Total : 12 books (6 + 5 + 1 = 12)

Number of ways to choose 3 books from 12 (whereby order is not important)

= 12C3
= 12! / 3!(12-3)!
= 12! / 3!9!
= 220

There is only 1 computer book.


Number of ways to choose 1 computer book and 2 other books (total of 3 book

= 1 × 11C2 (set the computer book aside and just pick 2 books out of the remaining 11)

= 11! / 2!(11-2)!
= 11! / 2!9!
= 55

Required probability = 55/220 = ¼

So if I had used the method for independent events to solve this, I would have to write a long working?

Like how?

P(A) = P(A∩B) / P(B) ?

Oh wait not independent events. Yeah something like your equation with the sketch of the tree diagram

This could be another way to get ¼ :

If we select the books one by one,

Probability that the first book is the computer book and the next two can be any book

= 1/12 × 11/11 × 10/10
= 1/12 × 1 × 1
= 1/12

(You can just write 1/12 since if the first book is the computer book, the other two must be/are guaranteed to be other books)


Probability that the second book is the computer book

= 11/12 × 1/11 × 10/10
= 1/12


Probability that the third book is the computer book

= 11/12 × 10/11 × 1/10
= 1/12


This can be summarised as :

1/12 × 3 = ¼

If you use my first method, it will just be :

11C2 / 12C3

= 55/220

= ¼

Ah I see. It's just that whenever I see a question asks to find the probability of an event, I would immediately use your second method to solve it. Thanks!

Welcome.

Probability is just = number of required outcomes ÷ total number of outcomes