Tree diagram: probability of a car needing new brake pads

Question

the probability of a car need new brake pads is 0.3. If it does need brake pads the probability of it also needing new brake discs is 0.8. If not the probability of needing new brake discs is 0.3.

Find the probability that the car needs new brake pads given it needs new brake discs.


I have drawn the tree diagram but don't know how I answer the question

PCmaster said:

Question

the probability of a car need new brake pads is 0.3. If it does need brake pads the probability of it also needing new brake discs is 0.8. If not the probability of needing new brake discs is 0.3.

Find the probability that the car needs new brake pads given it needs new brake discs.


I have drawn the tree diagram but don't know how I answer the question

Click to expand...

I'm not that good at sadistics - whoops statistics - so in trying to understand what is happening I fall back on the "take 1000 cases" scenario: You have a 1000 trials and (on average) you will have 30% need pads so \(\displaystyle P\) = 300, of those 80% also need disks so \(\displaystyle PD\)=0.8*300=240 and thus those who don't need disks given they need pads is \(\displaystyle P\overline{D}\)=300-240=60, ... So what do we have given
\(\displaystyle P\) = 300
\(\displaystyle PD\) = 240
\(\displaystyle P\overline{D}\) = 60
\(\displaystyle \overline{P}\) = ?
\(\displaystyle \overline{P}D\) = ?
\(\displaystyle \overline{P}\overline{D}\) = ?

Needs disks is \(\displaystyle PD\)+\(\displaystyle \overline{P}D\)=240+?=?, needs disks and pads is \(\displaystyle PD\)=240. So needs pads given disks is \(\displaystyle DP\)=240/?

From that demonstration and others, hopefully you now see where the formulas for various compound probabilities come from.