- #1
Dell
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student A comes to 80% of the classes while student B comes to 60%, (A,B are independant of one another), on a given day exactly one student arrived, what is the probability that it was student A
P(A)=0.8
P(B)=0.6
P([tex]\bar{A}[/tex])*P(B)=0.12
P([tex]\bar{B}[/tex])*P(A)=0.32
C=> only one student arrived
P(C)=P([tex]\bar{A}[/tex])*P(B) +P([tex]\bar{B}[/tex])*P(A)=0.12 + 0.32 =0.44
P(C/A)=P([tex]\bar{B}[/tex])*P(A)=0.32
P(A/C)=[tex]\frac{P(A)*P(C/A)}{P(C)}[/tex]=[tex]\frac{0.8*0.32}{0.44}[/tex]=0.58
this is wrong the correct answer is meant to be 0.73, any help??
P(A)=0.8
P(B)=0.6
P([tex]\bar{A}[/tex])*P(B)=0.12
P([tex]\bar{B}[/tex])*P(A)=0.32
C=> only one student arrived
P(C)=P([tex]\bar{A}[/tex])*P(B) +P([tex]\bar{B}[/tex])*P(A)=0.12 + 0.32 =0.44
P(C/A)=P([tex]\bar{B}[/tex])*P(A)=0.32
P(A/C)=[tex]\frac{P(A)*P(C/A)}{P(C)}[/tex]=[tex]\frac{0.8*0.32}{0.44}[/tex]=0.58
this is wrong the correct answer is meant to be 0.73, any help??