Answer:
[tex]\frac{73,331}{75,516}\approx 97.11\%[/tex]
Step-by-step explanation:
The probability that she will find at least one student cheating is equal to the probability that she finds no students cheating subtracted from 1.
Each time she randomly chooses a student the probability she will catch a cheater is equal to the number of cheaters divided by the number of students.
Therefore, for the first student she chooses, there is a [tex]\frac{9}{35}[/tex] chance that the student chosen is a cheater and therefore a [tex]\frac{26}{35}[/tex] chance she does not catch a cheater. For the second student, there are only 34 students to choose from. If we stipulate that the first student chosen was not a cheater, then there is a [tex]\frac{9}{34}[/tex] chance she will catch a cheater and a [tex]\frac{25}{34}[/tex] chance she does not catch the cheater.
Therefore, the probability she does not catch a single cheater after randomly choosing ten students is equal to:
[tex]\frac{26}{35}\cdot \frac{25}{34}\cdot \frac{24}{33}\cdot \frac{23}{32}\cdot \frac{22}{31}\cdot \frac{21}{30}\cdot \frac{20}{29}\cdot \frac{19}{28}\cdot \frac{18}{27}\cdot \frac{17}{26}[/tex]
Subtract this from one to get the probability she finds at least one of the students cheating after randomly selecting nine students. Let event A occur when the professor finds at least one student cheating after randomly selecting ten students from a group of 35 students.
[tex]P(A)=1-\frac{26}{35}\cdot \frac{25}{34}\cdot \frac{24}{33}\cdot \frac{23}{32}\cdot \frac{22}{31}\cdot \frac{21}{30}\cdot \frac{20}{29}\cdot \frac{19}{28}\cdot \frac{18}{27}\cdot \frac{17}{26},\\\\P(A)=1-\frac{2,185}{75,516},\\\\P(A)=\boxed{\frac{73,331}{75,516}}\approx 0.97106573441\approx \boxed{97.11\%}[/tex]
