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1. Re: Maths help needed for questions below.
(Original post by aznkid66)
Yup, just draw a smooth curve. It doesn't have to be exact, as it's only used to help estimate the intersection between R and C.
oh! That's where I went wrong; you know I did IGCSE business studies and there we learnt that in a break even chart, total revenue and total costs line are straight lines so since this is a break even sum I never imagined R to be a curve, so it's okay if R is a curve right?
2. Re: Maths help needed for questions below.
Yup, it wouldn't be weird if TR or TC were curves instead of straight lines. And who cares about economics, this is maths! :P

Of course, maybe it would be better if someone else confirmed this...
3. Re: Maths homework help needed quickly!
(Original post by notnek)
This is one of those things in maths that needs to be thought about by the student for it to click and an explanation may or may not help. Since it hasn't clicked so far, I think you should try the Venn diagram approach which should hopefully make it clearer:

Attachment 157486

Fill in the blanks first and then the number of people who "is male or has a business qualification" is the sum of the numbers in the male circle and the qualified circle. The bottom number is 0 since every qualified person is either male or female (there are no e.g. children).

If this makes you more confused then I apologise in advance
Hey man,

I've been trying to use probability venn diagrams in probability question but I've got stuck in this question. Q8 (a) An inspection carried out on 300 electronic items manufactured by two companies
(Company A and Company B) found that 30 were faulty. Of the 150 electronic
items inspected that were manufactured by Company A, 10 were found to be faulty.
In contrast, 20 were found to be faulty out of the 150 inspected electronic items
manufactured by Company B.
If one electronic item is randomly selected from these 300 electronic items, fi nd the
probability that this electronic item is:
(i) Manufactured by Company B; (3 marks)
(ii) Faulty; (3 marks)
(iii) Not faulty and is manufactured by Company B; (4 marks)
(iv) Manufactured by Company A, given that it is faulty. (4 marks)

So (i) and (ii) are very easy and I'm okay there but (iii) and (iv) is giving me the problem. I like solving probability questions using venn diagrams, and I've attached my venn diagram, see below. This venn diagram is my answer to (iii) and (iv). From my venn diagram you can see that (P) of faulty and manufactured by company B (answer to (iii)) is 20/300, so (P) of not faulty and manufactured by company B will be 1-(20/300) which is 14/15. However, this answer seems to be wrong and the mark schemes shows (P) of this question as 13/30. Isn't my method correct. What's the problem. And then for (iv) (P) of manufactured by A and faulty, from my venn diagram is 10/300 which is 1/30 but in the mark scheme the (P) of this question is (1/3). I think for (iv) the mark scheme is wrong. What do you think? Please help me. Thank you for your time and co-operation.
Attached Thumbnails

4. Re: Maths help needed for questions below.
iii. You have said that P(Not faulty and manufactured by company B) = 1-P(faulty and manufactured by company B)

But this is not right. It is true that for two events A and B, P(A)=1-P(not A) but in general, P(A and B) 1-P(not A and B). This can be seen from this Venn Diagram:

Notice that the two regions do not add up to the total diagram.

Instead you can read off the number of elements in {Not faulty and is manufactured by Company B} directly from your Venn diagram.

Spoiler:
Show
130

And then divide this number by the total to find the probability.

iv. "Given that is faulty" - this means that we are only focusing on faulty items and ignoring the items that are not faulty so the total goes from 300 to 30 (your answer to ii). So the value on your denominator should be 30 instead of 300.

Make sure you understand everything that I've posted and ask me if you're not sure.
5. Re: Maths help needed for questions below.
(Original post by notnek)
iii. You have said that P(Not faulty and manufactured by company B) = 1-P(faulty and manufactured by company B)

But this is not right. It is true that for two events A and B, P(A)=1-P(not A) but in general, P(A and B) 1-P(not A and B). This can be seen from this Venn Diagram:

Notice that the two regions do not add up to the total diagram.

Instead you can read off the number of elements in {Not faulty and is manufactured by Company B} directly from your Venn diagram.

Spoiler:
Show
130

And then divide this number by the total to find the probability.

iv. "Given that is faulty" - this means that we are only focusing on faulty items and ignoring the items that are not faulty so the total goes from 300 to 30 (your answer to ii). So the value on your denominator should be 30 instead of 300.

