prashant wrote:

aps_can wrote:

23. 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 those 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 is 0.94

Here's the explanation:

Let the total no. of policies be T

Let no. of incorrect paper be x

Let no. of incorrect electric be y

Now 0.03*T = 0.6*x = 0.75*y

The required probability is [T-(x+y-0.03*T)]/T

Substituting and solving, we get the required prob. as 0.94