I'm always getting stuck on these questions, the first part is fine. It's applying the information in the second part, are there any ways to really get this?
Probablity
Announcements  Posted on  

Take our survey to be in with the chance of winning a £50 Amazon voucher or one of 5 x £10 Amazon vouchers  28052016 


I would draw a venn

(Original post by LifeIsGood)
I'm always getting stuck on these questions, the first part is fine. It's applying the information in the second part, are there any ways to really get this?
Next on your agenda though, should be understanding what the following questions are really asking in 'wordy' terms.
(b) = What is the probability of event B not happening (because of the '), given that event A has happened (because of the  prior to A).
(c) = What is the probability of event A not happening, or event B happening or event A not happening, and event B not happening.
First look at B... Now you know what the question is really asking, you now need to know your formulae:
The probability of A given B.. ( P(AB) ) = P(A∩B)/(B)
So for (b)... applying this formula, the probability is (B'∩A)/(A)
If you cannot answer the question from here..
I suggest reading the notes on probability in post #4, here:
http://www.thestudentroom.co.uk/show....php?t=1619124
What you will see is that to go any further, as TenOfThem suggested, you must familiarise yourself with Venn Diagrams. Once you have a Venn Diagram drawn, you can begin to use the formula I have just shown you to put numbers in and reach your answer.
The same is the case for (d), where you just need to know that the formula for P(A∪B) = P(A) + P(B) P(A∩B).. So you can manipulate your Venn once again to find P(A∩B) and you're away.
Finally, (d), events are independent if P(A∩B) = P(A)×P(B)... Use your Venn and the information given.. So.. Learn formulae, and how to use Venn Diagrams. 
(Original post by dslc)
You clearly understand that the first part is essentially saying what is the probability of event A and event B occurring, so this is a great start.
Next on your agenda though, should be understanding what the following questions are really asking in 'wordy' terms.
(b) = What is the probability of event B not happening (because of the '), given that event A has happened (because of the  prior to A).
(c) = What is the probability of event A not happening, or event B happening or event A not happening, and event B not happening.
First look at B... Now you know what the question is really asking, you now need to know your formulae:
The probability of A given B.. ( P(AB) ) = P(A∩B)/(B)
So for (b)... applying this formula, the probability is (B'∩A)/(A)
If you cannot answer the question from here..
I suggest reading the notes on probability in post #4, here:
http://www.thestudentroom.co.uk/show....php?t=1619124
What you will see is that to go any further, as TenOfThem suggested, you must familiarise yourself with Venn Diagrams
My Venn Diagram:

