Number of Arts Students : 90

P(exactly 2 out of 3 students chosen (without replacement) student Art)

= P(1st student Art, 2nd student Art, 3rd student non-Art)

+ P(1st student Art, 2nd student non-Art, 3rd Art)

+ P(1st student non-Art, 2nd student Art, 3rd Art)

=

90/300 x 89/299 x 210/298
+ 90/300 x 210/299 x 89/298
+ 210/300 x 90/299 x 89/298

= 3 x (90 x 89 x 210)/(300 x 299 x 298)

= 16821/89102

≈ 0.18878

≈ 0.189 (3 d.p)

Edited

Can it be [ (90C1 x 89C1 x 210C1)(3!/2!) ] / 300C3

No. There is no need to permute the numerator since you're doing it in the combinations way. The 3! should not be there.

There is also actually no need to separate the two arts students. We can just choose 2 out of 90 straight away instead of each one separately.


(Edited)

It would just be (90C2 x 210C1) / (300C3)

If you try to compute the calculation you provided,


(90C1 x 89C1 x 210C1 x 3!/2!) / 300C3

= (90 x 89 x 210 x 3) / ( 300! / ( 297! 3!) )

= (90 x 89 x 210 x 3) / (300 x 299 x 298 ÷ 6)

= (90 x 89 x 210 x 3 x 6) / (300 x 299 x 298)

= 1.13270185


The probability is already more than 1, which is not possible.

I see, thanks!

Welcome


If you choose to pick the Arts students separately,

it would be 90C1 x 89C1 instead of 90C2


So there are 90 choices for the 1st one and 89 choices for the second one. But, we will end up having repeated cases.

Let's say we label each of the 90 Arts students from A1 to A90

And we have these two cases:

①1st student is A1, 2nd is A90
②1st student is A90, 2nd is A1

This two cases are permutations of each other. But since we are doing combinations, they are considered the same as order does not matter.

They both indicate choosing student A1 and A90.


Likewise, something like :

①1st student is A55, 2nd is A63
②1st student is A63, 2nd is A55

These are also repeated cases.


So for every one case, there is a permutation that gives the same combo.
Basically, this means that half of all the combos are repeated ones.

We would have over counted if we only do 90C1 x 89C1. We need to divide by 2! or 2 to eliminate these repeated cases since we only want half of them.



90C1 x 89C1 ÷ 2!

= (90 x 89) / 2

This is actually the same as doing 90C2

90C2
= 90!/((90-2)!2!)
= 90!/(88!2!)
= (90 x 89)/2

So actually we can just choose 2 arts students out from 90 immediately. If you choose separately, you must remember to divide by 2!