Statistics 1 - Arrangements and Selections

Hi,

I need some help in doing questions involving Arrangements and Selections, particularly the probabilities involved in these types of questions.

So, I need to know when to use which function, for example: nCr or nPr (or n!).
I used nCr when the question asks for the number of possibles selections, and nPr when it asks for how many different arrangements are possible.

However, when probability comes into the questions I find it much more difficult. I guess, really, I just need another lesson doing arrangements and selections!

6b) Find the probability that the seven cards include exactly one card showing the letter A.

(The diagram below shows 5 white cards and 10 grey cards, each with a letter painted on it.)

A | C | E | J | T - White cards

A | B | G | L | M
N | P | Q | R | Z - Grey cards

It is these types of questions which I find difficult. Looking at the mark schemes allow me to answer the question, as I know how to start.

But, for the real exam, I can't whip out a mark scheme and see how to start. So, I need guidance on how to start these kinds of questions involving probabilities! How to decipher the question, and use the information to answer the question.

Any tips for improving at selections and arrangements?
Thanks! :smile:

Reply 1

ghostwalker

17

Original post by Konflict

Any tips for improving at selections and arrangements?
Thanks! :smile:

Can't think any obvious tips, but here's to ways to do the specific question which might shed some light.

You need to choose 7 cards so that just one is an A.

One way.
So, you can choose A followed by 6 other cards.

Probability A is the first card is 2/15.
Then probability second card is not an A is 13/14.
Third card is not an A is 12/13.
Etc.
Giving 8/(15x7).
But the A could have occurred in any of 7 positions so multiply by 7 to get 8/15.

Alternatively:
You can consider the number of choices.
Number of ways of choosing an A, is 2C1.
Number of ways of choosing the remaining 6 cards is 13C6
Total possible number of choices with no restrictions is 15C7.

Hence probability of choosing just one A is (2C1 x 13C6)/(15C7)= 8/15.

Reply 2

Konflict

OP

7

Original post by ghostwalker

Can't think any obvious tips, but here's to ways to do the specific question which might shed some light.

You need to choose 7 cards so that just one is an A.

One way.
So, you can choose A followed by 6 other cards.

Probability A is the first card is 2/15.
Then probability second card is not an A is 13/14.
Third card is not an A is 12/13.
Etc.
Giving 8/(15x7).
But the A could have occurred in any of 7 positions so multiply by 7 to get 8/15.

Alternatively:
You can consider the number of choices.
Number of ways of choosing an A, is 2C1.
Number of ways of choosing the remaining 6 cards is 13C6
Total possible number of choices with no restrictions is 15C7.

Hence probability of choosing just one A is (2C1 x 13C6)/(15C7)= 8/15.

Thanks.

I figured how to do it after looking at the mark scheme. :tongue:

Originally, my post was just going to be on hints on how to complete such arrangement and selection questions, and what method to use. However, after reading it (and, as you mentioned), it's quite hard to ask for tips when you don't have a set example! :tongue:

Hardest part of S1 and C2 for me - these permutations... :frown:

Statistics 1 - Arrangements and Selections

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