I have literally done as many probability questions as I can but I STILL seem to get them wrong :/ I don't know what I'm doing wrong
How do you know whether to use a Venn diagram or the rules of probability. I know sometimes they go hand in hand but sometimes they just dont work out.
Example: Q7, January 2014 (IAL) paper.
I got everything right in this paper except for this probability question where I got no marks. I tried to use a Venn diagram :/ why was I wrong for doing that?
S1  Probability
Announcements  Posted on  

Last day to win £100 of Amazon vouchers  don't miss out! Take our quick survey to enter  24102016 

 Follow
 1
 11062016 12:25

 Follow
 2
 11062016 14:27
(Original post by Someboady)
I have literally done as many probability questions as I can but I STILL seem to get them wrong :/ I don't know what I'm doing wrong
How do you know whether to use a Venn diagram or the rules of probability. I know sometimes they go hand in hand but sometimes they just dont work out.
Example: Q7, January 2014 (IAL) paper.
I got everything right in this paper except for this probability question where I got no marks. I tried to use a Venn diagram :/ why was I wrong for doing that?
Is it possible that you're drawing or misinterpreting things for the Venn diagram? 
 Follow
 3
 11062016 14:31
(Original post by SeanFM)
Please post a link or picture of the question  saves people hassle
Is it possible that you're drawing or misinterpreting things for the Venn diagram?
Also I interpreted 1/5 of male students are left handed as P(M AND L) rather than P(LM) 
 Follow
 4
 11062016 14:34
(Original post by Someboady)
Yes I didn't draw the diagram correctly
Also I interpreted 1/5 of male students are left handed as P(M AND L) rather than P(LM) 
 Follow
 5
 11062016 14:37
(Original post by SeanFM)
Yes.. there you go P(M and L) = 1/5 would mean that 1 fifth of all students are male and left handed, whereas P(L given M) = 1/5 means that.. well, you know what it means : 
 Follow
 6
 11062016 14:39
(Original post by Someboady)
Ah yes thank you. Problem is I'd make that mistake in an exam _ ... I keep losing marks to Probability questions _ Sometimes I get them and sometimes I dont.... 
 Follow
 7
 11062016 14:42
(Original post by SeanFM)
It is difficult.. just underline the information given, write it down in terms of probability notation, and.. practice as many questions as you can to get the hang of it. It's definitely the most difficult part of S1.
Quick question:
Is P(A n B) ' (i.e. not A and B) the same as P (A' n B' (i.e not A and Not B)
If not, what is it the same as?
EDIT: And how can you figure it out in an exam (i.e. derive it)Last edited by Someboady; 11062016 at 14:44. 
 Follow
 8
 11062016 14:54
(Original post by Someboady)
Thanks! I'll try my best
Quick question:
Is P(A n B) ' (i.e. not A and B) the same as P (A' n B' (i.e not A and Not B)
If not, what is it the same as?
EDIT: And how can you figure it out in an exam (i.e. derive it)
P(A u B)' = P(a' n b') 
 Follow
 9
 11062016 14:56

 Follow
 10
 11062016 14:57
(Original post by Someboady)
Is this derivable in any way? or just something you can figure out via common sense and a venn diagram lol. thats how I figured it out but idk if i can do it in an exam
P(aub) is the entirety of what's in both a and b, right? So p(aub)' is everything not in a and not in b, so the outside bit
P(a' n b') is the intersection between not in a, and not in b, which again is just the outside bit 
 Follow
 11
 11062016 14:58
(Original post by Someboady)
Thanks! I'll try my best
Quick question:
Is P(A n B) ' (i.e. not A and B) the same as P (A' n B' (i.e not A and Not B)
If not, what is it the same as?
EDIT: And how can you figure it out in an exam (i.e. derive it)
I would say is everything but the intersection of A and B, so but that's all a bit messy, and you'll find that you can express it as (draw it and see for yourself).
There's something called De Morgan's law which is very easy to memorise  if you take the compliment of something, flip the bit in the middle (the U or the N) and add a complement to each event inside.

