As we are given in the problem
Total number of students=7
Number of martial arts master=2
Teacher chooses 2 students randomly.
Two Students can be chosen in 7C2 ways
[tex] 7_C_{_2} \ ways=\frac{7!}{2!5!}=\frac{6*7}{2}=21 [/tex]
and two martial arts masters out of 2 can be chosen in [tex] 1 \ ways\\ [/tex]
Then the probability that both of them are martial arts masters can be given by
[tex] P(E)=\frac{n(E)}{n(S)}\\
\\
n(E)=1, n(s)=21\\
\\
P(E)=\frac{1}{21}\\ [/tex]
In a class of 7, 2 students are martial arts masters. The probability that both of them are martial arts masters would be 1/21.
Probability refers to a possibility that deals with the occurrence of random events. The probability of all the events occurring need to be 1.
The formula of probability is defined as the ratio of a number of favorable outcomes to the total number of outcomes.
P(E) = Number of favorable outcomes / total number of outcomes
It is given that
Total number of students = 7
Number of martial arts masters = 2
Two Students can be chosen in [tex]7c_{2}[/tex] ways
[tex]7c_{2}[/tex] = 7! / 2! 5!
= 21
Two martial arts masters out of 2 can be chosen in 1 way.
Then the probability that both of them are martial arts masters
P(E) = Number of favorable outcomes / total number of outcomes
= 1/21
Thus, The probability that both of them are martial arts masters would be 1/21.
Learn more about probability here;
https://brainly.com/question/11234923
