I am working on an example of a statistics course and would like to know where I go wrong. Here is the problem:
Assume that 1% of the population are musicians and that 10% of the total population is left-handed. A recent survey of musicians reveals that 60% of them are left-handed. What is the probability that a left-handed toddler will be a musician?
my thinking so far:
1) I think I have to use Bayes theorem because "Bayes' theorem describes the probability of an event, based on prior knowledge of conditions that might be related to the event" which seems to be the case here. Right or wrong?
2) I translate into proper probability syntax: A = musicians B = left handed people C = left handed musicians
$P(musician)= 0.01$
$P(left-handed)= 0.1$
$P(left handed musician)= 0.6$
So I am looking for the probability for left-handed toddler and musician $P(A|B)$
3) now I apply Baye's theorem:
$P(left-handed|musician) = \frac{P(left-handed|musician)*P(left-handed)}{P(musician)} $
$P(A|B)= (0.6 * 0.1)/0.01$
and that's the point where it stops making sense because the result is 6. Since I put in decimalized percent this would mean 600% and that just makes no sense. Where did I go wrong?
First define convenient and intuitive symbols for your events.
$M^+, M^-, L^+, L^-$ respectively mean musician, non-musician, left-handed, not left-handed.
Then get the order of the conditional probability you need to find correct. You can read the conditional symbol as "given that", I find that most intuitive and least confusing.
If you need the probability that a left-handed toddler will be a musician, then your given condition is left-handedness.
So what you really need to find is $P(M^+|L^+)$. This was your first error.
Next apply Bayes' theorem. Your next error was in getting the form of your theorem mixed up, which caused a nonsensical result (probability over one).
The formula here is $\displaystyle P(M^+|L^+) = \frac{P(L^+|M^+)\cdot P(M^+)}{P(L^+)}$
Understanding that is one thing, but how do you easily recall the form so you can quickly and accurately use it? This is how I do it.
I first realise I use Bayes' theorem to find conditional probabilities where I already know the "reverse" conditional. So the first term on top in the RHS has to be the reverse of what I want to find (on the LHS). Then, I think of the conditional operator almost as a fraction with a division symbol and think of cancelling out the "denominator", which means multiplying out by $P(M^+)$ in this case. Finally, I use the fraction analogy one last time to recognise that since the thing on the right side of the conditional operator on the LHS is $L^+$, that needs to occur on the bottom of the RHS.
Note that this is a very informal and non-rigorous way of looking at it, meant only for you to be able to remember the form easily. You can also think of the denominator on the RHS as being the "given condition" always, which is $L^+$ here.
Anyway, last step is plugging in numbers, and you get $\displaystyle P(M^+|L^+) = \frac{0.6\cdot 0.01}{0.1} = 0.06$, which is $6 \%$.