Answer:
The probability that one red chip was selected is 0.0053.
Explanation:
Let the random variable X be defined as the number of red chips selected.
It is provided that the selections of the n = 5 chips are done with replacement.
This implies that the probability of selecting a red chip remains same for each trial, i.e. p = 6/9 = 2/3.
The color of the chip selected at nth draw is independent of the other selections.
The random variable X thus follows a binomial distribution with parameters n = 5 and p = 2/3.
The probability mass function of X is:
![P(X=x)={5\choose x}\ ((2)/(3))^(x)\ (1-(2)/(3))^(5-x);\ x=0,1,2...](https://img.qammunity.org/2021/formulas/mathematics/college/cwxa5y7rhbi2msjirw530w1idopnmulbgt.png)
Compute the probability that one red chip was selected as follows:
![P(X=1)={5\choose 1}\ ((2)/(3))^(1)\ (1-(2)/(3))^(5-1)](https://img.qammunity.org/2021/formulas/mathematics/college/izblwi28wjjfkg4k5hoed820y4s5acch52.png)
![=5*(2)/(3)* (1)/(625)\\\\=\farc{2}{375}\\\\=0.00533\\\\\approx 0.0053](https://img.qammunity.org/2021/formulas/mathematics/college/jp97t6cbbgooubme2w7q41pemvf5j280jc.png)
Thus, the probability that one red chip was selected is 0.0053.