(Original post by LifeIsGood)
I get up to this point and that's where I get stuck. I don't know how to do (B'∩A)/(A). It's been really confusing me as I don't know what it's really asking for
You can use a Venn diagram to find (B'∩A)/(A) directly.
Take a look at the notes I linked to, and its the left hand part of the circle.
∴ (B'∩A)/(A) = (3/16)÷(3/16+1/8) 
(Original post by dslc)
Well, I suggest it would be beneficial for you to take some time to learn how to draw out Venn diagrams from information you are given in questions.
You can use a Venn diagram to find (B'∩A)/(A) directly.
Take a look at the notes I linked to, and its the left hand part of the circle. 
[
(Original post by LifeIsGood)
I'm always getting stuck on these questions, the first part is fine. It's applying the information in the second part, are there any ways to really get this?
Is this A Level S1 or GCSE? I ask because I've done S1 and never done such questions. 
(Original post by LifeIsGood)
Thanks! That's helped so much The marking scheme said to do it another way I'm not familiar with. I'm going to try the others
But you should find that "my way"/every other mathematicians way works just fine. Please post back if not, as I'm sitting S1 too!
I've attached the notes on probability in case you had trouble finding them too! 
(Original post by Azland)
[
Is this A Level S1 or GCSE? I ask because I've done S1 and never done such questions. 
(Original post by dslc)
Looks like an S1 paper to me. 
(Original post by dslc)
I have to say, if you're referring to Edexcel mark schemes, they make it far more complicated than these type of questions need be with a good Venn  and I'm not good enough at math to understand why...
But you should find that "my way"/every other mathematicians way works just fine. Please post back if not, as I'm sitting S1 too!
I've attached the notes on probability in case you had trouble finding them too!
It baffles me because I don't know what formula to apply where. As in the first example, it's easy to figure out as all the information for the formula is there but for this one it's not. Can I try to use Venn Diagrams to solve all these problems?
And thanks so much, you managed to explain something in 5 minutes which taken ages to figure out 
(Original post by Azland)
Oh, I guess my board covers different topics then.. never mind :/ 
(Original post by LifeIsGood)
I got it correct! The Venn Diagrams really do help but when I go to questions like this:
It baffles me because I don't know what formula to apply where. As in the first example, it's easy to figure out as all the information for the formula is there but for this one it's not. Can I try to use Venn Diagrams to solve all these problems?
And thanks so much, you managed to explain something in 5 minutes which taken ages to figure out
P(A∪B) = P(A) + P(B)  P(A∩B)
P(A∪B) = P(A) + P(B)  P(A∪B) (just the first equation rearranged)
P(AB) = P(A∩B)÷P(B)
Mutually Exclusive if
P(A∩B) = 0
P(A∪B) = P(A) + P(B)
Independent if
P(A∩B) = P(A) × P(B)
So long as you know these formulae, not only should you be able to construct Venns in most situations, but you shouldn't really face a probability question that will cause you too much trouble (unless you hate tree diagrams) 
(Original post by dslc)
You need to know the following:
P(A∪B) = P(A) + P(B)  P(A∩B)
P(A∪B) = P(A) + P(B)  P(A∪B) (just the first equation rearranged)
P(AB) = P(A∩B)÷P(B)
Mutually Exclusive if
P(A∩B) = 0
P(A∪B) = P(A) + P(B)
Independent if
P(A∩B) = P(A) × P(B)
So long as you know these formulae, not only should you be able to construct Venns in most situations, but you shouldn't really face a probability question that will cause you too much trouble (unless you hate tree diagrams) 
(Original post by LifeIsGood)
I still don't get how you do the first bit without P(B)?
(a) this is a really easy question.
You are told that (in wordy form) the probability of event A occurring or event B occurring or event A and event B occurring is 0.6. Therefore if the probability of everything occurring is always 1, then the probability of event A and B not occurring  what (a) is asking for  is 10.6 = 0.4 (this is the space outside the Venn diagram)
(b) this is asking for event B occurring but not event A occurring. So on a Venn diagram, event B occurring would refer to the middle section and the B section.. But, given that they don't want event A occurring, we have to ignore the middle part.. Following this logic it must be P(A∪B).. event A or B or both.. P(A) = 0.60.2=0.4
(c) we know P(A∩B) = P(A) + P(B)  P(A∪B) therefore.. P(A∩B) the probability of A and B = 0.2 + P(B)  0.6...
we also know that P(A∩B) = 0.2 × P(B) or 0.2P(B)
So... Putting the second part into the first: 0.2P(B) = P(B)  0.4
Therefore 0.4 = 0.8P(B).. therefore P(B) = 0.4÷0.8
Because we do not know P(B), we effectively have to use simultaneous equations. 
I'm stuck on the last part of this one:
a) I applied the formula:
P(AB) = P(A∩B)/P(B)
Rearranged it and got 3/8 which is correct.
b) I drew the Venn Diagram out:
I applied the addition formula: where P(A) + P(B)  P(A∩B)
And I got that correct which is 17/24
c) I'm stuck on this part because I've attempted the question and got it wrong:
Here's what I did:
P(B  A′ ) = P(B∩A')/P(A')
The part I got wrong is the P(B∩A') because I assumed on the Venn Diagram it'll be just the B side, I don't understand why that is incorrect if you've given me the labels of the Venn Diagram 
(Original post by LifeIsGood)
I'm stuck on the last part of this one:
a) I applied the formula:
P(AB) = P(A∩B)/P(B)
Rearranged it and got 3/8 which is correct.
b) I drew the Venn Diagram out:
I applied the addition formula: where P(A) + P(B)  P(A∩B)
And I got that correct which is 17/24
c) I'm stuck on this part because I've attempted the question and got it wrong:
Here's what I did:
P(B  A′ ) = P(B∩A')/P(A')
The part I got wrong is the P(B∩A') because I assumed on the Venn Diagram it'll be just the B side, I don't understand why that is incorrect if you've given me the labels of the Venn Diagram
Your Bside should be (3/8)(1/4) 
I've done the first and 3rd part correctly. There's one thing I'm stuck on:
For part b:
P(A′ ∪ B)  I looked at the Venn Diagram and took the value as B as it says in the notes thing above in the thread. It's wrong but I don't understand how you know when to use the values they've given you and the ones in the venn diagram 
(Original post by LifeIsGood)
I've done the first and 3rd part correctly. There's one thing I'm stuck on:
For part b:
P(A′ ∪ B)  I looked at the Venn Diagram and took the value as B as it says in the notes thing above in the thread. It's wrong but I don't understand how you know when to use the values they've given you and the ones in the venn diagram
It's asking, "what is the probability of event A not occurring or event B occurring or both event A not occurring and event B occurring.
So, start with the first bit:
Probability of event A not occurring = P(B∩A') + P(B'∩A')
You are ignoring P(B'∩A')
This equals 0.26+0.24... Where the 0.24 is 1(everything inside the Venn circles, in this case: 0.26+0.34+0.16)
The next bit,
Probability of B occurring = 0.26 +0.16 but we have already considered the 0.26, P(B∩A'), in the first bit, so we leave this out.. We do not want to double count probabilities.
Now the third bit,
Probability of A not occurring and B occurring = 0.26.. But, again this has already been considered.
So, we look at all the components we have now and that comes to:
0.26+0.24+0.16=0.66 
(Original post by dslc)
What is it asking for in the question?
It's asking, "what is the probability of event A not occurring or event B occurring or both event A not occurring and event B occurring.
So, start with the first bit:
Probability of event A not occurring = P(B∩A') + P(B'∩A')
You are ignoring P(B'∩A')
This equals 0.26+0.24... Where the 0.24 is 1(everything inside the Venn circles, in this case: 0.26+0.34+0.16)
The next bit,
Probability of B occurring = 0.26 +0.16 but we have already considered the 0.26, P(B∩A'), in the first bit, so we leave this out.. We do not want to double count probabilities.
Now the third bit,
Probability of A not occurring and B occurring = 0.26.. But, again this has already been considered.
So, we look at all the components we have now and that comes to:
0.26+0.24+0.16=0.66
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: