Question

Let A and B be two events in a sample with P(A)=0.4 and P(AuB)=0.7.Let P(B)=p

a)i For what value of p are A and B mutually exclusive?

ii for what value of p are A and B independent?

b) Assume that P(A)=0.4 and P(B)=0.3

i find P(B')

ii if A and B are mutually exclusive , what is P(A or B)

ii given that P(A or B)=0.6, Find the P(B/A)

c) suppose events A and B are such that P(A)=0.25,P(B) =0.33 and P(A|B)=0.5, Compute,

P(AnB) and P(AuB)

Answer 1

P(A) = 0.4     
P(A U B) = 0.7     
P(B) = p     
      
a)         
i) When A and B are mutually exclusive, P(A ∩ B) = 0     
We know      
P(A U B) = P(A) + P(B) - P(A ∩ B)     
0.7 = 0.4 + p + 0     
Hence, p = 0.7 - 0.4 = 0.3     
p = 0.3     
      
ii) When A and B are independent, P(A ∩ B) = P(A) * P(B)     
We know      
P(A U B) = P(A) + P(B) - P(A ∩ B)     
P(A U B) = P(A) + P(B) - P(A)*P(B)     
0.7 = 0.4 + p + 0.4*p     
0.3 = p + 0.4p     
1.4p = 0.3     
p = 0.3/1.4     
p = 0.2143     
      
b) P(A) = 0.4     
P(B) = 0.3     
       
i)   P(B') = 1 - P(B)      
           = 1 - 0.3      
           = 0.4      
P(B') = 0.4      
       
ii) When A and B are mutually exclusive, P(A ∩ B) = 0      
P(A or B) = P(A) + P(B) - P(A ∩ B)      
                   = 0.4 + 0.4 - 0      
                   = 0.8      
P(A or B) = 0.8    

iii) P(A or B) = 0.6     
P(B|A) = P(A ∩ B) / P(A)   …Baye's conditional theorem  

       

                = 0.25     
P(B|A) = 0.25    

c) P(A) = 0.25   
P(B) = 0.33   
P(A | B) = 0.5

        = 0.25 + 0.33 - 0.5*0.25  
       = 0.455  
P(A ∩ B) = 0.455   
P(A U B) = P(A) + P(B) - P(A ∩ B)   
                   = 0.25 + 0.33 - 0.455   
                   = 0.125   
P(A U B) = 0.125