• Welcome to DOSBODS

     

    DOSBODS is free of any advertising.

    Ads are annoying, and - increasingly - advertising companies limit free speech online. DOSBODS Forums are completely free to use. Please create a free account to be able to access all the features of the DOSBODS community. It only takes 20 seconds!

     

Roger_Mellie

Stats question got me stumped

Recommended Posts

A A A

A A B = Boom!

A A C

A B A = Boom!

A B B = Boom!

A B C

A C A

A C B

A C C

B A A = Boom!

B A B = Boom!

B A C

B B A = Boom!

B B B

B B C

B C A

B C B

B C C

C A A

C A B

C A C

C B A

C B B

C B C

C C A

C C B

C C C

=6/27

Share this post


Link to post
Share on other sites

AAA 
AAB - explosion
ABB - explosion
BBB 
AAC
ACC
CCC
ABC
BB.C

2/9 ?
 

Share this post


Link to post
Share on other sites
1 hour ago, Joxer said:

15/63

I expected it to be higher but there you go. Maths is never intuitive

 

Actually the answer is 3/14 which is even less.

15/63 is the probability of none of the three broken bottles containing element C, but that doesn't cause a bang because you need a mixture of both A and B so AAA and BBB need to be discounted

Share this post


Link to post
Share on other sites

I'm with Joxer at 3/14 chance of explosion.

This comes from:

Chance of selecting only A & B:

6/9 X 5/8 X 4/7

Chance of selecting AAA or BBB:

2 X (3/9 X 2/8 X 1/7)

Cancelling within each takes it to:

(20 - 2) / 84

Divide top and bottom by 6 gives:

3/14

 

Share this post


Link to post
Share on other sites

I got 15/63 and did realise it needed to be adjusted for all As or Bs and could work out the odds of that but couldn’t remember what to do after with the two sets of different odds - I thought divide rather than multiply but assume Frank’s working is right - what is year nine age, I thought pre-GCSE, this seems probably harder than O-level as I recall more like first year of A-level actually.

Edited by SNACR

Share this post


Link to post
Share on other sites

Well this has been keeping me up as I couldn't prove 11/14 as the chance of no explosion so had to take the nuclear approach and do the probabilities of all 27 combinations; which is not as bad as it sounds as it patterned nicely.

The first jar has a 1/3 chance of being anything.

The second has a 3/8 chance if not the first picked or 2/8 if picked.

The third then has a 1/7, 2/7 or 3/7 chance depending on the two prior picks.

Denominator is 3 X 8 X 7 = 168

There are only six combinations that lead to an explosion and all with probability 6 / 168 (1 X 2 X 3 or 1 X 3 X 2).

These are: AAB ABA ABB BAA BAB BBA.

These total 36 out of 168.

There are 21 combinations that lead to no explosion with probability out of 168 being 2 (all the same, three examples), 6 (two the same, twelve examples) or 9 (all different, six examples).

These total 132 out of 168.

And 36 + 132 = 168. Hurrah.

Cancelling down 36/168 (boom) and 132/168 ( no boom) gives 3/14 and 11/14 respectively.

I would appreciate seeing a more sophisticated proof of the 11/14 as I could get close by adding probabilities but not quite there.

Edited by Frank Hovis
Probability each three the same should have said 2 (out of 168) not 3.

Share this post


Link to post
Share on other sites

The question is not a standard year 9 task but is part of a competition for Year 10s and below promoted by Mathematical Education on Merseyside. As such it is aimed to stretch the most able. Some more nice (imaginative) questions on there too.

http://www.mathsmerseyside.org.uk/challenge/19sch-questions.pdf

Edited by man o' the year
spelling

Share this post


Link to post
Share on other sites
28 minutes ago, man o' the year said:

The question is not a standard year 9 task but is part of a competition for Year 10s and below promoted by Mathematical Education on Merseyside. As such it is aimed to stretch the most able. Some more nice (imaginative) questions on there too.

http://www.mathsmerseyside.org.uk/challenge/19sch-questions.pdf

I seem to recall doing something similar in physics but it was under exam conditions - they shut me in the lab assistant's room for an hour - I can't remember much about it other than I couldn't get very far with it but there was a level of assumed knowledge I didn't have a bit like when Mensa questions need pythag or basic trig.

