

(e) Let W_A and W_B denote the waiting times for counter A and B, respectively. Then W_A and W_B are exponentially distributed with

parameters λ_A = 1/10 and λ_B = 1/8, respectively. Let W be the waiting time for the first available counter, which is the minimum of W_A and W_B. Then the probability that W is greater than or equal to 7 minutes is:

P(W ≥ 7) = P(W_A ≥ 7, W_B ≥ 7) = P(W_A ≥ 7) * P(W_B ≥ 7)

Using the cumulative distribution function of the exponential distribution, we get:

P(W_A ≥ 7) = e^(-λ_A * 7) ≈ 0.498 P(W_B ≥ 7) = e^(-λ_B * 7) ≈ 0.420

Therefore, the probability that the waiting time for the first available counter is at least 7 minutes is:

P(W ≥ 7) ≈ 0.498 * 0.420 ≈ 0.209

The assumption made in this computation is that the waiting times for counter A and B are independent exponentially distributed random variables.

Note: In part (b), there was an error in the variance calculation. The variance of C+T should be the sum of the variances of C and T, not their sum squared. The correct probability is:

P(C+T = 45) = e^(-λ) * λ^(45) / 45! ≈ 0.099

Sorry for any confusion caused. Here are the corrected solutions for parts (b) and (c):

(b) If C and T are Poisson distributed, then the mean and variance of C are:

mean(C) = λ_C = 3 cups/5 minutes × 12 × 1 hour = 36 cups/hour var(C) = λ_C = 36 cups/hour

Similarly, the mean and variance of T are:

mean(T) = λ_T = 1 cup/5 minutes × 12 × 1 hour = 12 cups/hour var(T) = λ_T = 12 cups/hour

The total number of cups of beverages sold in one hour is C + T, which is also Poisson distributed with mean λ = mean(C) + mean(T) = 48 cups/hour and variance var(C+T) = var(C) + var(T) = 48 cups/hour.

The probability of selling a total of 45 cups of beverages in a randomly chosen one hour interval is:

P(C+T = 45) = e^(-λ) * λ^(45) / 45! ≈ 0.099

The assumptions made in this computation are that the random variables C and T are Poisson distributed, and that they are independent of each other.

(c) From 15:00 to 16:00, a total of 45 cups of beverages are sold. Let X denote the number of cups of coffee sold in this time period. Then X is a binomial random variable with parameters n = 45 and p = 3/4, since 3 out of 4 cups sold are coffee.

The probability of selling between 30 to 33 cups of coffee, inclusive, is:

P(30 ≤ X ≤ 33) = ∑_(k=30)^(33) (45 choose k) * (3/4)^k * (1/4)^(45-k) ≈ 0.168

The assumption made in this computation is that the number of cups sold in a time interval follows a binomial distribution.