Answer:
Step-by-step explanation:
Suppose
A be the Event :Incorrect Paper record
B be the Event: Incorrect Electronic Record
P(A/B)=Probability of occurring A given that B already happened
From Question
Probability that all the policies have both Incorrect Paper record and Incorrect Electronic record  is 3%
[tex]P(A\ and\ B)=P\left ( A\cap B\right )=0.03[/tex]
[tex]P(A/B)=0.75[/tex]
[tex]P(B/A)=0.60[/tex]
[tex]P(A/B)=\frac{P\left ( A\cap B\right )}{P(B)}[/tex]
[tex]0.75=\frac{0.03}{P(B)}[/tex]
[tex]P(B)=\frac{0.03}{0.75}=0.04[/tex]
similarly [tex]P(A)=\frac{0.03}{0.6}=0.05[/tex]
[tex]P\left ( A\cup B\right )=P(A)+P(B)-P\left ( A\cap B\right )[/tex]
[tex]P\left ( A\cup B\right )=0.05+0.04-0.03=0.06[/tex]
Probability that the one having both correct paper and correct electric records
[tex]P=1-P\left ( A\cup B\right )=1-0.06=0.94[/tex]