Make sure you understand everything that I've posted and ask me if you're not sure.
Okay, so for (iii) does that mean that we can not use the 1-P(not A and B) rule to find a certain probability if it has the words and or or in them? We can only use the 1 minus rule for simple single probabilities? I still don't get (iv) why does the denominator go down to 30? I thought that the product is taken at random from 300 products, so the probability of faulty and manufactured by A should be 10/300 because the product is chosen from 300 products and not 30 products. This doubt is confusing me. Please elaborate on this.
Last edited by gabriel 41; 23-07-2012 at 13:01.
6. Re: Maths help needed for questions below.
(Original post by gabriel 41)
Okay, so for (iii) does that mean that we can not use the 1-P(not A and B) rule to find a certain probability if it has the words and or or in them? We can only use the 1 minus rule for simple single probabilities? I still don't get (iv) why does the denominator go down to 30? I thought that the product is taken at random from 300 products, so the probability of faulty and manufactured by A should be 10/300 because the product is chosen from 300 products and not 30 products. This doubt is confusing me. Please elaborate on this.
Yes, the 1-P(A)=P(not A) should only be used for a single event. Do not use it when "and", "or" are involved.

I'll try to explain iv. with a different example.

A couple are about to have a baby. The probability of the sex of the baby is 50-50. If it's a girl, theres a 50-50 chance it will be called either Jane or Alice. If it's a boy, theres a 50-50 chance it will be called Chris or Peter. What is the probability that the baby will be called Peter?

Since all of the outcomes are equally likely, we can use simple y7 probability theory to find that P(Peter) is the number of outcomes of the event (1) divided by the total amount of outcomes (4) so it's 1/4.

The baby is born and it is a boy. What is the probability it will be called Peter?

Since we now know it's a boy, we can disregard the names Jane and Alice so we're left with 2 total outcomes (Peter and Chris). So P(Peter if it's a boy) is 1/2. Notice that the denominator has changed from 4 to 2.

"The baby is born and it is a boy" is equivalent to saying" Given that the baby is a boy...". Given that X has happened implies that we are now assuming X has happened and anything that's outside X is not relevant to the question.

So in your question, given that it is faulty means that we're now only interested in the 30 faulty items so the total number of outcomes (the thing that goes on the denominator) is now 30.
Last edited by notnek; 23-07-2012 at 13:56.
7. Re: Maths help needed for questions below.
(Original post by notnek)
Yes, the 1-P(A)=P(not A) should only be used for a single event. Do not use it when "and", "or" are involved.

I'll try to explain iv. with a different example.

Since all of the outcomes are equally likely, we can use simple y7 probability theory to find that P(Peter) is the number of outcomes of the event (1) divided by the total amount of outcomes (4) so it's 1/4.

Since we now know it's a boy, we can disregard the names Jane and Alice so we're left with 2 total outcomes (Peter and Chris). So P(Peter if it's a boy) is 1/2. Notice that the denominator has changed from 4 to 2.

"The baby is born and it is a boy" is equivalent to saying" Given that the baby is a boy...". Given that X has happened implies that we are now assuming X has happened and anything that's outside X is not relevant to the question.

So in your question, given that it is faulty means that we're now only interested in the 30 faulty items so the total number of outcomes (the thing that goes on the denominator) is now 30.
Okay so it was the wording that was causing the confusion. I now understand that whenever they say "given x" we just take the total of outcomes for that X. But I would like to ask for (iv) is it the same as P(manufactured by A and faulty) so products manufactured by A are 150/300 and faulty are 30/300 so can we do (150/300) * (30/300) but then with this way the answer doesn't come. Why?
8. Re: Maths help needed for questions below.
(Original post by gabriel 41)
Okay so it was the wording that was causing the confusion. I now understand that whenever they say "given x" we just take the total of outcomes for that X. But I would like to ask for (iv) is it the same as P(manufactured by A and faulty) so products manufactured by A are 150/300 and faulty are 30/300 so can we do (150/300) * (30/300) but then with this way the answer doesn't come. Why?
P(A given B) is not the same thing as P(A and B).

P(A and B) is the probability of A happening and then B happening. P(A given B) is the probability of A happening if we assume that B has happened and the outcomes outside B have been discounted.

Another way to think of about it: By doing the calculation (150/300)*(30/300), you are including in your working items that are not faulty (since the total: 300 includes faulty and not faulty items). Your denominators should not be 300 since we are assuming in iv that the items are all faulty.

If a question ever asks for P(A given B), just pretend in your mind that the outcomes that are not in B don't exist. That should help you.
9. Re: Maths help needed for questions below.
(Original post by notnek)
P(A given B) is not the same thing as P(A and B).

P(A and B) is the probability of A happening and then B happening. P(A given B) is the probability of A happening if we assume that B has happened and the outcomes outside B have been discounted.

Another way to think of about it: By doing the calculation (150/300)*(30/300), you are including in your working items that are not faulty (since the total: 300 includes faulty and not faulty items). Your denominators should not be 300 since we are assuming in iv that the items are all faulty.