Share this post


Link to post
Share on other sites
4 minutes ago, Roger_Mellie said:

So, Swampy has 6/27 and laid it out... Which looks good to me.

Frank has a slightly lower chance at 3/14 (I.e. 6/28)...

Which is it?

I worked it out using the same logic as Frank but made a daft mistake.

Frank's because there is more than one A B C. Swampy would need combinations with A(bottle 1) A( bottle 2) A (bottle 3) B (bottle 1) etc .

eg

A (bottle 1) A (bottle 2) B (bottle 1)

A (bottle 1 A (bottle 3) B (bottle 1)

Edited by SNACR

Share this post


Link to post
Share on other sites
5 minutes ago, SNACR said:

I seem to recall doing something similar in physics but it was under exam conditions - they shut me in the lab assistant's room for an hour - I can't remember much about it other than I couldn't get very far with it but there was a level of assumed knowledge I didn't have a bit like when Mensa questions need pythag or basic trig.

In physics we had the "Physics Olympiad" - still going = https://www.bpho.org.uk/

Share this post


Link to post
Share on other sites
17 minutes ago, Roger_Mellie said:

Aaah, indeed. 

I don't think that's relevant as there are three of each type of bottle, they all break at the same time. How can you have 64 different ways of smashing 3 bottles?

 

Share this post


Link to post
Share on other sites
2 minutes ago, XswampyX said:

I don't think that's relevant as there are three of each type of bottle, they all break at the same time. How can you have 64 different ways of smashing 3 bottles?

 

There's nine different bottles on the tray and three smash and six remain intact so it's all the different ways of smashing three bottles and not smashing six.

Share this post


Link to post
Share on other sites
7 hours ago, Roger_Mellie said:

So, year 9 homework and this stats question has me stumped, so I'm putting it out to the dosboderatti.

My answer is 1/14, but I'm struggling to get the opposite answer to 13/14, so I think its wrong.

What say ye?

15512268308361597111805.jpg

The probability is 100% that the white male is to blame.

The feminazi teacher will claim early retirement and a pension boost for the stress and go home to her cats.

The student will end up renting a BTL from her.

Edited by VeryMeanReversion
cyncism

Share this post


Link to post
Share on other sites
5 minutes ago, Napoleon Dynamite said:

I think Swampy's got it.

3 choices (A, B, C) in 3 positions gives 3^3 = 27 combinations

If there's no C involved it's 2 (A, B) choices in 3 positions 3^2 = 6 combinations.

So a 6/27 chance.

 

Imagine alternatively it's Scrabble letters from a bag, full of only the letters A, B and C, Swampy's method takes no account of how many of each letter are in the bag.

Share this post


Link to post
Share on other sites

It's pretty straightforward.

Number of combinations in choosing 3 bottles from 9 i.e. 9C3 = 84

Number of combinations in choosing 3 bottles from the 6 that don't include C (the neutralising chemical) and hence will not be neutralised, 6C3 = 20

The answer would then be 20/84 but we must consider that an explosion needs a mixture of both A and B. Of the 20 ways of picking 3 bottles from the 6 potentially explosive combinations there are only 2, AAA and BBB that don't involve a mix of the two chemicals, A and B. Thus we actually only have 18 (I.e. 20-2) combinations of choosing an explosive mixture when choosing 3 bottles from the 6 potentially explosive chemicals.

Thus the answer is 18/84 or 3/14.

 

 

 

Share this post


Link to post
Share on other sites
36 minutes ago, SNACR said:

Imagine alternatively it's Scrabble letters from a bag, full of only the letters A, B and C, Swampy's method takes no account of how many of each letter are in the bag.

It doesn't need to as you pick 3 tiles in one go.

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.


  • Recently Browsing   0 members

    No registered users viewing this page.