Answer:
The probabilities are 0.480;0.077 and 0.443 respectively
Step-by-step explanation:
This is a conditional probability exercise.
Let's define conditional probability :
Given two events A and B :
[tex]P(A/B)=\frac{P(A,B)}{P(B)} \\P(B) > 0[/tex]
P(A,B) = P(A∩B) = P(B∩A) = P(B,A) : Is the probability that event A and event B occur at the same time.
We define the following events :
S1 : ''Specialist 1 processes requisitions''
S2 : ''Specialist 2 processes requisitions''
S3 : ''Specialist 3 preocesses requisitions''
I : ''Incorrect entered requisitions''
In our exercise :
[tex]P(S1)=0.42\\P(S2)=0.27\\P(S3)=0.31\\P(I/S1)=0.04\\P(I/S2)=0.01\\P(I/S3)=0.05[/tex]
We are ask to find
[tex]P(S1/I) ;P(S2/I);P(S3/I)[/tex]
We write the conditional equations :
[tex]P(I/S1)=\frac{P(I,S1)}{P(S1)} \\0.04=\frac{P(I,S1)}{0.42} \\P(I,S1)=0.0168[/tex]
[tex]P(I/S2)=\frac{P(I,S2)}{P(S2)} \\0.01=\frac{P(I,S2)}{0.27} \\P(I,S2)=0.0027[/tex]
[tex]P(I/S3)=\frac{P(I,S3)}{P(S3)} \\0.05=\frac{P(I,S3)}{0.31} \\P(I,S3)=0.0155[/tex]
We also define
P(A∪B) = P(A) + P(B) - P(A∩B)
P(I) = P [(I,S1)∪(I,S2)∪(I,S3)]
[tex]P(I) =P(I,S1) +P(I,S2)+P(I,S3)\\P(I)=0.0168+0.0027+0.0155\\P(I)=0.035[/tex]
There is no intersection between (I,S1);(I,S2) and (I,S3) because they are mutually exclusive events.
[tex]P(S1/I)=\frac{P(I,S1)}{P(I)} =\frac{0.0168}{0.035} =0.480\\P(S2/I)=\frac{P(I,S2)}{P(I)} =\frac{0.0027}{0.035} =0.077\\P(S3/I)=\frac{P(I,S3)}{P(I)} =\frac{0.0155}{0.035} =0.443[/tex]
