Probability of profit when the number of buyers is a random variable

Question

Just started working with random variables, would appreciate if anyone could check if the solution is correct :)

The problem: A kid runs a lemonade stand. He has 64 drinks up for sale in total. Each sold drink nets him a profit of 21 units, each unsold drink nets him a loss of 20 units. The total profit of his business during a period of time is a random variable X. The number of purchases is a random variable with equally likely values of 0, 1, ..., 64. Find:

1. The value of distribution function Fₓ(x) when x = 27.38
2. The probability that the kid suffers no loss (profit ≥ 0).

Let Y be the number of drinks sold. Then the profit from the lemonade stand can be expressed as

$$X = 21Y - 20(64 - Y) = 41Y - 1280$$

--- 1 ---

In order to calculate Fₓ(27.38), we need to get $$P(X ≤ 27.38)$$

Let's rewrite this inequality:

$$41Y - 1280 ≤ 27.38$$

$$Y ≤ \frac{(27.38 + 1280)}{41} ≈ 31.88$$

So, in order for the kid to get a profit of 27.38 or less, he needs to sell 31 drinks or less.

Since the probability of selling a drink is 0.5, the probability of selling i drinks is

$$P(Y = i) = C(64, i) * (0.5)⁶⁴$$

Then, where 0 ≤ i ≤ 31,

$$∑P(Y = i) ≈ 0.4503266$$

will give the answer for Fₓ(27.38) which is ≈0.4503266.

--- 2 ---

In order to calculate the probability that the kid is going to end up with profit ≥ 0, we need to get $$P(0 ≤ X)$$

Let's rewrite this inequality:

$$0 ≤ 41Y - 1280$$

$$Y ≥ \frac{1280}{41} ≈ 31.21$$

So, in order to end up with profit of more than 0, the kid needs to sell at least 32 drinks.

Like in the 1st question, since the probability of selling a drink is 0.5, the probability of selling i drinks is

$$P(Y = i) = C(64, i) * (0.5)⁶⁴$$

Then, where 32 ≤ i ≤ 64,

$$∑P(Y = i) ≈ 0.5496734$$

So the probability of profit being positive or zero is 0.5496734.

Answer 1

The number $Y$ of drinks sold is not a binomial random variable. You are told

The number of purchases is a random variable with equally likely values of $0, 1, \ldots, 64$.

This means $Y$ is a discrete uniform random variable, specifically $$\Pr[Y = y] = \frac{1}{65}, \quad y \in \{0, 1, \ldots, 64\}.$$ Each of the $65$ possible outcomes is equally likely, analogous to rolling a fair $65$-sided die.

Hence you will need to recalculate the relevant probabilities. Your model for $Y$ is that of a binomial random variable, which represents the outcome of a fair coin toss for each of the $64$ drinks, and purchase of a given drink is contingent on, say, the coin toss being heads. In such a situation, it is much more likely that $32$ drinks are sold than $0$ drinks sold, and this violates the assumptions of the problem.