 Follow
 12
 11062016 16:11
(Original post by SeanFM)
Venn diagram to derive it. First shade the bit without the ' and then everything else is your ...'.
I would say is everything but the intersection of A and B, so but that's all a bit messy, and you'll find that you can express it as (draw it and see for yourself).
There's something called De Morgan's law which is very easy to memorise  if you take the compliment of something, flip the bit in the middle (the U or the N) and add a complement to each event inside.
I had another question
Why is P(BC) = 5/11 and not 0.2 as shown in the diagram o.0 
 Follow
 13
 11062016 16:13
(Original post by Someboady)
This was very useful thank you!
I had another question
Why is P(BC) = 5/11 and not 0.2 as shown in the diagram o.0 
 Follow
 14
 11062016 16:17
(Original post by SeanFM)
Because 0.2 = P(B intersect C). 
 Follow
 15
 11062016 16:25
(Original post by Someboady)
Yes but by observation.. if its B given C should it be the same as P(B n C) ? How is P(BC) different to P( B n C) in this context? I understand that if B and C were independent P(BC) = P(B n C) and in this case this is not true. BUT, by observation I would have thought P(BC) was 0.2. Could you explain why its not?
It should not be the same, for the very reason you stated. Otherwise there would be no point or use in conditional probability! The definition of P(BC) = P(B N C) / P(C), which is like saying 'what percentage of C does B contain'?) 
 Follow
 16
 12062016 11:12
(Original post by SeanFM)
P(B given C) means, given that C has happened, what is the probbability of B having happened?
It should not be the same, for the very reason you stated. Otherwise there would be no point or use in conditional probability! The definition of P(BC) = P(B N C) / P(C), which is like saying 'what percentage of C does B contain'?)
Also I had some trouble with these questions
Part f
In part f I don't understand how you know that P(X<5) is 0.8. I got the answer correct but by "cheating" and using the show that part. But what I found is that the Probability of S<5 = 0.8 so is this why X<5 is 0.8? How does that work o.0
Reference to question = June 2013 (R), Q7
EDIT: Sorry I figured it out!. Because the first probability is S1 = 1, therefore S2 has to be < 4 for X to be <5.
In part c, I don't understand why the answer is multiplied by 3.
Reference to paper: June 2008 Q6
Sorry to trouble you.. I'm kind of panicking because it seems I suck at Stats lol and the exam is in 2 days :/
Attachment 548359548361Last edited by Someboady; 12062016 at 11:17. 
 Follow
 17
 12062016 11:17
(Original post by Someboady)
Ah thank you, so as a general rule I shouldn't rely on observation for Given that probabilities?
Also I had some trouble with these questions
Part f
In part f I don't understand how you know that P(X<5) is 0.8. I got the answer correct but by "cheating" and using the show that part. But what I found is that the Probability of S<5 = 0.8 so is this why X<5 is 0.8? How does that work o.0
Reference to question = June 2013 (R), Q7
In part c, I don't understand why the answer is multiplied by 3.
Reference to paper: June 2008 Q6
Sorry to trouble you.. I'm kind of panicking because it seems I suck at Stats lol and the exam is in 2 days :/
Attachment 548359548361
Multiplied by 3? At a glance, it would be p^2(1p) where p is the P(X>53).Post rating:1 
 Follow
 18
 12062016 11:19
(Original post by SeanFM)
Yes, that is why. If you can't see it, think about the definition of X, and what values S2 has to take for X < 5.
Multiplied by 3? At a glance, it would be p^2(1p) where p is the P(X>53).
The mark scheme for the second one shows:
P(2 weigh more than 53kg and 1 less) = 3 x 0.0668^2 x (1  0.0668) 
 Follow
 19
 12062016 11:33
(Original post by Someboady)
Yes thank you!
The mark scheme for the second one shows:
P(2 weigh more than 53kg and 1 less) = 3 x 0.0668^2 x (1  0.0668)
Posted from TSR MobilePost rating:1 
 Follow
 20
 12062016 11:35
(Original post by iMacJack)
Because there are three ways of choosing it, could have more more less, more less more, less more more
Posted from TSR Mobile
Write a reply…
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:
Sign in
Not got an account? Sign up now
Updated: June 12, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.