## Ask Singapore Homework?

Upload a photo of a Singapore homework and someone will email you the solution for free.

## Question

secondary 4 | E Maths

** One Answer Below **

Anyone can contribute an answer, even non-tutors.

**Answer This Question**

I need help with part a

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

P(A) = P(A∩B) / P(B) ?

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

11C2 / 12C3

= 55/220

= ¼

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

### See 1 Answer