Need pro statistician help to answer question

[dowie]

I'm on a mobile at the mo and so a longer explanation for problem two I'll post later... but the posts on the initial problem you posted should help in thinking about it. You've got two events requiring treatment one with a 0.05 chance the other with a 0.0004 chance.

Re problem one - the chance of both teeth developing a problem is 0.05 * 0.05

The chance of one tooth (out of two possible teeth) is 2 * 0.05 * 0.95 (i.e tooth 1 has a problem tooth 2 doesn't or tooth 1 doesn't but tooth 2 does)

and obviously chance of no problem is 0.95 * 0.95

 

[jpod]

Understood, thank you.

Ok problem 1 (two teeth)
Chance of both teeth developing a problem is 0.05*0.05 = 0.025 which is 2.5% chance

Chance of one tooth developing a problem is 2 * 0.05 * 0.95 = 0.005 which is a 9.5% chance

That explanation is brilliant I get it, ty.

Chance of no problem is 0.95 * 0.95 = 0.9025 which is a 90% chance


Ok Problem 2, going to have a think about that one.

 

[jpod]

The probabilities are per tooth by the way. There is one patient, one individual.

 

[estebanrey]

One individual has 4 teeth that are injured.

i There are 2 rare complications you can develop over time, each has a chance of 5% of developing in a lifetime

ii There is one other rare complication, the risk is 7.5% over a lifetime.

Q/ What is the percentage chance of this individual developing a complication?

Is it 7.5% x 4 = 28% chance over their life

Please explain the correct answer for this maths numpty so I can understand the field of statistics we are referring to. And thank you.

Haven't read any of the previous but probabilities are a strong point of mine so here goes.

This is an "or" question like if I flipped two coins what is the chance at least one of them will land on heads. With these you turn it around into and "and" question/opposite outcome and calculate the chance as you were taught in school , the answer is then the remainder. So....

The chance of both coins landing as tails = 0.5 * 0.5 = 0.25 or 25%.

Therefore the chance of either landing as a head is 75%.


So, let's do this problem. It is asking given probabilities stated, what is chance one of his four injured teeth develops a complication over his lifetime. It's an "or" question so like before we reverse it and ask "what is the chance he will have no complications at all".

So let's ask, on each tooth what is the chance he will get no complications... Well

0.95 x 0.95 x 0.725 (the chance he gets none of three complications) = 83.4%

He has four teeth so the chance he gets no complications on any of them are then

0.834 x 0.834 x 0.834 x 0.834 = 48.5%

Therefore, by definition, the probability of him getting any complication is 51.5%

Hope that helps

 

[jpod]

Howdy, thanks for posting, have a look at the new problem, post #38

They are all independent risks, problem one has the individual patient has 2 teeth, problem 3 the patient has 3 teeth.

 

[dowie]

0.95 x 0.95 x 0.725 (the chance he gets none of three complications) = 83.4%

He has four teeth so the chance he gets no complications on any of them are then

0.834 x 0.834 x 0.834 x 0.834 = 48.5%

Therefore, by definition, the probability of him getting any complication is 51.5%

Hope that helps

actually 51.4% (you've made a rounding error doing it that way)

1 - (.95 * .95 * .925) ^4 = 0.51431378948 or roughly 51.4%

(obvs doesn't matter for the sake of the OP but just highlighting the potential issue with these calculations)

 

[jpod]

Lol

I have had a go and I can't figure it out. I am too confused, sorry.

Problem 1
2 teeth one patient, independent risks of 5%
So is this correct

Chance of no problems 1 - 2*(0.95 * 0.95)

Chance of one tooth out of two with problems 1 - 2*(0.05 * 0.95)

Chance of two teeth with problems 1 - 2*(0.05 * 0.05)


(Can't understand why there is no to the power of 2)

 

[estebanrey]

Problem 1[/B]
There is a complication risk of 5%, independent complication, there are two teeth:

Chance = 1 - (0.95x.0.95)/\2 = 0.81

The chance of complication in a single tooth is 19%

Q/ What is the chance of complications in both teeth?

Is it 1/2 of 19%? So 9.5%

The first calculation is irrelevant (and wrong) as the chance of him getting it in both teeth is quicker to calculate than either tooth.

Chance of complications in both teeth is simply 0.05 x 0.05 = 0.25%

With such a small number of outcomes, why not do them all to check they add back to 100.

Both Teeth = 0.05 x 0.05 = 0.25%
Neither Teeth = 0.95 x 0.95 = 90.25%
Either Tooth (remaining option) = 100 - 90.25 - 0.25 = 9.5%

Not sure how you made it 19%?

 

[jpod]

Both Teeth = 0.05 x 0.05 = 0.25%
Neither Teeth = 0.95 x 0.95 = 90.25%
Either Tooth (remaining option) = 100 - 90.25 - 0.25 = 9.5%

Understand that. Thank you, appreciated.

So problem 2
3 teeth, same patient, independant risks
Risk 1 5% chance = 0.05
Risk 2 4% chance, 0.04, of those 1% require treatment, chance = 0.0004

I need to figure out the % chance of the complications with the need for treatment:
No tooth complication
1 tooth comp
2 teeth comp
3 teeth comp

Is no tooth complication 1 - (0.95*0.9996)/\3 = 0.14 or 14% chance?

 

[estebanrey]

Understand that. Thank you, appreciated.

