# Dependent chances using Bayes' rule

The chance that somebody get's mad cow disease is 0.01 (1%). If someone visits the USA this chance becomes 0.05 (5%). The chance that somebody goes to the USA is 0.01 (1%). If someone goes to the USA, he'll most likely buy the flag as a souvenir, the chance that somebody does this is 0.7 (70%). Because the flag is well known, people who never have been to the USA have a chance of 0.3 (30%) to have the flag. Now given that somebody has the mad cow disease and he has the flag of the USA, what are the chances he has visited the USA?

Thus:
P(USA) = 0.01
P(MCD) = 0.01
P(MCD|USA) = 0.05
P(flag|!USA) = 0.3
P(flag|USA) = 0.7

Following bayes' rule we can calculate the chance that somebody has been to america given the knowledge this person has MCD:
P(USA|MCD) = P(MCD|USA)*P(USA)/P(MCD) = 0.05 *0.01 / 0.01 = 0.05.

Now the problem is how do we calculate the chance that someone has visited the USA given he has the american flag, and after that given somebody has and mad cow disease and the american flag the chance he has been to america.

How does one does this?

Cheers!

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You need information about the joint probability of $MCD$ and $flag$ (given $USA$) in order to solve this problem. I'll assume they're conditionally independent, but that may not necessarily be true.

If $MCD$ and $flag$ are conditionally independent, then $P(MCD \cap flag|USA) = P(MCD | USA) \times P(flag | USA) = 0.035$. We also need to calculate $P(MCD \cap flag|!USA)$, for which we need $P(MCD | !USA)$. We can calculate this as $$\frac{P(MCD \cap !USA)}{P(!USA)} = \frac{P(MCD) - P(MCD \cap USA)}{P(!USA)} = \frac{0.01-0.0005}{0.99} \approx 0.009596.$$ Then $$P(MCD \cap flag | !USA) = P(MCD | !USA) \times P(flag|!USA) \approx 0.009596 \times 0.3 \approx 0.002879.$$

Now Bayes' rule tells us that $$P(USA|flag \cap MCD) = \frac{P(flag \cap MCD | USA) \times P(USA)}{P(flag \cap MCD)}.$$ By the law of total probability we can expand the denominator as $$P(flag \cap MCD) = P(flag \cap MCD | USA)\times P(USA) + P(flag \cap MCD | !USA)\times P(!USA).$$ We already calculated the two conditional probabilities on the right side as 0.035 and (approximately) 0.002879, respectively, so we have

$$P(USA|flag \cap MCD) = \frac{0.035 \times 0.01}{0.035 \times 0.01 + 0.002879 \times 0.99} \approx 0.109.$$ So there is approximately a $10.9\%$ chance that someone has been to the USA given that they have a flag and have MCD. The crucial parts to the solution are (1) assuming flag and MCD are independent conditional on USA, which may not be true, and (2) using the law of total probability to get the denominator.

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Shouldn't P(MCD ∩ flag|USA)=P(MCD|USA) × P(flag|USA) be 0.05 x 0.7 = 0.035? – Devos50 Nov 4 '12 at 14:13
Ah yes, I forgot to mention they are indeed independent. Althought the flag en mcd are both related to USA (as their chances increase) – Gooey Nov 4 '12 at 18:17
@Devos50 Right, good catch. – Jonathan Christensen Nov 6 '12 at 0:23
@Gooey Technically the assumption I am making is that they are independent conditional on USA; I'll add a note to that effect. – Jonathan Christensen Nov 6 '12 at 0:23