Proabity formulas or rules

What would be the correct approach to solving these problems:

Sixth grade students are randomly put into classrooms every year. There are 100 students and 4 classes. Three students, are best friends and want to be in the same class.
a) What is the probability that they are all in the same class?
= 1/4 * 1/4 * 1/4 = .016
b) What is the probability that they are all in Mr. Collins’ class?
= 25/100 * 24/100* 23/100 = .0138
c) What is the probability that they are all in different classes?
= ?

Try (a) and (b) again.

Suppose the classes have the same number of students, 25 in each
and there is an equal chance that a student can end up in any of the 4 classes.

Then, the chances of the 3 friends ending up in Mrs Collins' class, if all 4 classes have different
teachers, is 0.25(0.25)(0.25)

as this is the chance they all end up in the same particular class.
Mrs Collins' class is one of the 4 classes the friends could be together in.

Then, as that is the answer to (b), the answer to (a) is how many times that ?

For (c) you can find out the chances of 2 of them being together, add the answer to the answer for (a)
and subtract the total from 1, though you can do it in alternative ways.

Think through the logic first before trying (c) and complete (a) and (b).

so a) would be .25*.25*.25*4= .0625?

yes, that's the chance they could be in any of the 4 classes together,
the probabilty they are all in the same class.

shouldn't the answer to c) be 3/4 (3 students in four classes)=0.75 ?? what am I missing?

The 3 students will be in 3 different classes out of the 4 classes for part (c)

We could list these possibilities as follows:

Student A in class 1, Student B in class 2, Student C in class 3
A in 1, B in 2, C in 4
A in 1, B in 3, C in 2
A in 1, B in 3, C in 4
A in 1, B in 4, C in 2
A in 1, B in 4, C in 3

That is 6 or 3!

The same procedure for A being in any of the other 3 classes gives 4(3!) = 4! possibilities altogether.

The total number of possibilities open to the friends being in whatever classes is 4[sup:242fofgb]3[/sup:242fofgb].
This is because the possibilities are:

A in 1, B in 1, C in 1
A in 1, B in 1, C in 2
A in 1, B in 1, C in 3
A in 1, B in 1, C in 4

A in 1, B in 2, C in 1
A in 1, B in 2, C in 2,
A in 1, B in 2, C in 3,
A in 1, B in 2, C in 4

if you keep counting, that will be 4+4+4+4 with A in class 1.
Then the same procedure with A in classes 2, 3 and 4.
The total is 4(4[4])=4[sup:242fofgb]3[/sup:242fofgb].

Therefore the probability of the students being in different classes completely is 4!/(4[sup:242fofgb]3[/sup:242fofgb])
= 6/(4[sup:242fofgb]2[/sup:242fofgb]) = 3/8.

That is one of the 2 main ways to find the probability.

All probabilities in this situation (students in various classes) sum to 1.

If we calculate the chances of 2 of the 3 students being in the same class, we get 9/16.
That covers all 3 cases and the sum of the probabilities is (1+9+6)/16 = 1.