Question

Bob has recently been hired by a shop downtown to help customers with various computer-related problems. Lately, two different viruses have been bugging many customers: Dummy and Smarty. It is estimated that about 65% of the customers with virus problems are bothered by Dummy and the remaining 35% by Smarty. If the computer is infected by Dummy, Bob has a 90% chance of fixing the problem. However, if the computer is infected by Smarty, this chance is only 70%. If a virus-infected computer is randomly selected from the shop, and we know it was fixed by Bob, what is the probability that it was infected with Dummy

Answer 1

Answer:

70.48% probability that it was infected with Dummy

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Fixed by Bob

Event B: Infected with Dummy

65% of the customers with virus problems are bothered by Dummy

This means that $P(B) = 0.65$

If the computer is infected by Dummy, Bob has a 90% chance of fixing the problem.

This means that $P(A|B) = 0.9$

Probability of Bob Fixing

90% of 65%(fixing a Dummy-infected virus)

70% of 35%(fixing a Smarty-infected virus).

So

$P(A) = 0.9*0.65 + 0.7*0.35 = 0.83$

What is the probability that it was infected with Dummy

$P(B|A) = \frac{0.65*0.9}{0.83} = 0.7048$

70.48% probability that it was infected with Dummy