part (b) of this statistics problem

I got 5/18 but the answer is 5/9.

Why do I still have to times 2?
Or the answer is wrong?

just write out the 7th row of pascals triangle and ignore the first and last element.
7 21 35 35 21 7
So you have 70/126 = 5/9

Obviously you can express this In terms of partial sums of 7Cr.

I have to say that I don't quite understand then.

I used the simple sample space diagram, so that either friend A gets 1 to 6 or the other person gets 1 to 6, with total of 7.
There are only 36 ways of doing so.
And 10 ways that one friend gets exactly 1 more then the other.

Why is that method wrong???

The bracelets are selected or given at random. You need to count all the occouurences.

So if friend A gets 2 and friend B gets 5, there are 7C2 ways for this to happen. So overall you want:
7C3/(7C1+7C2+7C3)
Where Ive used symmetry to halve the number of terms on the numerator and denominator.

If friend A gets 2 bracelets, you get 5 as the total number of selections? I get 21 combinations of bracelets. If a bracelet is given to one of the friends randomly, you'd expect a binomial distribution?

hmmmm. I kind of see your thinking now.

I guess that's just another WORDY issue .......
I won't really bother the "7C2 ways for this to happen" as the question should be focusing on the NUMBERs the friends actually gets.....

Thanks for clariying this though.

BTW: this is another CIE textbook problem students asks ...never had so much "difficult" problems due to how they actually ask the questions...

If they focussed on the number, the "at random" wouldn't hold. Also the question states that she shares the remainder with her two friends. So if one friend gets 2, the other must get 5. For equal probabilities you factor the 1/2^7 out of the binomial distribution and are just left with 7Cr where r=0 and 7 are discounted because both friends get at least one.

Exactly.

For what I can understand,
So if one friend gets 2, the other must get 5
That's why we do it the way I posted in the OP.
You get 36 outcomes, which are equally likely.

It's just that WHICH 2 you get, or WHICH 5 the other get are random......


Also, part (a) is poorly phrased. It's tempting to lead pupils to use 7C2....

I really dont see how you could represent it the way you have. You seem to leave some bracelets unallocated and only the leading diagonal [5 3 1 1 3 5] represents all being allocated, its not 36. Of these, the two 1s represent one more than the other friend. The counts in the different cells would not have the same weighting?

Also I dont really have a problem with A) either, sorry. Nine is clearly mentioned but seven is not. Its a one marker 9C2 which is a bit of a hint for B) involving 7Cr.

Nevermind part (a)......


For part (b), it's like you are saying there is a bag, which has 7 items, taking out items randomly. What's the probability for the difference of number of items in the bag, and outside the bag to be exactly 1.....(And you must take at least 1 item out)

For b) yes, and there must be at least one ball remaining in the bag. Taking r balls out, you have 7Cr different scenarios, and these can be summed over r as each probability is the same.

Terribly sorry I had been a dumb ass.................

No problem, we've all been there ...

I blame that COVID 19.....

Just being glad that I realised how silly I had been on this question before I explain it to the pupils..........
Not having exams this year made the stress worse in a way, there is no pressure on them to keep up with the online lessons....