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# AQA S1 probability help please

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1. Hi all,

I don't understand why 3(d) requires the "*6" at the end of the calculation?:

Calculation:
0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

0.084

Attached Images

2. (Original post by londoncricket)
Hi all,

I don't understand why 3(d) requires the "*6" at the end of the calculation?:

Calculation:
0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

0.084

There are ways of arranging those days, which is the same as 3!, which is 6.

It's just like having 3 balls RBG in a bag, each of a different colour, and there are 6 ways of taking them out (RBG, RGB, GBR, GRB, BRG, BGR).
3. (Original post by SeanFM)
There are ways of arranging those days, which is the same as 3!, which is 6.

It's just like having 3 balls RBG in a bag, each of a different colour, and there are 6 ways of taking them out (RBG, RGB, GBR, GRB, BRG, BGR).

So would it be x!, where x corresponds to the number of possible combinations?

Thank you!
4. (Original post by londoncricket)

So would it be x!, where x corresponds to the number of possible combinations?

Thank you!
Not quite

If you have n objects, r of one type and n-r of the other, then the number of ways that you can arrange them is

More generally, if you have n objects, a of one type, b of another, c.... such that a + b + c + .... = n, then the number of ways you can arrange them is and in this case, it's just a,b and c with a = b = c = 1.
5. (Original post by SeanFM)
Not quite

If you have n objects, r of one type and n-r of the other, then the number of ways that you can arrange them is

More generally, if you have n objects, a of one type, b of another, c.... such that a + b + c + .... = n, then the number of ways you can arrange them is and in this case, it's just a,b and c with a = b = c = 1.
Thank you for that.

I am not sure as to whether I fully understand your response. In this case, why does a = b = c = 1? Also, why would I do this in this question: where are the combinations here?

Also, do I simply stick (n!)/(a!)/(b!)/(c!) on to the end of the 0.16/0.8 * 0.56/0.8 * 0.08/0.8? Or should this be done in replacement of the calculation from above?

Thank you!
6. (Original post by londoncricket)
Thank you for that.

I am not sure as to whether I fully understand your response. In this case, why does a = b = c = 1? Also, why would I do this in this question: where are the combinations here?

Also, do I simply stick (n!)/(a!)/(b!)/(c!) on to the end of the 0.16/0.8 * 0.56/0.8 * 0.08/0.8? Or should this be done in replacement of the calculation from above?

Thank you!
The three different types of objects, a, b and c in this case, is that it arrives early (a), arrives on time (b) or arrives late (c) and from the formula, there are 6 different ways of ordering them.

To see this, let's look at some orders. The first one could be early, second on time, third late. The probability of this is
0.16/0.8 * 0.56/0.8 * 0.08/0.8 as you say. (The order in which you multiply them may be different but it does not matter).

Another order could be the reverse - first day is late, second is on time, third is early. The probability is the same as the first one. So combine this one and the first one and that gives you 2 * 0.16/0.8 * 0.56/0.8 * 0.08/0.8

And you can get four more orders (from the order or just trying RBG as in a previous post), and their probabilities will be equal so it gives you 6 * ..., and that 6 can come directly from the formula.

You can stick the formula on at the end, or calculate it elsewhere and put in 6. (If you have never seen the formula before, ask your teacher about it).
7. Have you heard of the binomial distribution?
8. (Original post by SeanFM)
The three different types of objects, a, b and c in this case, is that it arrives early (a), arrives on time (b) or arrives late (c) and from the formula, there are 6 different ways of ordering them.

To see this, let's look at some orders. The first one could be early, second on time, third late. The probability of this is
0.16/0.8 * 0.56/0.8 * 0.08/0.8 as you say. (The order in which you multiply them may be different but it does not matter).

Another order could be the reverse - first day is late, second is on time, third is early. The probability is the same as the first one. So combine this one and the first one and that gives you 2 * 0.16/0.8 * 0.56/0.8 * 0.08/0.8

And you can get four more orders (from the order or just trying RBG as in a previous post), and their probabilities will be equal so it gives you 6 * ..., and that 6 can come directly from the formula.

You can stick the formula on at the end, or calculate it elsewhere and put in 6. (If you have never seen the formula before, ask your teacher about it).
Ahh I got it! Thank you so much!

In the attachment, why would you not use the factorial formula on (b). I understand that you would use it on (c).

(Original post by Zacken)
Have you heard of the binomial distribution?
Yes I have. I understand that this and the factorial formula have something to do with Pascal's triangle and the nCr function on the calculator.
Attached Images

9. (Original post by londoncricket)
Ahh I got it! Thank you so much!

In the attachment, why would you not use the factorial formula on (b). I understand that you would use it on (c).

Yes I have. I understand that this and the factorial formula have something to do with Pascal's triangle and the nCr function on the calculator.
Think of it like tossing a coin. There is only one way of getting HH, one way of TT but two ways of HT. (In this case, think of both labour as both heads).

(you're also only looking at one object, so it would be 2!/2! = 1 way.
10. (Original post by SeanFM)
Think of it like tossing a coin. There is only one way of getting HH, one way of TT but two ways of HT. (In this case, think of both labour as both heads).

(you're also only looking at one object, so it would be 2!/2! = 1 way.
I think I understand. You multiply it by the number of ways that the combination can be made. For example in the P(Labour, Labour) one, it is still being multiplied by 1!, which is just 1.

Am I right in thinking this?

Thank you.
11. (Original post by londoncricket)
I think I understand. You multiply it by the number of ways that the combination can be made. For example in the P(Labour, Labour) one, it is still being multiplied by 1!, which is just 1.

Am I right in thinking this?

Thank you.
Correct it is really 2!/2!*0! Where 0! = 1.
12. (Original post by SeanFM)
Correct it is really 2!/2!*0! Where 0! = 1.
Great! So would I do this for every question in which there are a possible number of combinations that is greater than one?
13. (Original post by londoncricket)
Hi all,

I don't understand why 3(d) requires the "*6" at the end of the calculation?:

Calculation:
0.16/0.8 * 0.56/0.8 * 0.08/0.8 *6

0.084

This 6 is needed because the 3 events (A-B-C) could happen in any combination: ABC or ACB or BAC or BCA or CAB or CBA. I hope this is helpful.
14. (Original post by MARIOUSKA)
This 6 is needed because the 3 events (A-B-C) could happen in any combination: ABC or ACB or BAC or BCA or CAB or CBA. I hope this is helpful.
Thank you for that!
15. (Original post by SeanFM)
Correct it is really 2!/2!*0! Where 0! = 1.

5(b)(ii)

0.85*0.60*0.45 + 0.85*0.55*0.40 + 0.60*0.55*0.15 * 2!
= 0.516

0.85*0.60*0.45 + 0.85*0.55*0.40 + 0.60*0.55*0.15
= 0.466

Why would I not put the "* 2!" here? Is there not a combination here? Rhys could purchase those two items in two different orders?

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