Question

If not distracted, the probability of success if 0.9. If distracted, the probability of success if 0.6. And the probability of distraction is 0.8.

1) Find the probability of success.

2) If do not succeed, what is the conditional probability of being distracted?

Answer 1

We are given here that:
P( success | not distracted ) = 0.9,
P( success | distracted ) = 0.6

Also, we are given here that:
P( distracted ) = 0.8

a) The probability of success using the law of total probability here is computed as:
P( success ) = P( success | not distracted )P(not distracted ) + P( success | distracted )P( distracted )

= 0.9*(1 - 0.8) + 0.6*0.8 = 0.18 + 0.48 = 0.66
Therefore 0.66 is the required probability here.

b) P(not success ) = 1 - P(success ) = 1 - 0.66 = 0.34

Given that there is success, the conditional probability of being distracted is computed using Bayes theorem here as:
P( distracted | not success ) = P(not success | distracted ) P(distracted ) / P(not success )

= (1 - 0.6)*0.8 / 0.34

= 0.32 / 0.34

= 0.9412

Therefore 0.9412 is the required probability here.