If a question ever asks for P(A given B), just pretend in your mind that the outcomes that are not in B don't exist. That should help you.
Oh yes! I now get it. We can't multiply the probabilities together because our denominator must be 30 and not 300. Thank you! I really appreciate your help. You've been patient with me and given me so much help over this last few months. But I wanted to ask one more thing: is there a way we can answer these "given X" questions through the venn diagram technique you taught me before, or are these "given X'' questions straightforward and there's no need to draw a venn diagram?
10. Re: Maths help needed for questions below.
(Original post by notnek)
Yes, the 1-P(A)=P(not A) should only be used for a single event. Do not use it when "and", "or" are involved.

So this is wrong?
11. Re: Maths help needed for questions below.
(Original post by smalltalk)

So this is wrong?
No that is correct. My post wasn't very clear but I meant that you can't use the 1-P(A)=P(A') in the same way for union and intersection (like the OP was doing) e.g.

There are a few probability identities that I didn't want to mention so as not to add to the confusion.
12. Re: Maths help needed for questions below.
(Original post by gabriel 41)
Oh yes! I now get it. We can't multiply the probabilities together because our denominator must be 30 and not 300. Thank you! I really appreciate your help. You've been patient with me and given me so much help over this last few months. But I wanted to ask one more thing: is there a way we can answer these "given X" questions through the venn diagram technique you taught me before, or are these "given X'' questions straightforward and there's no need to draw a venn diagram?
To use a Venn Diagram for "A given B", only consider the circle B and ignore everything else.

e.g. Let the Venn Diagram below show the number of equally likely outcomes of two events A and B:

Then for P(A given B) we ignore everything outside B and so it is equal to 12 (the number of outcomes in A ignoring everything outside B) divided by 25 (the number of total outcomes in B). So P(A given B) = 12/25

There is actually a formula that you can use which tells you the same thing:

Here, P(A|B) means P(A given B).
Last edited by notnek; 24-07-2012 at 04:38.
13. Re: Maths help needed for questions below.
(Original post by notnek)
To use a Venn Diagram for "A given B", only consider the circle B and ignore everything else.

e.g. Let the Venn Diagram below show the number of equally likely outcomes of two events A and B:

Then for P(A given B) we ignore everything outside B and so it is equal to 12 (the number of outcomes in A ignoring everything outside B) divided by 25 (the number of total outcomes in B). So P(A given B) = 12/25

There is actually a formula that you can use which tells you the same thing:

Here, P(A|B) means P(A given B).
Okay so there's a formula like the one you gave on my formula sheet which I don't quite understand. Please find it attached as "probs formula 2005" and please fully explain this formula to me.

So there's another small question I have. Please find the question attached as "probs question 7 2005". In this question (i), (ii) (iii) and (v) are very clear to me; however, (iv) is giving me a problem. The question says" (iv) is manufactured by company A or is defective so this is a or question, so I decided to add up the two probabilities together. I.e. P(A) = 150/300 AND P(DEFECTIVE) = 30/300. When I add the two, I get an answer of 0.6. But, this answer appears to be false and the correct answer is 0.57. Why is this so? Could you please explain this to me. Thank you very much for your help
Attached Thumbnails

Last edited by gabriel 41; 24-07-2012 at 13:57. Reason: gave only half the explanation
14. Re: Maths help needed for questions below.
(Original post by gabriel 41)
So there's another small question I have. Please find the question attached as "probs question 7 2005". In this question (i), (ii) (iii) and (v) are very clear to me; however, (iv) is giving me a problem. The question says" (iv) is manufactured by company A or is defective so this is a or question, so I decided to add up the two probabilities together. I.e. P(A) = 150/300 AND P(DEFECTIVE) = 30/300. When I add the two, I get an answer of 0.6. But, this answer appears to be false and the correct answer is 0.57. Why is this so? Could you please explain this to me. Thank you very much for your help
15. Re: Maths help needed for questions below.
(Original post by gabriel 41)
Okay so there's a formula like the one you gave on my formula sheet which I don't quite understand. Please find it attached as "probs formula 2005" and please fully explain this formula to me.
I tried to explain it in my last post by showing you the Venn diagram. The probability was 12 (A and B) divided by 12+13 (B) so the formula is:

P(A given B) = P(A and B) / P(B)

So there's another small question I have. Please find the question attached as "probs question 7 2005". In this question (i), (ii) (iii) and (v) are very clear to me; however, (iv) is giving me a problem. The question says" (iv) is manufactured by company A or is defective so this is a or question, so I decided to add up the two probabilities together. I.e. P(A) = 150/300 AND P(DEFECTIVE) = 30/300. When I add the two, I get an answer of 0.6. But, this answer appears to be false and the correct answer is 0.57. Why is this so? Could you please explain this to me. Thank you very much for your help
Smalltalk showed you the formula which is true for all events A and B. In you working you used this:

P(A or B) = P(A) + P(B)

This formula is only true when P(A and B) is 0 i.e. when A and B are mutually exclusive.

