S1 - Probability

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?

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Reply 1

Kevin De Bruyne

Original post by Someboady

Please post a link or picture of the question - saves people hassle :tongue:

Is it possible that you're drawing or misinterpreting things for the Venn diagram?

Reply 2

Someboady

Original post by SeanFM

Please post a link or picture of the question - saves people hassle :tongue:

Is it possible that you're drawing or misinterpreting things for the Venn diagram?

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(L|M)

Reply 3

Kevin De Bruyne

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(L|M)

Yes.. there you go :tongue:

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 :

Reply 4

Someboady

Original post by SeanFM

Yes.. there you go :tongue:

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 :

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....

Reply 5

Kevin De Bruyne

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....

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.

Reply 6

Someboady

Original post by SeanFM

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' :wink:

(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)

(edited 7 years ago)

Reply 7

iMacJack

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' :wink:

(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)

No

P(A u B)' = P(a' n b')

Reply 8

Someboady

Original post by iMacJack

No

P(A u B)' = P(a' n b' :wink:

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

Reply 9

iMacJack

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

It is common sense

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

Reply 10

Kevin De Bruyne

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' :wink:

(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)

Venn diagram to derive it. First shade the bit without the ' and then everything else is your ...'.

I would say P(A \cap B)^c is everything but the intersection of A and B, so P(A \cap B^c) + P(B \cap A^c) + P(A^c \cap B^c) but that's all a bit messy, and you'll find that you can express it as

A^c \cup B^c p

(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.

Reply 11

Someboady

Original post by SeanFM

Venn diagram to derive it. First shade the bit without the ' and then everything else is your ...'.

I would say P(A \cap B)^c

is everything but the intersection of A and B, so P(A \cap B^c) + P(B \cap A^c) + P(A^c \cap B^c) but that's all a bit messy, and you'll find that you can express it as

A^c \cup B^c p

This was very useful thank you!

I had another question

Why is P(B|C) = 5/11 and not 0.2 as shown in the diagram o.0

Reply 12

Kevin De Bruyne

Original post by Someboady

This was very useful thank you!

I had another question

Why is P(B|C) = 5/11 and not 0.2 as shown in the diagram o.0

Because 0.2 = P(B intersect C).

Reply 13

Someboady

Original post by SeanFM

Because 0.2 = P(B intersect C).

Yes but by observation.. if its B given C should it be the same as P(B n C) ? How is P(B|C) different to P( B n C) in this context? I understand that if B and C were independent P(B|C) = P(B n C) and in this case this is not true. BUT, by observation I would have thought P(B|C) was 0.2. Could you explain why its not?

Reply 14

Kevin De Bruyne

Original post by Someboady

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(B|C) = P(B N C) / P(C), which is like saying 'what percentage of C does B contain'?)

Reply 15

Someboady

Original post by SeanFM

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

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 :/

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(edited 7 years ago)

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Reply 16

Kevin De Bruyne

Original post by Someboady

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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(1-p) where p is the P(X>53).

Reply 17

Someboady

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(1-p) where p is the P(X>53).