I am having difficulty grasping conditional probability. This seemingly innocuous problem, or rather its solution, is confusing me:
The probability of student A solving an assignment is 0.7 and the probability of student B solving an assignment is 0.9. If an assignment is solved, what is the probability that student B solved it?
Here is how I tried solving the problem. First, the probability of an assignment being solved is:
$P(S)=P(A\bar{B})+P(\bar{A}B)+P(AB)=0.97
$
Using that, I would calculate $P(B|S)$ as:
$P(B|S)=\frac{P(B\cap S)}{P(S)}=\frac{P(\bar{A}B)+P(AB)}{P(S)}=\frac{0.9}{0.97}\approx0.93$
I've seen the problem solved on the website and used the same method to solve this problem but it's not adding up to the solution given in my workbook:
Probability of success given another success (conditional probability problem)
The solution in my workbook for this question simply states:
$P=\frac{1}{2}\cdot0.7+\frac{1}{2}\cdot0.9=0.8$
Can someone explain why my solution was wrong and why is the solution in my workbook correct?