the probability that applicants A, B and C are successful are 1/3, 2/3, 1/4 respectively. if the three were invited what is the probability that only two of them were successful
P(A)=13,P(B)=23,P(C)=14P(only two successful)=P(A)P(B)(1−P(C))+P(A)(1−P(B))P(C)+(1−P(A))P(B)P(C)==13⋅23⋅34+13⋅13⋅14+23⋅23⋅14=0.305556P\left( A \right) =\frac{1}{3},P\left( B \right) =\frac{2}{3},P\left( C \right) =\frac{1}{4}\\P\left( only\,\,two\,\,successful \right) =P\left( A \right) P\left( B \right) \left( 1-P\left( C \right) \right) +P\left( A \right) \left( 1-P\left( B \right) \right) P\left( C \right) +\left( 1-P\left( A \right) \right) P\left( B \right) P\left( C \right) =\\=\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{3}{4}+\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{4}+\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{4}=0.305556P(A)=31,P(B)=32,P(C)=41P(onlytwosuccessful)=P(A)P(B)(1−P(C))+P(A)(1−P(B))P(C)+(1−P(A))P(B)P(C)==31⋅32⋅43+31⋅31⋅41+32⋅32⋅41=0.305556
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