So problem 2
3 teeth, same patient, independant risks
Risk 1 5% chance = 0.05
Risk 2 4% chance, 0.04, of those 1% require treatment, chance = 0.0004

I need to figure out the % chance of the complications with the need for treatment:
No tooth complication
1 tooth comp
2 teeth comp
3 teeth comp

Is no tooth complication 1 - (0.95*0.9996)/\3 = 0.14 or 14% chance?

No tooth complication = 85.63%
1 tooth comp = 13.63%
2 teeth comp = 0.72%
3 teeth comp = 0.01%


Complication A is fairly simple as it always requires treatment. So in any given tooth there is 95% chance he won't need treatment X for it.

Complication B we will reverse as explained before and ask what is the chance he will need treatment Y (therefore the remainder is the chance he won't).

So that is 0.04 * 0.01 = 0.04% and thus a 99.96% chance he won't need treatment y for any given tooth.

So now we multiply both non-treatment chances (99.96 x 95) which gives you a 94.962% chance of not needing either treatment X or treatment Y per tooth. Therefore, minus from 100 to get the chance he will need either treatment X or treatment Y for a given tooth which is 5.038%

Now we know these we can go through all the options.

Needs Either Treatment
0 Teeth = 0.94962 * 0.94962 * 0.94962 = 85.634...%
3 Teeth = 0.05038 * 0.05038 *0.05038 = 0.0127...%
1 Tooth = 0.05038 * 0.94962 * 0.94962 = 0.04543 * 3 [possible affected teeth] = 13.629%

...therefore...

2 Teeth = 100 - (85.634... + 0.0127... + 13.629...) = 0.723%

I made this up quickly to show you all the possibilities (and to check my working LOL)

They all add back up to 100 so we know we've covered everything.

 

[estebanrey]

actually 51.4% (you've made a rounding error doing it that way)

1 - (.95 * .95 * .925) ^4 = 0.51431378948 or roughly 51.4%

(obvs doesn't matter for the sake of the OP but just highlighting the potential issue with these calculations)

I rounded at each stage as I was using pen, paper and a calculator at the time, the working is what matters which was correct. Besides what is 0.1 of a single percentage point between friends? LOL

However, criticism noted and made sure next two explanations are accurate to 3 decimal places just for your Royal Pedanticness

 

[dowie]

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

 

[dowie]

I rounded at each stage as I was using pen, paper and a calculator at the time, the working is what matters which was correct. Besides what is 0.1 of a single percentage point between friends? LOL

However, criticism noted and made sure next two explanations are accurate to 3 decimal places just for your Royal Pedanticness

I wasn't being pedantic I was pointing out there was a simpler way of calculating it that removed the potential for errors

 

[estebanrey]

Ohhh...we disagree by slightly different decimals on problem 2 Dowie (13.53 plays 13.62), so who's correct now?

...not asking that arrogantly, genuine question, my head hurts now and can't be arsed to double check so can you have a look at my working above yours?

 

[dowie]

we don't disagree on this answer, I've just pointed out the results of only considering event A too

chance of one tooth needing treatment when considering event A and B

3 * 0.05038 * 0.94962^2 = approx 13.63% (which agrees with your spreadsheet)

chance of one tooth needing treatment when considering just event A

3 * 0.05 * 0.95^2 = 13.53%

I'm just pointing out that considering B is a bit futile as it is so negligible and I'd wager the 5% chance for event A isn't really all that precise

 

[estebanrey]

we don't disagree on this answer, I've just pointed out the results of only considering event A too

chance of one tooth needing treatment when considering event A and B

3 * 0.05038 * 0.94962^2 = approx 13.63% (which agrees with your spreadsheet)

chance of one tooth needing treatment when considering just event A

3 * 0.05 * 0.95^2 = 13.53%

I'm just pointing out that considering B is a bit futile as it is so negligible and I'd wager the 5% chance for event A isn't really all that precise

My bad, was being lazy so just skipped to your final three answers...coolio

 

[jpod]

(Shh...he is very strict, don't upset him or he will make it 4 decimal placemats).

Fellas that is exactly what I needed, thank you for taking the time to post such crystal clear proofs. I am very grateful indeed. It may interest you to know that once again your uber maths prowess has benefited my professional practice and patient care.

PS
I got the point about the irrelevant consideration, 5% v the very small %, made perfect sense.

Once I have had a look if I get stuck may I post again.

 

[estebanrey]

(Shh...he is very strict, don't upset him or he will make it 4 decimal placemats).

Fellas that is exactly what I needed, thank you for taking the time to post such crystal clear proofs. I am very grateful indeed. It may interest you to know that once again your uber maths prowess has benefited my professional practice and patient care.

PS
I got the point about the irrelevant consideration, 5% v the very small %, made perfect sense.

Once I have had a look if I get stuck may I post again.

If you want to make it easier and do some revision there are plenty of good sites out there that show you how to calculate lottery probabilities.

The calculations used to ask "What are the chances I get 2 numbers on my ticket?" is the same as asking "What are chances 2 of these 6 teeth will fall out"

 

[Heofz]

Easiest to calculate the probability of not getting any complications at all, on any of your teeth, and subtract it from 100%.

The result is: 51.4%

So it's pretty much a coin toss on whether you get one or more complications over a lifetime with at least 1 of your teeth.

 

[jpod]

Hi Heofz, thanks for that. I have been schooled on the second two maths problems starting post #38, there are two elegant solutions posted there, I am studying them today! Feel free to join in.