SOLUTION: Memory 80 GB 120 GB 2 GB 15 55 4 GB 10 20 A single computer is selected at random from the shipment. Let A be the event that

Algebra ->  Probability-and-statistics -> SOLUTION: Memory 80 GB 120 GB 2 GB 15 55 4 GB 10 20 A single computer is selected at random from the shipment. Let A be the event that       Log On


   

Question 1100861: Memory 80 GB 120 GB
2 GB 15 55
4 GB 10 20
A single computer is selected at random from the shipment.
Let A be the event that the computer has an 80 GB hard drive.
Let B be the event that the computer has a 120 GB hard drive.
Let C be the event that the computer has 2 GBs of memory.
Let D be the event that the computer has 4 GBs of memory.

What is the conditional probability P(C|B)?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

It's rather hard to interpret the information as you present it. Perhaps some formatting got lost in cyberspace....?

Apparently we have a shipment of computers....
15 with 80GB hard drive and 2GB memory
55 with 120GB hard drive and 2GB memory
10 with 80GB hard drive and 4GB memory
20 with 120GB hard drive and 4GB memory

The probability P(C|B) is the probability that C is true (the computer has 2GB memory), GIVEN THAT B is true (it has a 120GB hard drive).

With this conditional probability, the sample space is only those computers that have a 120GB hard drive; that is 55+20 = 75 computers.

The desired condition is that it is a computer with a 120GB hard drive AND a 2 GB memory; there are 55 of those computers.

So P(C|B) is 55/75 = 11/15.

Formally, the probability P(C|B) is P%28C+and+B%29%2FP%28B%29

I personally find this hard to work with. I find it much easier to think in terms of the sample space consisting of only those computers that satisfy condition B.