To find the theoretical probability that Nadir will be chosen as one of the four members of the student council, we first need to determine the total number of combinations for selecting 4 students from a group of 6 students.
The number of ways to choose \( k \) students from \( n \) students is given by the binomial coefficient, which is calculated as follows:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
In our scenario, we have \( n = 6 \) and \( k = 4 \):
\[ \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4! \cdot 2!} \]
Calculating \( 6! \), \( 4! \), and \( 2! \):
- \( 6! = 720 \)
- \( 4! = 24 \)
- \( 2! = 2 \)
Now, substitute these values in:
\[ \binom{6}{4} = \frac{720}{24 \cdot 2} = \frac{720}{48} = 15 \]
So, there are 15 different combinations of choosing 4 students out of 6.
Next, we need to find out how many of these combinations include Nadir. If Nadir is chosen, we then need to select 3 more students from the remaining 5 students (Michelle, Olivia, Parvi, Quinn, and Richard). The number of ways to choose 3 students from 5 is given by:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} \]
Calculating \( 5! \), \( 3! \), and \( 2! \):
- \( 5! = 120 \)
- \( 3! = 6 \)
- \( 2! = 2 \)
Now, substitute these values in:
\[ \binom{5}{3} = \frac{120}{6 \cdot 2} = \frac{120}{12} = 10 \]
Thus, there are 10 combinations that include Nadir as one of the selected students.
Finally, the probability that Nadir is chosen is the number of favorable outcomes divided by the total number of outcomes:
\[ P(\text{Nadir is chosen}) = \frac{\text{Number of combinations including Nadir}}{\text{Total number of combinations}} = \frac{10}{15} \]
To simplify the fraction:
\[ \frac{10}{15} = \frac{2}{3} \]
Therefore, the theoretical probability that Nadir will be chosen is:
\[ \boxed{\frac{2}{3}} \]