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I'm doing some work on the binomial distribution but currently finding it difficult to know whether my attempt of the question is the actual correct way to do it?

At a party, it is discovered that a renegade guest has drawn a controversial image on the bathroom wall. There are 40 people at the party, and in the absence of other evidence, each is equally likely to have drawn the controversial image. The drawing style of the controversial image is analysed and found to have characteristics unique to 8% of the general (innocent) population. A suspect whose drawing style is analysed is found to match the controversial image. Assuming the guests are representative of the general population, what is the probability that the suspect is guilty given the drawing evidence?

So for this question I decided to use the binomial distribution because of the properties of the distribution:

  • The number of trials n is fixed
  • The probability, p is constant throughout

Where X denotes the number of guilty suspects. I.e X ~ (40,0.08) From there on we know that the number of guilty suspects is 1 so by finding P(X = 1) we should find the probability that a chosen suspect is guilty?

So I calculated the probability using the formula

${^nC_x}$ $0.08^x$ $(1 - 0.08)^{n-x}$

= ${^{40}C_1}$ $0.08^1$ $(1 - 0.08)^{39}$

= ~0.124 (3.sf)

However I am not satisfied with my method of approach to this question. I feel as if I've misunderstood something here. Should I have used the binomial distribution here? At first I considered conditional probability because the binomial distribution seems a little excessive and possibly wrong.

Throughout the process I was really unsure of what I was doing. Can someone please point out how I should approach this problem?

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  • $\begingroup$ If you have $n$ independent trials and the probability of sucess is $p$ in each trial, you can always use the binomial distribution to find the probability of $k$ successes. But this problem seems to be a bit more complex. $\endgroup$ – Peter Nov 20 '15 at 20:14
  • $\begingroup$ The last sentence of the problem statement suggests to me that you’re meant to examine conditional probabilities. $\endgroup$ – amd Nov 20 '15 at 21:44
  • $\begingroup$ I also thought it was this too but I wasn't sure. P(X is guilty | the evidence)? $\endgroup$ – Nubcake Nov 20 '15 at 21:48
  • $\begingroup$ Yeah, something like that. It’s not a pure binomial trial since you have the additional information that the drawing style is a match. $\endgroup$ – amd Nov 20 '15 at 21:58
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There are three types of guests at the party: the guilty one, $X$ innocent matches, $39-X$ innocent non-matches.   The guilty one is (considered) certainly a match; and as for the rest we assume $X$ has a Binomial Probability Distribution. $$X\sim\mathcal{Bin}(39, 0.08)$$

Then we have:

$$\begin{align} \mathsf P(G\mid E) & = \dfrac{\mathsf P(G\cap E)}{\mathsf P(E)} \\[1ex] & = \dfrac{\mathsf P(G)}{\mathsf P(E)} \\[1ex] & = \dfrac{\sum\limits_{x=0}^{39} \mathsf P(G\mid X{=}x)\mathsf P(X{=}x)}{\sum\limits_{x=0}^{39} \mathsf P(E\mid X{=}x)\mathsf P(X{=}x)} \\[1ex] & = \frac{\sum\limits_{x=0}^{39} \frac{1}{40}\mathsf P(X{=}x)}{\sum\limits_{x=0}^{39} \frac {x+1}{40}\mathsf P(X{=}x)} \\[1ex] & =\frac{1}{39(0.08)+1} \\[1ex] & \approx 0.242{\small 7} \end{align}$$

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The binomial distribution isn’t really appropriate here. We’re given that exactly one of the guests is guilty, so the events “guest X is guilty” and “guest Y is guilty” are mutually exclusive, which is about as far as you can be from independence. (Remember that the binomial distribution applies to a set of independent Bernoulli trials.) The key to the problem is in the last sentence: the phrase “given that” indicates that you’ll be working with conditional probabilities and that Bayes’ Theorem is likely to come into play.

Let $A$ be the event “suspect is guilty” and $B$ “suspect draws in the same style.” We have the prior probability $P(A)=2.5\%$ and we’re asked to compute the posterior probability $P(A\,|\,B)$. We know that $P(B)=8\%$ for the general population and, assuming that the perpetrator didn’t deliberately mask his style, we can take $P(B\,|\,A)=1$. Bayes’ Theorem then gives the posterior probability as $$ P(A\,|\,B) = {P(A)\,P(B\,|\,A)\over P(B)} = {0.025 \times 1 \over 0.08} = 0.3125. $$

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  • $\begingroup$ Sorry I don't really understand why P(B|A) is 1? Does it mean because we know that person is guilty then it's obvious that he drew it? $\endgroup$ – Nubcake Nov 21 '15 at 16:55
  • $\begingroup$ Not quite. $B$ is “the person drew in this style.” We’re told that the drawing is in a particular style, so whoever is guilty has to have drawn it in that style. There’s an implicit assumption in the problem setup that the guilty party used her normal drawing style, though. $\endgroup$ – amd Nov 21 '15 at 18:38
  • $\begingroup$ I see what you mean, thanks I understand it much better now. $\endgroup$ – Nubcake Nov 21 '15 at 18:44

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