OT BBC 'Maths' Quiz

Full quiz at:
http://news.bbc.co.uk/1/hi/programmes/more_or_less/3258627.stm

Will somebody please explain the answer number three?

"A test for an illness is 90% accurate and 10% of the population have
that illness. If you are tested at random and your test is positive,
how likely are you to have the illness? "

Possible answers are 50%, 80% or 90%

 

A. 90 %

The 10 % of the population is irrellevant and designed to throw you because
the question assumes you gave a positive therefore you must be in the 10 %
of the population. The question then relates to the accuracy of the test.
**** Me I amazed myself at getting that right[1]

Mike Cook

[1] Please dont tell me I'm wrong

 

<snip wrong answer>

Shit , Should have checked the webpage first before inserting foot in mouth
!

 

The correct answer according to the webpage page is 50%

Mike Cook

 

Just emailed the beeb , Im sure the answer is 90% and they have screwed up.

Mike Cook

 

I thought exactly the same... the answer (apparently) is 50%

Shit, 3 out of 15...

 

It’s been a long time since I did stats but if you test 100 people for
the disease 10 of them will have it but the test is only 90% accurate so
it will tell 1 of the 10 sufferers he’s clear giving 9 positive results.
In your sample you have 90 non suffers it will tell you that 9 of them
are ill due to the inaccuracy of the test, so on a test of 100 people 18
would test positive, 9 from the ill group and 9 from the well group,
this gives a 50% chance of any positive test being correct.
HTH


--
Bob
Currently borrowing a black and red Yamaha XJ750 with fuel injection
Present: Honda XL125RF (FS)
Past: Honda CG125
bob at homeurl tomato dot co dot uk
remove the red fruit if you’d like to email me.

 

9 but then most of it isn't really maths


--
Ian
"reorganising, a wonderful method for creating an illusion of progress"
znvygb: (ROT13 all of it to mail me)
The FAQ is here http://www.ukrm.net/faq/index.html
98 FZS600, 72 T120R MIB#21 TWA#6

 

<g> See Bob123's answer below. The BBC is right.

You're not alone - most people find statistics the hardest part of
maths. Here's a very simple question I once used when someone boasted
of their ability with statistics.

You hear two children playing in the garden next door. One of them
sticks their head over the fence and you see it is a girl. What is the
probability that the other child is also a girl? No tricks in the
question, assume boy/girl is 50/50.

Choose from 3/4, 2/3, 1/2, 1/3.

Bob123 is not allowed to answer!

 

Which is why I'm not going to try it.

 

Very odd - I can't figure how they got their answer.

Rick

 

Like what I said

--
Bob
Currently borrowing a black and red Yamaha XJ750 with fuel injection
Present: Honda XL125RF (FS)
Past: Honda CG125
bob at homeurl tomato dot co dot uk
remove the red fruit if you’d like to email me.

 

Agreed. They might just have been able to bamboozle me with some
mathematical sleight of hand into believing it was 80%, but 50%??

Rick

 

Colin Irvine wrote:

Another badly worded question. You are ignoring the
psychology of children. Girls play together.
One sex may be more curious than the other...

Now if the question was:
Assuming 50% each of boys and girls.
You have two dead babies in a sack and,
without looking, pull one out....

1/3.

 

Now I know why I hated statistics 100% of the time. ;-)

Rick

 

Crikey - cheers for that, Ivan. I got 10 out of 15 (tho the test
really measures ones abilities to make lucky guesses) but I'd never
have got that one.

 

"A test for an illness is 90% accurate and 10% of the population have

If you want the really mathmatical solution, take a look at
http://faculty.vassar.edu/lowry/bayes.html - it's all to do with Bayes'
theorem - which some people reckon governs the entire universe ! The
example is nearly identical.

Read P(A/B) as the Probability of A happening, given that B has happened
already); and P(~B) as The probability of B not happening.

Bayes theorem states:
P(A/B) = P(B/A) x P(B) divided by P(B)

So
P(Being ill given that Test is +ve) = P(test is +ve given that you are ill)
x P(being ill) divide by P(Test is +ve)

P(Test is +ve given that you are ill) = 0.9
P(Being ill) =0.1

P(Test is +ve) = P(Test is +ve and you ARE ill) + P(Test is +ve and you are
not ill)
= 0.9x0.1 + 0.1x0.9 =0.09+0.09=0.18

So answer is 0.9x0.1 divided by 0.18
=0.5 - ie 50%


But Dr Ivan's answer is much easier to understand !

 

I think you’ll find that you’re in the majority there. TBH I never
studied it except for at the very basic level needed to understand
Alevel physics, I rather wish I had now as it’s quite fun (compared with
pure maths) and one of the few mathematical areas that you can get a job
in fairly easily, massaging pole and so on.

--
Bob
Currently borrowing a black and red Yamaha XJ750 with fuel injection
Present: Honda XL125RF (FS)
Past: Honda CG125
bob at homeurl tomato dot co dot uk
remove the red fruit if you’d like to email me.

 

<g> I should have offered 1/4 as an option, as that's what most people
choose. Oh, and you're right of course.

 

Colin Irvine wrote:

A lot of practice.