Business Statistics help (probability) ITS URGENT!!

PlayaFlow

Im in DCU 2yr.

I have 5 Questions on probability that i cant do ( i can try them , but im not sure if im doing them right or using the right method )

Basically , I need to be sure they are 100% right, So is there anybody who is familiar with statistics "Probability and Random Sampling" that can help me out.

Maybe somebody who has studied stats before or just somebody who is sure of what they are doing. IT's really urgent , I NEED TO PASS THIS!!
Any help greatly appreciated

PlayaFlow

this is Q1:

Box A contains 8 items of which 3 are defective, and Box B contains 5 items of which 2 are defective.
(a) An item is drawn at random from each box:
(i) What is the probability that both items are defective?
(ii) What is the probability that one item is defective and one is not?

(b)If one item is defective and one is not, what is the probability 'p' that the defective item came from Box A?


Anybody have any idea how to go about doing this, i'm not sure of which probability rule to use here, can anyone help?

ecksor

Probability is something that people notoriously get wrong, so bear that in mind as I do this and make sure that my explanation makes sense to you (partly so that you'll be able to do it in the exam and partly in case I make a mistake).

(a) There are 8 ways of selecting the first item and 5 ways of selecting the second item. Therefore there are 8*5 = 40 possible combinations.

(i) There are 3 ways of selecting the first defective item and 2 ways of selecting the second defective item so there are 3*2 = 6 ways of selecting 2 defective items. So the probability of selecting two defective items is 6/40 = 3/20 = .15

(ii) Two scenarios, either the defective item came from the first box or it came from the second box. If it came from the first box then there are 3 ways of drawing that, and 3 ways of selecting the non-defective one from the second box, so there are 9 ways of drawing the required combination in that way. If it came from the second box then there are 5 ways of choosing the non-defective item from the first box and 2 ways of drawing it from the second box so there are 10 ways of drawing it there. Altogether that makes 19 combinations out of 40 so the probability is 19/40.

(b) From (ii) above there are 19 ways of this happening, and 9 of those are situations where it came from the first box, so I make that 9/19.

Pherekydes

PlayaFlow wrote:

Im in DCU 2yr and have to hand in my assignment on monday...
I NEED TO PASS THIS!!
If u are sure u can get them right ill PAY some decent money.

Ecksor is right.

What you're doing is dishonest and cheating.

You should fit right in with Ireland's business community.

ecksor

What the hell ...

I thought you were preparing for an exam! **** sakes, I'm sorry I answered now.

PlayaFlow

Dont get me wrong im not trying to cheat here. its like this- i passed my stats exam but the assesments i failed , as the other member of my group wasnt pulling his weight.
The lecturer showed me the assesment an said those 5 questions were ver poorly attempted , and if i can get them 5 right and submit them by monday he'll pass me.
i am ok with probability but just wasnt sure of which methods to use , and i dint want to risk using the wrong rule and getting a big fat zero.

and i dont want to end up having to repeat the year because of some twat not doing his work and me answering 5 questions wrongly

PlayaFlow

heres Q5: i Made an attempt at it but again im not sure if its right.

Q5: in an experiment , you have three attempts to throw 4 sixes. First all four dice are thrown and any sixes put aside. Any dice without a six (if any) are thrown a second time and once again any sixes are put aside. Finally , the remaining dice (if any) are thrown.
What is the probability of having a six on each of the 4 dice after the three throws have been completed?

I got this answer : P(having six on each 4 dice) = 671/1296 (51.77%)

any help greatly appreciated