Answer:
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
Step-by-step explanation:
Let 'M' be the event of selecting males n(M) = 12
Number of ways of choosing 3 students From all males and females
[tex]n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276[/tex]
Number of ways of choosing 3 students From all males
[tex]n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220[/tex]
The probability that all are male of choosing '3' students
[tex]P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }[/tex]
[tex]P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}[/tex]
P(E) = 0.067 = 6.71%
Final answer:-
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
