Math: can anyone possibly do this probability question

http://www.admissionstests.cambridgeasse…

question 23 please

i know this:

2-9

12-15

1 won't work.
11 won't work.

1-9 and 12-15

how to work out the probability , and get a neat set of numbers so i do not make gross calculation errors due to the magnitude of calculations i must perform, i do not know

can anyone shed light in this matter?

thank you

> "Total of 3 digits..."

That means you must pick a 1-digit ball plus a 2-digit ball

> "All different"

That means the possible combo's are:

10, with any of {2,3,4,5,6,7,8,9} (8 possibilities)
12, with any of {3,4,5,6,7,8,9} (7 possibilities)
13, with any of {2,4,5,6,7,8,9} (7 possibilities)
14, with any of {2,3,5,6,7,8,9} (7 possibilities)
15, with any of {2,3,4,6,7,8,9} (7 possibilities)

That's 36 total possible ways to pick three all-different digits.
The number of ways to pick 2 balls is C(15,2) ("15 choice 2") which is 15!/(2!13!), or 105.

So the probability is 36/105, or 12/35.

The sample space is the collection of all pairs (m, n) of numbers such that:

1 <= m <= 15, 1 <= n <= 15, and m =/= n.

Each outcome (m, n) in the sample space has the same probability, which I will denote p. So, you need to calculate how many of the outcomes give three different digits, and multiply that number by p.

To understand how to approach these problems in principle, see chapter 3 of this book (which includes a medical example): http://books.google.com/books?id=PS8lQQ8…