OCR S1 Combinations and Permutations Exam paper question.

Obviously I know the answer to this question now because I have looked at the mark scheme but I am still unsure of how you do it and would be grateful if someone could show me how:

'An examination paper consists of two parts. Section A contains questions A1, A2, A3 and A4. Section B contains questions B1, B2, B3, B4, B5, B6 and B7.'

(ii) Assuming that all selections are equally likely, find the probability that a particular candidate chooses question A1 but does not choose question B1.

I know how to work out the total number of arrangements but that's it.

^4\mathrm{C}_3 \times ^7\mathrm{C}_4 = 140

Reply 1

gangsta316

13

Choosing 3 from section A and four from section B?

Is it this?

3C2 \times 6C4 = 45

\frac{45}{140}

=

\frac{9}{28}

Reply 2

122025278

15

The answer is just 6/7, its easy

Reply 3

Maths Buster

7

Superacid

Obviously I know the answer to this question now because I have looked at the mark scheme but I am still unsure of how you do it and would be grateful if someone could show me how:

'An examination paper consists of two parts. Section A contains questions A1, A2, A3 and A4. Section B contains questions B1, B2, B3, B4, B5, B6 and B7.'

(ii) Assuming that all selections are equally likely, find the probability that a particular candidate chooses question A1 but does not choose question B1.

I know how to work out the total number of arrangements but that's it.

^4\mathrm{C}_3 \times ^7\mathrm{C}_4 = 140

Chance of choosing A(1) = 1/4; chance of not choosing B(1) = 6/7.
You want both of these events to occur and so you multiply the probabilities to get [1/4] X [6/7] = 6/28.
Alternatively draw a tree with branches 1/4 and 3/4 and then with 1/7 and 6/7.

Reply 4

XEROX

Have a look at http://www.schoolworkout.co.uk/a_level.htm at the bottom, theres solutions for most of S1 papers, believe that ones off Jan 2006. Has been helping me!

:smile:

Reply 5

ofolion

I find it easiest just to consider the actual probabilities where i can rather than getting into combinations. Probability for choosing A1 is 3/4 (OP left a bit of the question out) as there is 4 questions and 3 must be chosen. Probability of not choosing 7 is 3/7 as 4 out of 7 must be chosen. 3/4 * 3/7 = 9/28 as it says in the markscheme

Reply 6

MANNESMANN

Has anyone got this paper's mark scheme?

Reply 7

spyderr

2

Original post by gangsta316

Choosing 3 from section A and four from section B?
Is it this?

3C2 \times 6C4 = 45

\frac{45}{140}

=

\frac{9}{28}

I know this is a 15 year old thread, but could anyone explain to me how to do this question? I don't understand why its 3c2 and 6c4

Reply 8

mqb2766

21

Original post by spyderr

I know this is a 15 year old thread, but could anyone explain to me how to do this question? I don't understand why its 3c2 and 6c4

Its a very old thread and best to start a new one, but you choose 3 from 4 in section A and 4 from 7 in section B. You want the case where you choose A1 and dont choose B1, so they must choose 2 from the remaining 3 in section A and 4 from the remaining 6 in section B.

OCR S1 Combinations and Permutations Exam paper question.

Quick Reply

Related discussions

Articles for you