OK back home so I can type out a more long winded answer here
Problem 2
There are three teeth, two complications, all independent
Complication A has a 5% risk and requires treatment X
Complication B has a 4% risk but does not require treatment, however there is thereafter a 1% risk of treatment Y
What is the % chance of requiring treatment in 1 tooth, 2 teeth and all the teeth? I definitely cannot figure that out.
Please explain like I am 10, thanks Pod
so considering B alongside A is a bit pointless as B occurring and requiring treatment is so negligible...
OK lets make event B be the event that complication B occurs AND it requires treatment and let event A simply be complication A that always requires treatment
so event A has a probability of 0.05 so notA is 0.95
event B has a probability of 0.0004 so notB is 0.9996 (as you can see event B is negligible)
this is just another variation on the first problem you posed only now you've got three teeth and two complications
so lets break it down
chance of event A and event B not happening in all three teeth = (0.95 * 0.9996)^3
approx 86%
so like your original question the chance of at least one tooth (out of the three) requiring treatment is simply 1 - (0.95 * 0.9996)^3 so approx 14%
However now you've asked about the individual chances of one, two and all three of the three teeth requiring treatment - this just gets slightly more long winded but isn't really any more complicated in principle (yet potentially rather futile in this instance as we'll see)
so lets say we have tooth 1, 2 and 3
Think about how you might need treatment for just one tooth... well tooth 1 could have event A occur or event B occur or both event A and B occur AND tooth 2 and three have both events NOT occur... or alternatively the same scenarios for tooth 2 or tooth 3 (so three different ways)
3* A * notB * notA * notB * notA * notB
3* notA * B * notA * notB * notA * notB
3* A * B *notA * notB * notA * notB
which all becomes a bit of a ball ache... and then when you consider treatments for two teeth and treatments for three teeth it becomes even more of a pain
fortunately we can just do what we did for your original question chance of not needing treatment is simply 1 - notA * notB = 1 - (0.95 * 0.9996) = 0.94962
so chance of needing treatment is 0.05038
this simplifies things
chance of one tooth needing treatment out of three is then simply
3 * 0.05038 * 0.94962^2 (as there are three ways this can occur)
= 13.63%
chance of two teeth needing treatment is:
3 * 0.05038^2 * 0.94962 (as there are three ways this can occur again)
= 0.72%
chance of three teeth needing treatment is:
0.05038^3
= 0.01%
these three probabilities should sum to the probability of one or more teeth requiring treatment (1 - (0.95 * 0.9996)^3) which they do
right and now back to why considering B is futile as mentioned earlier
imagine event B doesn't exist you'd have
for one tooth:
3 * 0.05 * 0.95^2 = 13.53%
for two teeth
3 * 0.05^2 * 0.95 = 0.71%
and for three teeth
0.05^3 = 0.01%
much the same probabilities as we already had when considering both events... I'd wager the 5% risk for A isn't all that precise to the point considering B becomes in anyway meaningfull