A repair shop uses a two step diagnostic process to repair?

Question

computers. Step 1 locates the problem in a system with probability 0.8. Step 2 (which is executed only if step 1 fails to locate the problem) locates the problem with probability 0.6. What is the probability that the entire procedure will fail to locate the problem?

Answer 1

Call the probabilities of success p1 and p2. p1 is applied to all cases. We can assume test 2 is used only if test 1 failed to locate the problam, so p2 is applied to 1-p1 of the cases. The total probability of success is the sum of the individual probabilities as applied: p1*1 + p2*(1-p1), or 0.8 + 0.12 = 0.92. So the probability of failure is 1 - 0.92 = 0.08.
Or you can calculate the probability of both tests failing as the product of each test failing: (1-p1) * (1-p2) = 0.2 * 0.4 = 0.08.
Think of 100 problems. 20 of them escape test 1 and of the remaining 20, 8 escape test 2. That's a 0.08 total failure rate.

Answer 2

My probability isn't very good. But I think this is right;

Step one has a .2 chance of failure, stage 2 has a .4 chance of failure. The chances of them both not working is .2*.8=.16

So not very likely.