Respuesta :
Answer:
18.59% probability that EXACTLY 10 will show that the above three food items were ordered
Step-by-step explanation:
For each customer, there are only two possible outcomes. Either they order the above food items, or they do not. The probability of a customer ordering these items is independent of other customers. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Sixty percent of the customers of a fast food chain order the Whopper, french fries and a drink.
This means that [tex]p = 0.6[/tex]
If a random sample of 15 cash register receipts is selected, what is the probability that EXACTLY 10 will show that the above three food items were ordered
This is P(X = 10) when n = 15. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 10) = C_{15,10}.(0.6)^{10}.(0.4)^{5} = 0.1859[/tex]
18.59% probability that EXACTLY 10 will show that the above three food items were ordered
The probability that EXACTLY 10 will show that the above three food items were ordered is 0.1859 or 18.59%.
Given information:
Sixty percent of the customers of a fast-food chain order whopper, french fries, and a drink.
A random sample of 15 cash register receipts is selected.
It is required to calculate the probability that EXACTLY 10 will show that the above three food items were ordered.
The events of choosing an item are independent. So, binomial distribution can be used to calculate the required probability.
So, the value of n is 15 and the value of x will be 10.
The probability of success will be 60% or 0.6 and that of failure will be 1-0.6=0.4. So, [tex]p=0.6[/tex] and [tex]q=0.4[/tex].
So, the required probability can be calculated as,
[tex]P=^nC_xp^xq^{n-x}\\P=^{15}C_{10}\;0.6^{10}\times 0.4^{5}\\P=3003\times 0.0060466176\times 0.01024\\P=0.1859\;\rm or\; 18.59\%[/tex]
Therefore, the probability that EXACTLY 10 will show that the above three food items were ordered is 0.1859 or 18.59%.
For more details, refer to the link:
https://brainly.com/question/17217746
