How do you do this problem and what are the conditions met that makes this a binomial distribution(all 3 conditions)?

The three conditions of a binomial distribution are:

Each trial has two outcomes (eg: "yes or no", "true or false", "red chip or not red chip", etc)There are a fixed number of trials. This is the value of n.Each trial has the same probability of success. This is the value of p.

In this case, all of the conditions of a binomial have been met because...

There are two outcomes. We either get a red chip, or we don't get a red chip. I'm focusing on the red chips because the question asks about the probability of getting 1 red chip.We have n = 5 trials, where each trial consists of selecting a chip and putting it back. It's important to put the chip back so that p is not altered.The probability of success is p = 6/9 = 2/3 This is the probability of selecting a red chip. This is because there are 6 red out of 6+3 = 9 total.

Since we only want one red chip, and there are 5 slots to fill (1 red+4 black), this must mean there are exactly 5 ways to get what we want.

We could have the following selections

With R = red and B = black. This value 5 will be used as the binomial coefficient. You can find this value through the use of the nCr combination formula or using Pascal's Triangle.

Note: we only have 3 black chips, but because we put the chip back each time, it means we can get more than 3 black selections.

We'll also be using x = 1 to find the probability of getting exactly one red chip

P(x) = (binomialCoefficient)*(p)^x*(1-p)^(n-x)

P(x) = 5*(p)^x*(1-p)^(n-x)

P(1) = 5*(2/3)^1*(1-2/3)^(5-1)

P(1) = 0.0411523

The probability of getting exactly one red chip is roughly 0.0411523

This does not make sense

These graph is cross the yaxis and x axis two ways.