You do agree that \(\mathcal{P}(Dd|Doc)\) is the question?View attachment 20820
I am stuck on part (c). I have made a start by writing down the formula for conditional probability but my answer is not correct. I think its 38/48 % 48/160. The answer is 38/60% 48/160. I am sure my answer is correct.
I think P (Db AND Doc) is 30/48 and P (Doc) is 48/160. That works out to be 25/12 obviously incorrect as it greater than one. Here is their answer.You do agree that \(\mathcal{P}(Dd|Doc)\) is the question?
Also \(\mathcal{P}(Dd|Doc)=\dfrac{\mathcal{P}(Dd\cap Doc)}{\mathcal{P}(Doc)}\).
Looking at the data distribution chart, it is easy to see the value of each if those.
Please tell us what you get.
That works. its 30/48. Correct.Since you are given that it is a documentary then your reduced sample space is just documentaries which has a total of 48. How many of those 48 are digital downloads? What is the desired probability.
I would not use any conditional probability formulas here. Just reduce the sample space to documentaries!
Can we see your work? It is hard to see your error if you fail to show us your work. Please post back.
Yes I got it now. A warm up ?. There is a typo in the answer book so I am allowed to feel confused. Thank you very much!From the chart \(\mathcal{P}(Dl\cap Doc)=\dfrac{30}{160}\) and \(\mathcal{P}(Doc)=\dfrac{48}{160}\)
Yes that is easier to see.... or you can say that of the 48 docs, that 30 are digital downloads giving us 30/48 or 5/8