Conditional Probability Word Problem?

Question

Supposed there are 3 men who are all owners of their own Smarties factories. Burr Kelly, being the brightest and most innovative of the men, produces considerably more Smarties than his competitors and has a commanding 45% of the market share. Yousef See, who inherited his riches, lags behind Burr and produces 35% of the world’s Smarties. Finally Stan Furd, brings up the rear with a measly 20%. However, a recent string of Smarties related food poisoning has forced the FDA investigate these factories to find the root of the problem. Through his investigations, the inspector found that one Smarty out of every 100 at Kelly’s factory was poisonous. At See’s factory, 1.5% of Smarties produced were poisonous. And at Furd’s factory, the probability a Smarty was poisonous was 0.02.

(a) What is the probability that a randomly selected Smarty will be safe to eat? I solved this one and got 0.98625

(b) If we know that a certain Smarty didn’t come from Burr Kelly’s factory, what is the probability that this Smarty is poisonous? not sure how to go about this one, please help

(c) Given this information, if a randomly selected Smarty is poisonous, what is the probability it came from Stan Furd’s Smarties Factory? not sure how to go about this one, please help

Answer 1

For the (b) part of your question you work similarly as with (a) but you must notice that from the smarties that don't come from Burr Celly's factory, the $\frac{35}{35+20}=\frac{7}{11}\simeq0.6364\%$ come from Yousef's factory and the other $\frac{4}{11}$ come from Stan's factory.

So now like you did in (a):

$P(poisoned|notfromBurrKelly)=\frac{7}{11}*0.015+\frac{4}{11}*0.02=0.0168181...$

For the (c) part of your question you need to use Bayes' Theorem.

Let's define as $A$ the event that a smarty is poisonous, and as $B$ the event that a smarty comes from Stan Furd.

We need to calculate $P(B|A)$ and using the Bayes Rule we have:

$P(B|A)=\frac{P(A|B)*P(B)}{P(A)}=\frac{0.02*0.2}{1-0.98625}=0.290909...$

Answer 2

Nice job with the first solution.I used casework and split it in 3 cases

$.45(\frac{99}{100})$ If Kelly's is the one that is picked and if safe

$.35(\frac{197}{200})$ If See's is the one that is picked and if safe

$.2(\frac{98}{100})$ If Furd's is the one that is picked and if safe

Adding these you'll get $.98625$ as you did.

$\textbf{b}$) If you're not aware of the formula $P(A∣B)=\frac{P(A\cap B)}{P(B)}$ ($P(A given B)=P(Both-A-and-B-occurring)/P(B)$) you should!! It's a really important formula and makes hard problems like this a lot easier.

Using it on this problem $P(Smarty-is-poison∣not-kelly's)=\frac{P(poison-and-its-kelly's)}{P(its-kelly's)}$

This should be solvable now! (use this same method for part $c$)

Answer 3

(a) should have been solved using the law of total probability. Let $A,B,C$ be the event that a smarty comes from each of the factories, and $D$ that the smarty is poisonous. $$\mathsf P(D) = \mathsf P(D\mid A)~\mathsf P(A)+\mathsf P(D\mid B)~\mathsf P(B)+\mathsf P(D\mid C)~\mathsf P(C)$$

(b) is solved similarly, and also uses the definition for conditional probability.$$\mathsf P(D\mid B\cup C)~=~\dfrac{\mathsf P(D\cap(B\cup C))}{\mathsf P(B\cup C)}~=~\dfrac{\mathsf P(D\mid B)~\mathsf P(B)+\mathsf P(D\mid C)~\mathsf P(C)}{\mathsf P(B)+\mathsf P(C)}$$

(c) is just Bayes' Rule. $$\mathsf P(C\mid D)~=~\dfrac{\mathsf P(D\mid C)~\mathsf P(C)}{\mathsf P(D)}$$