Assume a firm has different plants and in some of th eplants it
continues to offerno training programs for its workers (A), in some of
the plants it operates a training program that involves structured
classroom training run by top managers (B), and in the other plants
the firm operates a training program that is run by groups, with each
group responsible for training new workers (C).
The firm has measures of productivity and has grouped the productivity
figures into three categories: Low (L), MOderate (M), and High (H).

The results of the different plants are given below:

-----------------------------------------------------------
                  A              B                 C

(L)ow             32             40               21
(M)oderate        21             18               46
(H)igh             8              5               23

1) Calculate the marginal probabilites for each category of
productivity.
2) Calculate the following conditional probabilities: P (L/A), P(L/B),
P(L/C), P(H/A), P(H/B), and P(H/C).
3)What do the results tell us what the manager should do in terms of
training programs? Explain.

Hi k9queen!
In order to answer these questions, it's better to rewrite the table
you provide as follows. First we add the numbers for each column and
each row:

                  A              B                 C    Total
 
(L)ow             32             40               21      93
(M)oderate        21             18               46      85
(H)igh             8              5               23      36

Total             61             63               90     214

The column totals (61, 63, 90) show the quantity of people that was
trained by each group, irrespective of their productivity. The row
totals have an analogous interpretation. As you can see, there are 214
observations. We can then divide all the numbers by 214 to transform
these observations into probabilities. Thus, for example, 32/214=0.15
is the probability that someone had an A training AND has an L
productivity. Dividing each cell by 214 gives:

                  A              B                 C    Total
 
(L)ow             0.149         0.186           0.098     0.433
(M)oderate        0.098         0.084           0.214     0.397
(H)igh            0.037         0.023           0.107     0.168

Total             0.285         0.294           0.420       1

(the numbers are rounded to 3 digits, so the totals are not completely
accurate)

As you'll see, it's now easier to answer the questions:

1) Calculate the marginal probabilites for each category of
productivity.

This is basically the probability that someone has a specific
productivity, independently of the training they had. Therefore, we
simply use the numbers in the right margin:

The probability of productivity L is 0.433
The probability of productivity M is 0.397
The probability of productivity H is 0.168

These probabilities should clearly sum to 1, since everyone has either
L, M or H productivity. The fact that here they sum to 0.998 comes
from the rounding I did before.

Types of Probability
http://www.uwsp.edu/psych/cw/statmanual/probinter.html

2) Calculate the following conditional probabilities: P (L/A), P(L/B),
P(L/C), P(H/A), P(H/B), and P(H/C).

As you can see in the link I provided above, conditional probability
is defined in the following way. Let A and B be two events. Then:

P(A/B) = P(A and B)
         ----------
            P(B)

So, let's compute the probabilities in your question:

P(L/C) = P(L and C) / P(C)

P(L and C) is simply the probability that appears in row L, column C
(it's called the joint probability: the probability that these two
events -L and C- happen -both of them-). So this probability is 0.098.
P(C) is the marginal probability of C. It's the probability of
receiving C training, independently of the productivity. This is the
total probability in column C, which is 0.420.

Therefore, P(L/C)=0.098/0.420=0.233

Similarly, we can calculate the other probabilities:

P(L/B)=0.186/0.294=0.632
P(L/A)=0.149/0.285=0.522
P(H/A)=0.037/0.285=0.129
P(H/B)=0.023/0.294=0.078
P(H/C)=0.107/0.420=0.254

3) The conclusion is that the manager should implement the "groups"
training progran (program C). The reason is the following. P(L/C) is
smaller than P(L/A) and P(L/B). This means that the probability of
having a low productivity given that you were trained with C is lower
than the probability of having a low productivity if you were trained
with A or B. Thus it would appear that C traning reduces the
probability of having a low productivity. Similarly, if you look at
the probabilities P(H/A), P(H/B) and P(H/C), you'll see that C
training increases the probability of having a high productivity.
Therefore, the manager should consider implementing the C training
program.


Google search terms
"marginal probability"
://www.google.com.ar/search?q=%22marginal+probability%22&ie=UTF-8&oe=UTF-8&hl=es&meta=


I hope this helps! If you have any questions regarding my answer,
please let me know through a clarification request. Otherwise I await
your rating and final comments.

Best wishes!
elmarto