Answer with explanation:
Given : Number of individual needed to be selected to form jury = 4
Number of students = 13
Number of faculty members = 6
Total = 13+6=19
Using combinations , the number of combinations of choosing jury of 4 individuals
=
![^(19)C_(4)=(19!)/(4!(19-4)!)\\\\=(19*18*17*16*15!)/(4!15!)=3876](https://img.qammunity.org/2020/formulas/mathematics/high-school/gkt0lxksu5j449kfxcq0ncjdsd23k7kvr1.png)
a) Number of ways of selecting jury of all students
=
![^(13)C_(4)=(13!)/(4!(13-4)!)\\\\=(13*12*11*10*9!)/(4!9!)=715](https://img.qammunity.org/2020/formulas/mathematics/high-school/abg7bk9ybqd6ri3nmfyrdsg27vo246k801.png)
The probability of selecting a jury of all faculty=
![(715)/(3876)=0.1845](https://img.qammunity.org/2020/formulas/mathematics/high-school/4liein0xemzf1mvkgqxfm1vwop4fj6decu.png)
b) The number of ways of selecting jury of all faculty=
![^(6)C_(4)=(6!)/(4!(6-4)!)=15](https://img.qammunity.org/2020/formulas/mathematics/high-school/z8slwubcbf9fnblfd6zitn84qlfor7s2ua.png)
The probability of selecting a jury of all faculty=
![(15)/(3876)=0.0039](https://img.qammunity.org/2020/formulas/mathematics/high-school/sgmpq3kkbwntrg86t2ql072b776jztd9z1.png)
c) The number of ways of selecting jury of 2 students and two faculty:
![^(13)C_(2)* ^(6)C_2\\\\=(13!)/(2!(13-2)!)*(6!)/(2!(6-2)!)\\\\=1170](https://img.qammunity.org/2020/formulas/mathematics/high-school/ytjedoznzw6w5x4aee8tp6tuobe88zbbfm.png)
Now, the probability of selecting a jury of six students and two two faculty
![(1170)/(3876)=0.3019](https://img.qammunity.org/2020/formulas/mathematics/high-school/5x3h6wfww06df5kpws9hjk3waq0shkii99.png)