Answer:
The probability of selecting 5 female and 2 male students is 0.052.
Step-by-step explanation:
The class comprises of 7 female students and 10 male students.
Total number of students: 17.
Number of female students, 7.
Number of male students, 10.
The probability of an event E is:
[tex]P(E)=\frac{Favorable\ outcomes}{Total\ number\ of] outcomes}[/tex]
The number of ways to select 7 students from 17 is:
[tex]N ={17\choose 7}=\frac{17!}{7!(17-7)!}= 19448[/tex]
The number of ways to select 5 female students of 7 females is:
[tex]n(F) ={7\choose 5}=\frac{7!}{5!(7-5)!}= 21[/tex]
The number of ways to select 2 male students of 10 males is:
[tex]n(M) ={10\choose 2}=\frac{10!}{2!(10-2)!}= 45[/tex]
Compute the probability of selecting 5 female and 2 male students as follows:
P (5 F and 2 M) = [n (F) × n (M)] ÷ N
[tex]=\frac{21\times45}{19448} \\=0.05183\\\approx0.052[/tex]
Thus, the probability of selecting 5 female and 2 male students is 0.052.