Question

In an insurance company, each policy has a paper record and an electric record. For those policies having incorrect paper record, 60% also having incorrect electric record; For policies having incorrect electric record, 75% also having incorrect paper record.3% of all policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what’s the probability that the one having both correct paper and correct electric records?

Answer 1

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%

`P(A\ and\ B)=P\left ( A\cap B\right )=0.03`

`P(A/B)=0.75`

`P(B/A)=0.60`

`P(A/B)=(P\left ( A\cap B\right ))/(P(B))`

`0.75=(0.03)/(P(B))`

`P(B)=(0.03)/(0.75)=0.04`

similarly
`P(A)=(0.03)/(0.6)=0.05`

`P\left ( A\cup B\right )=P(A)+P(B)-P\left ( A\cap B\right )`

`P\left ( A\cup B\right )=0.05+0.04-0.03=0.06`

Probability that the one having both correct paper and correct electric records

`P=1-P\left ( A\cup B\right )=1-0.06=0.94`