Can anybody please help me out? My daughters maths (probability) homework has beaten me and I have always prided myself on being good at maths! (I do fly 380s for a living and number crunching has always been a point of pride with me!! This has thus rocked my world, although I am aware that I 'suck' at probability!)

The question is this:- To get a driving licence, you have to pass a theory test and a practical. The probability that joe passes his theory test is 0.8

If he passes his theory test, he has a 0.6 chance of passing his practical. Otherwise the chance of passing his practical is only 0.2

Giving your answers to 2 dp, find firstly, the probability that he gets his driving licence. Secondly, given that he fails to get his driving licence, find the probability that he failed because of his practical only. (please note the english - and correct spelling - of licence!)

Now we have drawn a tree diagram as her book suggests and the furst answer is easy, being 0.48 Am sure that the second part, where the correct answer alludes me, hinges on the words 'given that he fails......'. However, cannot work the answer out for neither love nor money........HELP!!!!

Also, please ignore the incorrect english -'furst'!!

By jove, I think that I have got it!

This is great, I have a thread all to myself.....what are the odds of that?!

0.62 BTW......am very pleased with myself....but none of you would know as I am on this thread all by myself! Anyone care to join in with a mathematical dunderhead?!

Absolutely correct! Not many people can solve that, I think. Well done!

/Tomas

Thank you!

Oh....and welcome to the thread, just the two of us. You have to answer your own questions here BTW!

And this post takes me to 400....

Having drawn the tree, you have all the information you need.

Let:

T = passing theory test
P = passing practical test
L = getting licence [sic]

You're given:

P(T) = 0.8
P(P |T) = 0.6
P(P |~T) = 0.2

So you can immediately deduce:

P(~T) = 0.2 (= 1 - 0.8)
P(~P |T) = 0.4 (= 1 - 0.6)
P(~P |~T) = 0.8 (= 1 - 0.2)

From your tree, you have:

P(T & P) = 0.48 (= 0.8 × 0.6)
P(T & ~P) = 0.32 (= 0.8 × 0.4)
P(~T & P) = 0.04 (= 0.2 × 0.2)
P(~T & ~P) = 0.16 (= 0.2 × 0.8)

First:

P(L) = P(T & P) = 0.48

Second:

P(~L) = 0.52 (= 0.32 + 0.04 + 0.16)
P((T & ~P) |~L) = 0.32 / 0.52 = 0.615

(It's an application of Bayes' Theorem.)

Thanks, yes that's the answers that I finally got with my tree and a little deduction.....thanks anyway