# Probability problem - Please help! :D

#### Serendipity02

##### New member
Hi all,
Probability is a new thing for me so bear with me please.
As the head of inventory for an IT company, you've had a tough week. A warehouse of your company was recently set on fire, and you had to report to all the computers stored there
to be recycled.
On the plus side, you must have been thrilled that you managed to ship two computers to a bigger customer.
last week. But, your assistant had not heard of the fire and had by mistake
transported a whole truck of computers from the damaged warehouse to the shipping center. It appears that
30% of all computers delivered last week were damaged. You don't know if your
larger customer received two damaged computers, two in good condition, or one of each ( one damaged and one not)
Computers were randomly selected from the dispatch center for delivery. If your client has
received two computers in good condition, everything is fine. If the customer receives a damaged computer, you will be
return at your expense - 1000 dollars - and you can replace it. However, if both computers
are damaged, the customer will cancel all his orders and you lose 100,000 dollars.

This problem asks us to elaborate the probability of each one of these scenarios to calculate the mean or the expected loss, and my response is not compatible with that of the professor.
The probability of losing a 1000 dollars was wrong, I responded by straightforwardly saying .3 the probability that one is damaged. I have the response attached to this thread as a picture. With X the probability of loss.

The thing which I don't understand is that the professor answered using a different logic or should I say calculation in the same scenario except that in here it's the probability of posession of a phone. Here's the problem:
15% of people in your travelling group have a phone on them. You are assigned randomly to two people of that group.

1- what is the probability that the first person will have a phone ( this one is similar to the one above, the correct answer would be .15 simply, why couldn't we answer the same as in the problem above in the Probability of 1000 bucks loss??)
2- that the 1st wont have a phone p=.85
3- both will have a phone p = .15 squared
4- 1 at most will have a phone ( is substracting the probability that both will have phone and both wont have a phone be correct from 1 be correct?)

And one last problem

At the exit of a fabrication chain, shoes are susceptible to present two flaws, a number of observations have established that:
.05 is the proportion of shoes presenting flaw A
.03 is the proportion of shoes presenting flaw B
.01 presenting flaw A and B

One of the questions asks to determine the probability that a shoe will present flaw A only. I dont know to approach this one.

Any elucidation would be welcome, thanks in advance.

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#### Dr.Peterson

##### Elite Member
I'm confused. Is the image your supposedly wrong answer, or the teacher's correct answer?

I would have done it differently, but I'd get the same answer, and the way shown there is valid.

If you're saying that your answer was 0.3, not 0.42, then you were clearly wrong, as your answer is the probability that the first was bad, ignoring the second.

The method in the image is P(exactly one bad) = 1 - P(both good) - P(both bad) = 1 - 0.7^2 - 0.3^2 = 0.42.

What I would normally do is 2*P(first bad, second good), since it could be in either order. (This is equivalent to the binomial distribution formula.) This gives 2*0.3*0.7 = 0.42.

#### Serendipity02

##### New member
The image I uploaded is the correct answer. My answer which is wrong is .3 . Thank you for the answer! How do I avoid to make the same mistake again? By using this binomial formula? How about the last problem?

#### Dr.Peterson

##### Elite Member
The image I uploaded is the correct answer. My answer which is wrong is .3 . Thank you for the answer! How do I avoid to make the same mistake again? By using this binomial formula? How about the last problem?
The main way to avoid errors in probability is to question everything you do. Sometimes I don't trust my own answer until I can get the same thing two ways (as I did in this problem!).

There are lots of subtleties. When they ask for the probability of one being bad, you have to ask, does that mean EITHER one, or THIS one? Always try to clarify in your mind exactly what the event is. Often that means visualizing a list of outcomes that must be included (e.g. GG, GB, BG, BB), and thinking about how to make sure you include exactly those and nothing else.

Eventually some things become familiar and you automatically use a particular formula (such as the binomial), but then you have to check whether you are doing it too automatically, and there is something different about this problem. In other words, never let down your guard.