Answer:
0.3891 = 38.91% probability that only one is a second
Explanation:
For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. 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.
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_(n,x) = (n!)/(x!(n-x)!)](https://img.qammunity.org/2021/formulas/mathematics/college/qaowm9lzn4vyb0kbgc2ooqh7fbldb6dkwq.png)
And p is the probability of X happening.
17% of its goblets have cosmetic flaws and must be classified as "seconds."
This means that
![p = 0.17](https://img.qammunity.org/2021/formulas/mathematics/college/dfctj2gzyffyae0s1b30o38n5gxdkk30qh.png)
Among seven randomly selected goblets, how likely is it that only one is a second
This is P(X = 1) when n = 7. So
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
![P(X = 1) = C_(7,1).(0.17)^(1).(0.83)^(6) = 0.3891](https://img.qammunity.org/2021/formulas/mathematics/college/tz78c0ia06895lwqpyp0xmj1hlhwhg0m14.png)
0.3891 = 38.91% probability that only one is a second