Define a probability distribution which satisfies independence conditions

Question

I have the following problem:

A certain hospital receives patients. 
Each patient could be healthy or sick. 
There are 2 doctors: Gabby, and Tully. 
Each doctor writes a report with his opinion.

There are 3 random variables: P, G, T with values {0,1} when 1 indicates sick and 0 indicates healthy.
P is the real status of the patients (Sick or healthy).
P[P=0] = P[P=1] = 0.5.

G, T indicates the doctors (Gabby, Tully) opinion.

Describe a distribution above P, G, T so that G, T will be independent, 
but G, T are not independent given P. 

The solution should be a table with all possible values of P, G, T
with the correct distribution for each combination. 

My question is how to approach such a question?

I've tried making a table like this:

P | G | T | P(G|P) | P(T|P) | P(G,T|P) 
0 | 0 | 0 |   0.3  |   0.6  |   0.5
0 | 0 | 1 |   0.3  |   0.6  |   0.5
0 | 1 | 0 |   0.3  |   0.6  |   0.5
0 | 1 | 1 |   0.3  |   0.6  |   0.5
1 | 0 | 0 |   0.3  |   0.6  |   0.5
1 | 0 | 1 |   0.3  |   0.6  |   0.5
1 | 1 | 0 |   0.3  |   0.6  |   0.5
1 | 1 | 1 |   0.3  |   0.6  |   0.5

And now I have P(G,T|P) != P(G|P) * P(T|P) as needed,

(Because of P(G|P) = 0.3, P(T|P) = 0.6, P(G|P) * P(T|P) = 0.5)

But I'm not sure how to calculate P(G), P(T), and check their independence.

Will be glad to know to approach this problem. Thanks.

Edit: This is my current solution:

Answer 1

As Misch said, in each row of the table you should write $\mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:

• All eight probabilities should sum to $1$ - they represent all possibilities.
• You calculate $\mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
• You calculate $\mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $\mathbb{P}(T = z )$ in the same way.
• You calculate $\mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.

To show that $G$ and $T$ are independent, you must show that $\mathbb{P}(G = x ) \mathbb{P}(T = z ) = \mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $\mathbb{P}( G = x | P = y) \mathbb{P}( T = z | P = y) \neq \mathbb{P}( G = x, T = z | P = y)$.

As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $\mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $\mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.