A teacher in a class of $30$ students randomly selects a group of $4$ students to be his assistants for the week. Tom, Peter, and Jana are friends and would like to be included in the selection together. What is the probability that at least two of these friends will be included in the group?
They calculated the probability as follows:
(1) Jana determined the total number of possible $4$-member groups that the teacher can create as unordered combinations of $4$ out of $30$, i.e.: $${30 \choose 4}=27\,405$$
(2) Petr calculated the number of groups in which all three friends are together, which he believed to be $27$.
(3) Tom calculated the number of groups in which exactly two of the friends are together: $${3\choose2}\cdot{28\choose2}=1\,134$$
(4) Jana quickly summed the number of favorable groups: $27+1134$. According to Jana, the teacher can create a total of $1\,161$ groups in which there are at least two of the friends.
(5) Tom completed the task by calculating the probability that at least two friends will be together: $$\frac{1\,161}{27\,405}=0.0424$$
Did they solve the problem correctly? If anyone made a mistake in their reasoning, who made it?
They solved the problem correctly together.
Jana made a mistake in step (1). She should have calculated the total number of $4$-member groups as ordered sets of four: $$30\cdot29\cdot28\cdot27=657\,720$$
Petr made a mistake in step (2). The number of groups in which all three friends are together is $30$.
Tom made a mistake in step (3). The number of groups in which exactly two of the friends are together is: $${3\choose2}\cdot{27\choose2}=1\,053$$
The teacher can create a total of $1080(= 27 + 1053)$ groups in which at least two of the friends are included. The probability that at least two friends will be together is: $$\frac{1\,080}{27\,405}\doteq0.0394$$