Question

How to use bionominal distribution on independent variables?

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

Answer 1

`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`