Aresearcher finds that, of 982 men who died in 2002, 221 died from some heart disease. also, of the 982 men, 334 had at least one parent who had some heart disease. of the latter 334 men, 111 died from some heart disease. a man is selected from the group of 982. given that neither of his parents had some heart disease, find the conditional probability that this man died of some heart disease.

The question may be solved using the laws of total probability, Bayes theorem, conditional probability, etc.However, it is most readily solved by constructing a contingency diagram.

Events:
H=died of heart disease
~H=did not die of heart disease
P=at least one parent has heart disease
~P=no parent have heart disease

From given information, we can construct contingency diagram as follows:
           H   ~H   Total
P       111           334
~P     
Total  221          982

From the above diagram, we calculate
|~P & H| = 221-111=110 
| P & ~H | = 334-111 = 223
Total | ~P | = 982-334=648
Total | ~H | = 982-221=761
| ~H & ~P | = 648-110 = 538
=>
          H   ~H   Total
P       111  223  334
~P     110  538  648
Total  221  761  982
=> 
P(H|~P)=110/648=55/324
=0.170 (approx.)