The "Correct" answer for one of the 'BBC - The Big Risk Quiz' questions?

• Apr 12th 2011, 03:59 PM
kwah
The "Correct" answer for one of the 'BBC - The Big Risk Quiz' questions?
Hi all,

I've just completed the BBC Big Risk Quiz and I'm confused about what the correct answer should be for this question (it was the only one of the 'understanding risk' questions I got wrong (Clapping))

https://www.bbc.co.uk/labuk/experiments/risk/

Note that this is a SPOILER ALERT!! for anybody who hasn't yet completed the quiz..

Quote:

5. A criminal hides in a room with 99 innocent people. You have a lie detector that correctly classifies 95% of people. You pick someone at random, wire them up to the machine, and ask them if they are a criminal. They say no, but the machine goes ‘ping’ and says the person is lying. What is the chance that you have caught the criminal?
Initially I answered 95% because the lie detector is 95% accurate and would hold true irrespective of whether the person is the criminal or not (see the mini rant below).

I can see why the answer of 95% might be incorrect, since there is a 1/100 chance of selecting the criminal (1x criminal + 99x innocents) in the first place and then a 95/100 chance of the lie detector being correct, but this results in a 9.5% chance of having caught the criminal and not 17%, surely?

Can anybody clarify why 17% is listed as the 'correct' answer? I do not recall 9.5% being any of the options.

Mini rant: I really dislike how imprecise the questions are.. It doesn't specify that the person selected is from the pool of 100 people mentioned at the start.. There's a whole bunch of other examples of it throughout the test too..
• Apr 12th 2011, 04:17 PM
pickslides
Sounds like conditional probability should be employed here. Are you aware of this method?
• Apr 12th 2011, 04:32 PM
Plato
Quote:

Originally Posted by kwah
Mini rant: I really dislike how imprecise the questions are.. It doesn't specify that the person selected is from the pool of 100 people mentioned at the start.. There's a whole bunch of other examples of it throughout the test too..

Well the question does say that the "person is selected at random".
I cannot not see the actual choices for this question.
But from the given question the actual answer is $16.1\%$.
I suppose that is 'close to' $17\%$ in some books.
That is a result of Bayes' Theorem. Do a web-search for that.
• Apr 13th 2011, 04:00 AM
kwah
Quote:

Originally Posted by Plato
Well the question does say that the "person is selected at random".
I cannot not see the actual choices for this question.
But from the given question the actual answer is $16.1\%$.
I suppose that is 'close to' $17\%$ in some books.
That is a result of Bayes' Theorem. Do a web-search for that.

Bayes' Theorem appears to be along the right lines and perhaps I'll realise it for myself after I'm finished reading but why are the two events (selecting the criminal/innocent & getting a true/false positive) not considered as independent events and the probability of both happening being just the product of both?

In my head I'm picturing a probability tree where the first set of branches are whether the criminal is selected at random (A). The second set (B) is that the lie detector result is a true positive. From this: $P(A\cap B) = 0.01 * 0.95 = 0.095$.

But then this confuses me because in my head I'm thinking 'the probability that they've caught the criminal will always be 0.01 surely? but on the other hand, the lie detector test would add to the level of confidence that the selected person is the criminal rather than detract from it'

I think I've stumbled onto something while typing this out though..

I guess an alternative way of summarising it is:
Given that there will be a 'criminal' outcome from the lie detector 5.9% of the time (sum of the $P(A\cap B)$ and $P(A'\cap B)$ branches), what proportion of these 'criminal' readings does $P(A\cap B)$ account for?

Would this be accurate?

The reason I said stumbled upon it is because I tried drawing out the tree into a spreadsheet and then making the steps toward an answer and 'stumbled' across the following whilst making the above mental leaps that gives 16.1% as an answer. It doesn't quite look like Bayes' Theorem however (though maybe its just an alternative form? idk)..

$\frac {P(A\cap B)}{P(A\cap B) + P(A'\cap B)}$
$\frac {0.0095}{0.0095 + 0.0495}$
$\frac {0.0095}{0.0590}$
$0.1610$

(nb: it has been a while since I studied statistics so I offer my apologies in advance if my notation is incorrect)