Answer:
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
Explanation:
Let 'M' be the event of selecting males n(M) = 12
Number of ways of choosing 3 students From all males and females
![n(M) = 28C_(3) = (28!)/((28-3)!3!) =(28 X 27 X 26)/(3 X 2 X 1 ) = 3,276](https://img.qammunity.org/2021/formulas/mathematics/college/e70o8ptju06gpn8tzhpfinlahimdq9knd0.png)
Number of ways of choosing 3 students From all males
![n(M) = 12C_(3) = (12!)/((12-3)!3!) =(12 X 11 X 10)/(3 X 2 X 1 ) =220](https://img.qammunity.org/2021/formulas/mathematics/college/fofymbicdfklb4nmeyo0gz501mz4ag3uc4.png)
The probability that all are male of choosing '3' students
![P(E) = (n(M))/(n(S)) = (12 C_(3) )/(28 C_(3) )](https://img.qammunity.org/2021/formulas/mathematics/college/67w3340onri9mkj5gnzjw2zsmd02bvvx7z.png)
![P(E) = (12 C_(3) )/(28 C_(3) ) = (220)/(3276)](https://img.qammunity.org/2021/formulas/mathematics/college/x9hx4fquy5slpuj65lhcdq5038keudmwv8.png)
P(E) = 0.067 = 6.71%
Final answer:-
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%