jan 05 OCR s1 past paper help!

the actual question (in shortened version):
4 questions including one on geometric distribution are worth 7 marks each and 4 questions including one on binomial distribution are worth 9 marks each. The 7-mark questions are the first four but are arranged in random order. the last four are the 9-mark questions but also arranged in any order.
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(1)the probability that the question on geo. distribution is next to the binomial one = 1/16

(2)the probability that the questions on geo. distribution and binomial distribution are seperated by AT LEAST 2 other questions... - how do i do this one? - the answers was 13/16. - just rmb to explain the working out etc as the markscheme has not helped..

any input on this is much appreciated (T^T)

Reply 1

Glutamic Acid

I'd:

- Find the total number of permutations.
- Find the number of permutations where they're not separated by at least 2 other questions. This can happen like:
X X X G B X X X
X X G X B X X X
X X X G X B X X
- Subtract the second from the first to give the required number of permutations.

Reply 2

Th0mas1337

Im really struggling to do this question aswell :/

Glutamic the answer asks for a probabilty, not a number of permutations :frown:

Im geussing you divide the number of perms by the total no. of perms?

Reply 3

Glutamic Acid

Yep.

Reply 4

Th0mas1337

Ok, so the toal number of permutations would be 8! = 40320
and the number of permuations were there not seperated is 6!2! x 3 = 4320

40320-4320 = 360000

4320/360000 = 3/250 ?!

GRagghhhhh!

Reply 5

Glutamic Acid

You need to make sure you're keeping the 7 mark questions and the 9 mark questions separate. The total number of permutations isn't 8! because they are restricted. And it's the same when finding the number of permutations for each of the 3 layouts where they aren't separated.

Reply 6

Th0mas1337

So the total is 6!2! and the number ermutations is??

I really can't see it, maybe im just stupid :/ Could you show me your working please? Or part of it?

Reply 7

Glutamic Acid

Permutations and combinations are generally considered to be the hardest part of OCR S1.

The number of ways the 7 mark questions can be arranged is 4!. The number of ways the 9 mark questions can be arranged is 4!. So the total is 4!4!.

Consider the layout:
X X X G B X X X
How many ways can the 7 mark questions be arranged? 3!
How many ways can the 9 mark questions be arranged? 3!
So the total for this layout is 3! x 3!.

Can you finish it off?

Reply 8

Th0mas1337

Ok,

Total = 4!4! Due to The 7 and 9 mark questions being seperate things.

Therefore the number of ways X X X G B X X X can be arranged is 3!3!1!1! taking in mind that the 7 and 9 questions are still seperate but im also ncluding a binomial and geometric question.

But, there are 3 ways in which they are less than 2 together so i multiply 3!3!1!1! by 3

Then 4!4! - 3!3!x3 = 468 (This is demonstrating that im actually looking for when they are at least two apart)
Divide it by the total to give 468/576 = 13/16