Sixty percent of the customers a fast food chain order hamburger french fries and drink If random sample 15 cash register receipts is selected what probability that 10 or more wi?

The probability of a customer ordering a hamburger, french fries, and drink is 0.6. Therefore, the probability of not ordering a hamburger, french fries, and drink is 1 - 0.6 = 0.4.

The number of customers ordering a hamburger, french fries, and drink in a random sample of 15 cash register receipts is a binomial random variable with parameters n = 15 and p = 0.6.

The probability of at least 10 customers ordering a hamburger, french fries, and drink is:

$$P(X ≥ 10) = 1 - P(X ≤ 9)$$

The random variable X follows a binomial distribution with parameters n = 15 and p = 0.6. Thus, the cumulative distribution function for X is given by:

$$P(X ≤ k) = \sum_{r=0}^k {15 \choose r} (0.6)^r (0.4)^{15-r}$$

Therefore,

$$P(X ≤ 9) = \sum_{r=0}^9 {15 \choose r} (0.6)^r (0.4)^{15-r} = 0.214$$

And,

$$P(X ≥ 10) = 1 - 0.214 = 0.786$$

Therefore, the probability that 10 or more of the customers in the random sample will order a hamburger, french fries, and drink is 0.786.