#### Serendipity02

##### New member

- Joined
- Jun 10, 2021

- Messages
- 2

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.

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.