Answer:
Probability that a smoker has lung disease = 0.2132 
Step-by-step explanation:
Let L = event that % of population having lung disease, P(L) = 0.07 
So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93
S = event that person is smoker 
% of population that are smokers given they are having lung disease, P(S/L) = 0.90 
% of population that are smokers given they are not having lung disease, P(S/L') = 0.25 
We know that, conditional probability formula is given by;
                         P(S/L) =  
  
                          = P(S/L) * P(L)
 = P(S/L) * P(L) 
                                       = 0.90 * 0.07 = 0.063
So,   = 0.063 .
 = 0.063 .
Now, probability that a smoker has lung disease is given by = P(L/S) 
       P(L/S) =  
 
P(S) = P(S/L) * P(L) + P(S/L') * P(L') 
        = 0.90 * 0.07 + 0.25 * 0.93 = 0.2955 
Therefore, P(L/S) =  = 0.2132
 = 0.2132 
Hence, probability that a smoker has lung disease is 0.2132 .