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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?

Tamara Todorovic
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  • The link you provide does not appear to be a solution to this problem. As to the "official" solution of $.8$ well, that's lower than the unconditional probability that $B$ solves the problem and, knowing that the problem was in fact solved could hardly lower our estimate of the probability that $B$ solved it. – lulu Jun 29 '21 at 22:08
  • Should say, I don't see any problem with the solution you provided. I agree with that method and that solution. Trusting, of course, that the two events $A,B$ are independent (which was not stated). Without making some assumption as to dependence, the problem can not be solved. – lulu Jun 29 '21 at 22:10
  • Yes, I agree, logically it doesn't make sense that it's lower than the unconditional probability of B. As for the link provided, it's not a direct solution indeed, it's just the closest thing I could find, I thought it'd be relevant to this one. I was assuming independence had to do with the difference in solutions, but I cited the problem word for word, there was no mention of them being independent so I assumed independence. – Tamara Todorovic Jun 29 '21 at 22:14
  • To be clear though, independence is unrealistic. After all, the ability to solve an assignment is related to the difficulty of the assignment. If the assignment is easy, both will solve it. If the problem is very hard, $B$ has a better shot than $A$. But, absent any other information, I do think you have to assume independence. – lulu Jun 30 '21 at 00:10
  • True, in a real world situation independence is not likely. But looking at this as a math problem I think it makes sense to assume independence. – Tamara Todorovic Jun 30 '21 at 08:23

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