Natalya Mayer

2022-09-14

Probability question regarding binomial distribution

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?

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?

Koen Henson

Beginner2022-09-15Added 17 answers

Step 1

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.

$$X\sim \mathcal{B}\mathcal{i}\mathcal{n}(39,0.08)$$

Step 2

Then we have:

$$\begin{array}{rl}\mathsf{P}(G\mid E)& ={\displaystyle \frac{\mathsf{P}(G\cap E)}{\mathsf{P}(E)}}\\ & ={\displaystyle \frac{\mathsf{P}(G)}{\mathsf{P}(E)}}\\ & ={\displaystyle \frac{\sum _{x=0}^{39}\mathsf{P}(G\mid X=x)\mathsf{P}(X=x)}{\sum _{x=0}^{39}\mathsf{P}(E\mid X=x)\mathsf{P}(X=x)}}\\ & =\frac{\sum _{x=0}^{39}\frac{1}{40}\mathsf{P}(X=x)}{\sum _{x=0}^{39}\frac{x+1}{40}\mathsf{P}(X=x)}\\ & =\frac{1}{39(0.08)+1}\\ & \approx 0.2427\end{array}$$

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.

$$X\sim \mathcal{B}\mathcal{i}\mathcal{n}(39,0.08)$$

Step 2

Then we have:

$$\begin{array}{rl}\mathsf{P}(G\mid E)& ={\displaystyle \frac{\mathsf{P}(G\cap E)}{\mathsf{P}(E)}}\\ & ={\displaystyle \frac{\mathsf{P}(G)}{\mathsf{P}(E)}}\\ & ={\displaystyle \frac{\sum _{x=0}^{39}\mathsf{P}(G\mid X=x)\mathsf{P}(X=x)}{\sum _{x=0}^{39}\mathsf{P}(E\mid X=x)\mathsf{P}(X=x)}}\\ & =\frac{\sum _{x=0}^{39}\frac{1}{40}\mathsf{P}(X=x)}{\sum _{x=0}^{39}\frac{x+1}{40}\mathsf{P}(X=x)}\\ & =\frac{1}{39(0.08)+1}\\ & \approx 0.2427\end{array}$$

cubanwongux

Beginner2022-09-16Added 4 answers

Step 1

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.

Step 2

Let A be the event “suspect is guilty” and B “suspect draws in the same style.” We have the prior probability $P(A)=2.5\mathrm{\%}$ and we’re asked to compute the posterior probability P(A|B). We know that $P(B)=8\mathrm{\%}$ for the general population and, assuming that the perpetrator didn’t deliberately mask his style, we can take $P(B\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}A)=1$. Bayes’ Theorem then gives the posterior probability as

$$P(A\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}B)=\frac{P(A)\phantom{\rule{thinmathspace}{0ex}}P(B\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}A)}{P(B)}=\frac{0.025\times 1}{0.08}=\mathrm{0.3125.}$$

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.

Step 2

Let A be the event “suspect is guilty” and B “suspect draws in the same style.” We have the prior probability $P(A)=2.5\mathrm{\%}$ and we’re asked to compute the posterior probability P(A|B). We know that $P(B)=8\mathrm{\%}$ for the general population and, assuming that the perpetrator didn’t deliberately mask his style, we can take $P(B\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}A)=1$. Bayes’ Theorem then gives the posterior probability as

$$P(A\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}B)=\frac{P(A)\phantom{\rule{thinmathspace}{0ex}}P(B\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}A)}{P(B)}=\frac{0.025\times 1}{0.08}=\mathrm{0.3125.}$$

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