How to use bionominal distribution on independent variables?

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Can someone please explain to me how to do question 11? Thanks!

1 Answer

1-((4),(4))(.25)^0(.75)^4~~1-(1)(1)(0.3164)~~0.6036

Explanation:

Let's first talk about binomial probability.

The relation I like to start with is

sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1

which essentially says that for a given probability (p) over a given number of trials (n), we can look at each trial (k). The sum of the results for each k over the entire n will be the entire field of probabilities for that given situation, and so will sum to 1.

Do we have a situation where we can use the relation? Yes.

We have n=4, p=.25. The situation where our customer won't buy anything, because all the cashiers are busy, is with k=4.

Before plugging in the figures, let's talk about the need for the cashiers to be independent of each other. The binomial probability only works if p doesn't change. The way p doesn't change is that the status of the cashiers is independent - if Cashier 1 is most likely to be busy, and then Cashier 2 is influenced by the activity/status of Cashier 1, and that then influences the probability that Cashier 3 is busy... p changes in that situation. And so independence is vital to allowing binomial probability to work.

The probability of our customer making a purchase can be found either by summing up 0<=k<4, or by subtracting the result of k=4 from 1. I'll do it that second way.

1-((4),(4))(.25)^0(.75)^4~~1-(1)(1)(0.3164)~~0.6036