# Conditional probability

1. Oct 13, 2007

1. The problem statement, all variables and given/known data

Assuming a comp is switched on, the probability that the monitor is not working is 0.005, the probability that the CPU is faulty is 0.02, and the probability that the keyboard cable has been damaged is 0.0025, and that there are no other faults.

Proceed to evaluate the probability that the computer will be operaitonal for a period of 100 days, if it is switched on and off once a day only, and that the faults have the same probability of occurence on each occasion

The question then says that the conditional probability of the monitor not working given that the keyboard cable has been damaged is 0.05, and askes how it affects the answers
2. Relevant equations

3. The attempt at a solution

For this :
( 1-(0.02+0.005+0.0025)^100)
which is 0.386 for the first part

A little stuck on the second part I assume:
(1 - (0.005 X 0.05 + 0.002 + 0.0025) ) ^100

but the correct answer is 0.390459

2. Oct 13, 2007

### CompuChip

OK, here are some steps:
1) on any given day, what is the chance that it will fail to boot?
2) then what is the chance that it crashes the first day
3) what is the chance that it happens the second day (warning! this is not the same as in 2, which is where you went wrong above)
4) now extend to 100 days

3. Oct 13, 2007

### Gib Z

What

4. Oct 14, 2007

### CompuChip

If it crashes the second day, this means that it did not give up the first day.
So it is the conditional probability
$$P(\text{crash on day 2}) = P(\text{no crash on day 1}) P(\text{crash day 1})$$
and similarly
$$P(\text{crash on day 3}) = P(\text{no crash on day 1}) P(\text{no crash on day 2}) P(\text{crash day 1})$$
etc., which is different from
$$P(\text{crash on day 1})$$

5. Oct 14, 2007

### Gib Z

But the boot ups are mutually exclusive events, they do not depend on previous results, why should they? Its like the chance of taking out a queen from a pack of cards, which is the boot up. Then at the end, you replace the card, the boot down. The next time you pull out a card, theres exactly the same chance!

6. Oct 14, 2007

### EnumaElish

Hint: Monitor fault and KB fault can no longer be assumed to be two independent events.

7. Oct 14, 2007

### Gokul43201

Staff Emeritus
The outcomes of boot-ups are not mutually exclusive, they are mutually independent (perhaps this is what you meant to say).

What compuchip was writing down may be clarified by a little rewording:

$$P(\text{crash only on day 2}) = P(\text{no crash on day 1}) P(\text{crash day 2}) = P(\text{no crash on day 1}) P(\text{crash day 1})$$

8. Oct 14, 2007

### CompuChip

And then I read the question as: what is the chance that I will draw a queen exactly the 100th time (and no queen all the times before it).

At least, I assumed from the word "operational for 100 days" that it shouldn't fail in any of those days, and then crash on either the 100th or the 101st (depending on the interpretation).

9. Oct 14, 2007

### learningphysics

Although this is true... I'm wondering for part a) why not just say the probability it is operational for 100 days =

(probability operation 1st day) AND (probability operational second day) AND etc...

= [(1-0.005)(1-0.02)(1-0.0025)]^100

I'm wondering if I misunderstood the question because I'm not getting 0.390459 for the second part.

10. Oct 14, 2007

### Sleek

The probability of monitor working or not given that keyboard cable is DAMAGED makes no difference to probability of computer booting or not, since its the same as the probability of keyboard cable not working. But this is true only if the probability of monitor not working given that keyboard cable is OK is the same as in first part, which may not be the case.

Thus if its possible to find the probability that monitor is not working given the keyboard cable is OK, we have a new value for monitor not working. Replacing this new value in the first part should give the relevant answer. Thats what I think, though I may be wrong :(.

Regards,
Sleek.