# Thread: Would anyone be so kind to help me calculate some probabilities?

1. ## Would anyone be so kind to help me calculate some probabilities?

Hi everyone!

I am modelling both eyes of a person, both of which can be categorized into 3 different states, depending on the visual acuity in each eye. These different states are VA1, VA2 and VA3 (where VA1 is very good vision, VA2 is average vision, and VA3 is bad vision).

So the probabilities for each eye falling into each VA state may be as follows:

 Eye 1 VA1 VA2 VA3 0.1 0.2 0.7

 Eye 2 VA1 VA2 VA3 0 0.2 0.8

Now my task is to calculate the probability that Eye 1 is the better-seeing eye (BSE). Just glancing at the distributions it is easy to see that Eye 1 will have the higher chance to be the BSE, but I am having trouble quantifying it hehe.

Can anyone help? Preferably I would like to calculate a table which basically says: if Eye 1 'falls' into VA1, then what is the probability that it is the BSE? and if Eye 1 'falls' into VA2, then what is the probability it is the BSE? and finally if Eye 1 'falls' into VA3, then what is the probability that it is the BSE?

Any help would be greatly appreciated. I have tried modelling this problem through different approaches and methods but I keep hitting a brick wall, and I don't have confidence in my results probabilities sometimes get me in a bind like this hehe

Thanks all!

2. ## Re: Would anyone be so kind to help me calculate some probabilities?

Mathematically speaking, we have two random variables X1 and X2 that assume values 1, 2, and 3 with the given probabilities. The first eye is better seeing than the second one if X1 ≤ X2 (or X1 < X2 for strictly better seeing). So we need to find P(X1 ≤ X2). The event X1 ≤ X2 can be broken into several disjoint events:

X1 = 1, X2 = 1
X1 = 1, X2 = 2
X1 = 1, X2 = 3
X1 = 2, X2 = 2
X1 = 2, X2 = 3
X1 = 3, X2 = 3

The probability of the union of the first three events is the same as P(X1 = 1). For the rest, assuming that vision acuity in different eyes is independent, we can multiply probabilities: P(X1 = m, X2 = n) = P(X1 = m)P(X2 = n). Therefore, the probability that the first eye sees at least as well as the second one is

0.1 + 0.2 * 0.2 + 0.2 * 0.8 + 0.7 * 0.8 = 0.86.

3. ## Re: Would anyone be so kind to help me calculate some probabilities?

Thank you very much for your help - I prefer calculating it using your method as I prefer maths to stats. However if there are 20 states per eye, then things get a lot more complicated