Tree Diagram

How do you answer 5iii A) using a tree diagram?

I have tired to do 1- (0.6 *0.8)=0.52 but the answer is 0.48, which is 0.6 *0.8. This doesn't make sense
Can someone tell me how to use to answer part A because it worked for part B of the question
I don't want to use the Venn Diagram

Q 5iii) A) and A: http://mei.org.uk/files/papers/s107ja_j38ctr.pdf

Reply 1

old_engineer

Original post by esmeralda123

Your method has broken down because you assumed that P(J’ n T’) = P(J’) x P(T’), but this is not the case as J’ and T’ are not independent.

If you want to use a tree diagram, you should use J and J’ as the first tier. You can then add T and T’ to the end of the J’ branch from the information given. Finally you will need to deduce what happens at the end of the J branch.

Reply 2

ryan02

I would work out part A by the following (which would be derived from the Venn diagram):

P(J u T) = P(J) + P(T) - P(J n T)

Have a go first on your own but here it is:

Spoiler

Reply 3

ryan02

Original post by ryan02

I would work out part A by the following (which would be derived from the Venn diagram):

P(J u T) = P(J) + P(T) - P(J n T)

Have a go first on your own but here it is:

Spoiler

Sorry! I read the exam paper and not your question, my apologies!

@old_engineer answered very well if you need a little more help deriving the end of the branches the below will help (though please try it yourself first):

Spoiler

Reply 4

debbie394

Original post by ryan02

Spoiler

But for P(J' and T) why do you have to do (0.2 -0.12). How would i know to do this from the tree diagram.
Why not 0.6 *0.8 because that is the same as no jacket and not tie, similar to the other probabilities

Reply 5

ryan02

Original post by esmeralda123

As @old_engineer said the two probabilities are not independent (they effect one another) and therefore you cannot multiply in this way. Therefore the only reason to use a tree diagram would be to work out the different combinations you can get and the derive their probabilities separately. It is not as straight forward as using the Venn diagram as the question suggests.

The problem with your approach is you have no information about the probability of wearing a tie when you are not wearing a jacket, the question simply doesn't provide you with it (it does provide you with info if you are wearing a jacket 0.3). Therefore I derive it:
The probability you wear a tie and not a jacket = the total probability you wear a tie - the probability you wear a tie and a jacket (info we have)

Reply 6

debbie394

Original post by ryan02

Ahh so you can't just put 0.2 after the probability of wearing no jacket because it is dependent. but why specifically subtract having a tie from (having a tie and jacket) how does that get you no jacket no tie. Sorry I just want to understand so that I can do these type of questions.
Why not p(tie)- p(tie with a jacket so 0.3)

(edited 5 years ago)

Reply 7

Prasiortle

Original post by esmeralda123

The post above was calculating P(tie and no jacket), hence why it's P(tie)-P(tie and jacket). In your case, for P(no tie and no jacket), you would need P(no tie)-P(no tie and jacket), which probably isn't any easier to work out. By far the best approach is to use a Venn diagram, which takes hardly any working; indeed, I'm not sure why in your original post you were so keen to do it using a tree diagram when that's just making life harder for you.