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View Full Version : need some math help guys, Conditional Probability



pi is exactly 3
2005-May-02, 09:18 PM
I missed my data management class on friday and now If find out I have a quiz tomorrow on the stuff I missed. I got fridays sheet but these questions seem really dumb and kind of confusing. Some of them seem so simple they are confusing.
here are some of the ones that confused me that maybe you guys can help out with
1) two dice are rolled . What is the probablility that the sum is 8 and the sum is even?
This seems kind of redundant to me. I mean if the sum is 8 then the sum is obviously even, so why bother saying that the sum is even? very confusing
2) You play a game of rilling two dice. You win if you roll a five before a seven. you continue to roll until you win or lose. what is probability that you win.
ok so the probablility that you roll a five is 4/36 and of rolling a seven is 6/36 but rolling a 5 before a 7. what do I do with that.
3) Chocolate bars are advertised to weigh 120g. It is found that there is a 65% probablility that the bar weighs 120g and a 20% probablility that it weighs less than 120g. what is the probablility that a bar, not weighing 120g is greater than 120g
It seems obvious that the answer is 15 chance. I have a feeling that this is trying to trick me in some way. (I just wish I knew which way it was)
4) a car manufacturing plant has 3 shifts working on the assembly line. THe morning shift produces 38% of the total production, the afternoon 34% and the evening 28%. Of their output 3%,2% and 1%, respectively, do not pass quality control. If a vehicle is selected randomly and found defective what is the probablilty that is was made by the a)morning b) afternoon c)evening shift.
This one I really have no idea on. obviously I only need help with one section a,b or c. I could figure the rest out for myself from that.

Well thanks for the help guys. I really need to ace this test tomorrow or my mark is gonna bomb.

jfribrg
2005-May-02, 09:31 PM
Some thoughts:

1) Maybe this is actually two questions.

2) How many rolls do you expect to have to make before rolling a five? How many rolls do you expect to have to make before rolling a seven?
Use that info to find your answer

3) 35% of the bars weigh 120g. Of the 65% that don't, what percentage are heavier? its not 15%.

4) This is related to the notion of expected value. Check the index of your text and read up on it.


Does your teacher have office hours between now and the test?

SeanF
2005-May-02, 09:44 PM
Some thoughts:

1) Maybe this is actually two questions.
Definitely two questions.


3) 35% of the bars weigh 120g. Of the 65% that don't, what percentage are heavier? its not 15%.
Not the way I read it. 65% weigh 120g, 20% weigh less than 120g, and thus 15% weigh more than 120g.

The question is, of only the bars that don't weigh 120g (35% of the total), how many of them weigh more than 120g.

Edited to add: I could pretty easily calculate the answers to these questions, but I don't know that I'd do it the "right way" for a Data Management class, so I don't want to say what I'd actually do . . . :)

JohnW
2005-May-02, 09:51 PM
I'm not going to do all your homework for you, but I think you're so close with the first two...


1) two dice are rolled . What is the probablility that the sum is 8 and the sum is even?
This seems kind of redundant to me. I mean if the sum is 8 then the sum is obviously even, so why bother saying that the sum is even? very confusing
Break this into two.
a. what's the probability of the sum being 8?
b. what's the probability of 8 being even? (hint: quite high).
So then the overall probability would be...?

<edited to add: I'm taking the question at face value here, and am assuming the wording you've given is correct.>


2) You play a game of rilling two dice. You win if you roll a five before a seven. you continue to roll until you win or lose. what is probability that you win.
ok so the probablility that you roll a five is 4/36 and of rolling a seven is 6/36 but rolling a 5 before a 7. what do I do with that.
At the first roll you have three outcomes:
5 (win)
7 (lose)
another number (roll again)
Ignore the other numbers. You will keep going until you hit 5 or 7. So you can think of this as "how likely is a 5, relative to a 7, on any one roll?"

One more point:

I have a feeling that this is trying to trick me in some way.
Looking for the trick has caused more time to be wasted and more tests to be failed than anything else. Just read the question carefully and answer what you think it's asking. That is probably what it is asking.

Nicolas
2005-May-02, 09:52 PM
NOrmally, I don't do math "homework" but I'll try to explain some things (as far as I can remember them).

1). there are 6 = 36 possible combinations, from which the pairs 6,2; 5,3; 4,4; 3,5; 2,6 = 5 solutions are 8. The probability of getting 8 as combination hence is 5/36.

Considering CONDITIONAL probability: the possiblity of getting 8 as a sum, given the sum is even is 5/18 if I'm not mistaken (18 even solutions possible). The chance of getting an even solution given the sum is 8, is 1 however: out of all possibilities X that the sum is 8, all X give even sums => the chance of an even sum given the sum is 8 is X/X = 1.

The chance for an even sum is 18/36 (18 even combinations/36 possible combinations) if I'm correct on that 18.

pi is exactly 3
2005-May-02, 10:18 PM
ok for the 5,7 dice one. you would expect a 7 every 6 rolls and a 5 every 9 rolls so would that mean there is a 6 in 9 chance of rolling a 7 before a 5 and therefore a 3 in 9 or 1/3 chance of getting a 5 before 7 ?

01101001
2005-May-02, 11:11 PM
ok for the 5,7 dice one. you would expect a 7 every 6 rolls and a 5 every 9 rolls so would that mean there is a 6 in 9 chance of rolling a 7 before a 5 and therefore a 3 in 9 or 1/3 chance of getting a 5 before 7 ?

You roll and roll and roll, maybe once, maybe 100 times, until you get a 5 or a 7. Ignore all those initial rolls. They don't matter. How likely is it -- given that you rolled a 5 or a 7 -- that you rolled a 5?

pi is exactly 3
2005-May-02, 11:30 PM
oh! 2/5 it makes sense now thanks 01101001
anyone got any help on that car assembly line one?

peter eldergill
2005-May-03, 12:14 AM
This is know as the Bayes process.

Make a tree diagram with the first stage being the shift percentages. The second stage is the probabilities that the item is defective or not. Notice that these probabilities are different for each shift.

Now calculate the total probability of an item being defective. Add them up for each of the three shifts.

The question is really asking what ratio of the defective items come from the day shift, afternoon shift and night shift.

For day shift: P(defective and day)/P(defective)

Hope this helps. I taught it to my Data Management class last week :P

Later

Pete