In your question, there are 10 events that are in company A and defective so the events are not mutually exclusive. In your working, you have counted these 10 events twice: in 150/300 and in 30/300 so you have to subtract 10/100 from your answer.
16. Re: Maths help needed for questions below.
(Original post by notnek)
I tried to explain it in my last post by showing you the Venn diagram. The probability was 12 (A and B) divided by 12+13 (B) so the formula is:

P(A given B) = P(A and B) / P(B)

Smalltalk showed you the formula which is true for all events A and B. In you working you used this:

P(A or B) = P(A) + P(B)

This formula is only true when P(A and B) is 0 i.e. when A and B are mutually exclusive.

In your question, there are 10 events that are in company A and defective so the events are not mutually exclusive. In your working, you have counted these 10 events twice: in 150/300 and in 30/300 so you have to subtract 10/100 from your answer.
I'm sorry but I didn't understand anything you said. I know mutually exclusive events are where the probability of one event occurring precludes the probability of another event occurring and in non mutually exclusive two events can occur together but why isn't this scenario mutually exclusive because the occurrence of a defective battery precludes the occurrence of a good battery so why is the rule wrong? Can you please explain.
17. Re: Maths help needed for questions below.
(Original post by gabriel 41)
I'm sorry but I didn't understand anything you said. I know mutually exclusive events are where the probability of one event occurring precludes the probability of another event occurring and in non mutually exclusive two events can occur together but why isn't this scenario mutually exclusive because the occurrence of a defective battery precludes the occurrence of a good battery so why is the rule wrong? Can you please explain.
Yes, producing a defective battery and a good battery are mutually exclusive since you can't have a battery that is both good and defective.

But your question is asking for P(In company A or defective) so the two events that you should be considering are "In company A" and "Defective". There are 10 batteries that are in company A and defective so these events cannot be mutually exclusive.

In probability terms: two events A and B are mutually exclusive if P(A and B)=0. And for all non-mutually exclusive events you can use this formula for P(A or B):

P(A or B) = P(A) + P(B) - P(A and B)

I find that a lot of this makes more sense by thinking about the Venn Diagram but it's quite hard to show you on an internet forum. I think a one-to-one explanation is needed for that.
18. (Original post by notnek)
Yes, producing a defective battery and a good battery are mutually exclusive since you can't have a battery that is both good and defective.

But your question is asking for P(In company A or defective) so the two events that you should be considering are "In company A" and "Defective". There are 10 batteries that are in company A and defective so these events cannot be mutually exclusive.

In probability terms: two events A and B are mutually exclusive if P(A and B)=0. And for all non-mutually exclusive events you can use this formula for P(A or B):

P(A or B) = P(A) + P(B) - P(A and B)

I find that a lot of this makes more sense by thinking about the Venn Diagram but it's quite hard to show you on an internet forum. I think a one-to-one explanation is needed for that.
OK so then I can give you my Skype/ google talk address and if you want we can meet up on these mediums. But this one-to-one explanation will be done at your timing arrangements, according to your wishes and convenience since you'll be helping me for free, so I'll only ask for help when you are comfortable to give it to me. Thanks a lot .

This was posted from The Student Room's iPhone/iPad App
19. Re: Maths help needed for questions below.
(Original post by gabriel 41)
OK so then I can give you my Skype/ google talk address and if you want we can meet up on these mediums. But this one-to-one explanation will be done at your timing arrangements, according to your wishes and convenience since you'll be helping me for free, so I'll only ask for help when you are comfortable to give it to me. Thanks a lot .

This was posted from The Student Room's iPhone/iPad App
I meant that you could get a teacher/lecturer to help you? I wasn't suggesting that I would give you one-to-one help.

Since I have my own teaching/tutoring commitments, I like to keep the help I give on TSR as casual forum-based volunteering and prefer not to give individual help to members outside TSR. Sorry

Please let me know what you don't understand in my last post(s) and I'll try to explain again.
20. Re: Maths help needed for questions below.
(Original post by notnek)
I meant that you could get a teacher/lecturer to help you? I wasn't suggesting that I would give you one-to-one help.

Since I have my own teaching/tutoring commitments, I like to keep the help I give on TSR as casual forum-based volunteering and prefer not to give individual help to members outside TSR. Sorry

Please let me know what you don't understand in my last post(s) and I'll try to explain again.
OK so I misunderstood you. No I'm okay I understood very nicely what you said in your last posts.