Hi, this is what i got for the practice final for these two problems and I just wanted to see if people got something similar. A5 = $\frac{(1/2)^3}{(1/2)^3+(1/2)(1/3)^2}$. I used a tree to solve this problem. as well as conditional probability as P(fair coin| get two heads) is P( it is the fair coin and 2 heads) / P( you get 2 heads).
For A6 I also used a tree format with branches for either having the disease or not, another branch for test A and another branch for test B. I also used conditional probability. In the end I got : $\frac{(1/10) (0.99)(0.8)+(9/10)(0.12)(0.02)}{(1/10)(0.99)+(9/10)(0.12)}$ Basically on the top, you go all the way to the branch for test B, but on the bottom you only go until branch for Test A.
Did anyone get anything similar/different? |