1 out of 3 will play soccer.
1. We can think about this problem as the probability of 3 events happening.
The first event is the teacher choosing one student who does not play soccer. The second event is the teacher choosing another student who does not play soccer, given that the teacher already chose someone who does not play soccer , and so on.
2. The probability that the teacher will choose someone who does not play soccer is the number of students who do not play soccer divided by the total number of students: 3/7.
3. Once the teacher's chosen one student, there are only 6 left.
4. There's also one fewer student who does not play soccer, since the teacher isn't going to pick the same student twice.
5. So, the probability that the teacher picks a second student who also does not play soccer is 2/6.
6. The probability of the teacher picking two students who do not play soccer must then be 3/7*2/6.
7. We can continue using the same logic for the rest of the students the teacher picks.
8. So, the probability of the teacher picking 333 students such that none of them play soccer is:
the probability of choosing a student who plays soccer is 57%meanwhile the chance of picking a non soccer player is 43%.
Students who do not play soccer = 7 - 4 = 3
Probability = 3/7 x 2/6 x 1/5 = 1/35
The probably that none of the 3 play soccer is 1/35.
Probability is defined as the likeliness of an event to occur,
Probability = # of favorable outcomes / # of total outcomes
As 4 students play soccer, that means the remaining 7 - 4 or 3 students do not play soccer. This would be considered our favorable event.
#of favorable outcomes = 3C3
Our number of total outcomes on the other hand would be choosing 3 from the 7 students in total.
# of total outcomes = 7C3
Substitute these values,
Probability = 3C3 / 7C3
Probability = [3! / 3!0!] / [7! / 3!4!]
Probability = 1 / [5 * 6 * 7 / 3!]
Probability = 1 / [5 * 6 * 7 / (1 * 2 * 3)]
Probability = 1 / 35
Therefore, the probability that none of them play soccer is 1/35.