Solution: We are given that a basket contains three green apples and six red apples.
Three apples are randomly selected from the basket.
Now the probability of selecting first apple green is
![(3)/(9) =(1)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/ckxgsubgqtltw8kn4h9g8ntqypt352dfc5.png)
The probability of selecting second apple green is
![(2)/(8) = (1)/(4)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/8bimt4q497buuf835bgc7e78e58us0cl9s.png)
The probability of selecting third apple green is
![(1)/(7)](https://img.qammunity.org/2019/formulas/mathematics/college/ududfdqvzfejyjcfp56onhqna053tkvett.png)
Therefore, the probability of selecting three green apples is:
or 1.19%
Now how many red apples must be added to the basket in order to make the above probability smaller than 0.1%
If we add 11 red apples to the basket, then there will be 17 red apples and 3 green apples in the basket. Therefore, the probability of selecting three green apples if three apples are randomly selected is:
or 0.09%
Therefore, we need to add 11 red apples to the basket to make this probability smaller than 0.1%