TITLE: need help with question on probability distribution table
QUESTION [1 upvotes]: In a computer vision application it is important to distinguish between different object put in from of a camera. In particular, it is important to distinguish between a box, a cylinder, and a sphere. To do this, the company developed a simple program that counts the number of corners the program finds in the image. Due to errors, the program sometimes finds too many or too few corners but never more than $5$. To evaluate the accuracy of the program, a number of experiments with different objects of the given types are made and the following likelihoods to
  find $0,...,5$ corners is determined:

$$\begin{array}{lccccc}\text{object/#corners found}&0&1&2&3&4&5\\
\text{sphere}&0.75&0.10&0.05&0.05&0.025&0.025\\
\text{cylinder}&0.05&0.15&0.30&0.35&0.10&0.05\\
\text{box}&0.05&0.05&0.10&0.15&0.30&0.35\\
\end{array}$$

a) Determine the probability of an object being of each of the different types if $3$ corners are found and all objects are equally likely to be the object presented.
b) Determine for each object the likelihood that it is identified correctly using this algorithm.

For part (A) we are given the algorithm has detected 3 corners. 
also they are equally likely to be present so P(S) = P(C) = P(B) = 1/3 
So P(S/3) = 0.05/(0.05+0.35+0.15)
Similarly we can find for the other two objects. 
in part (b) i am having doubt on what exactly the question is demanding me to find?

REPLY [1 votes]: This question seems incomplete or relying on unsaid assumptions for two reasons:

You need to know what the program would identify as a box, cylinder, or a sphere given the number of corners detected. (I assume that it picks the most likely shape, i.e. it picks a sphere if it detects zero corners.)
You need to know how often the shapes appear. (It seems that they assume that the shapes are equally likely to appear, but this wouldn't necessarily be the case in practice. Plus, this allows me to make a slight genetalization.)

In fact, the second reason can influence the first. For instance, if you knew that there were absolutely no spheres and equal numbers of boxes and cylinders in whatever you were using, then getting zero corners would have an equal probability of being a cylinder or a box, because there is no reason to guess that it would be a sphere.
However, assuming that the shapes all have equal probabilities of showing up and the program chooses the most likely shape given a number of corners, then all you need to do is add up the probabilities in each row that have the highest value in each column and take their average. In your example, you would have $75\%$ for the sphere, $15\%+30\%+35\%=80\%$ for the cylinder, and $30\%+35\%=65\%$ for the box. Their average is $73\frac13\%$.
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