Beginner Probability Question This is my second day of stats & probability and I'm a little confused.
Here's the question:

The Abigail Construction Company is determining whether it should
  submit a bid for the construction of a new shopping mall. In the past,
  its main competitor, the Jared Construction Company, has submitted
  bids 60% of the time. When Jared does not submit a bid, the
  probability the Abigail will win the job is 70%. However, when Jared
  does submit a bid, the probability that the Abigail will win the job
  is only 40%. If Abigail wins a job, what is the probability that Jared
  submitted a bid?

What I've solved for thus far (it's not much):
J not - .2
  does - .7
I think the .6 in the beginning is irrelevant, but I could be wrong. How do I move forward?
P.S. If you know of a good place online to learn probability, please don't hesitate to post it!
Thanks!
 A: We solve the problem informally, since it is early in the course. This will be done in two slightly different ways. Later in the course, one might write out a formal conditional probability argument. 
Imagine construction project opportunities occurring a large number of times, say $1000$. In about $600$ of these times, Jared submits a bid, and about $400$ times it doesn't. 
Ourt of the $400$ times that Jared does not submit a bid, Abigail wins about $70\%$ of the time, so $280$ times.
Out of the $600$ times Jared submits a bid, Abigail wins about $40\%$ of the time (we corrected a presumed typo). So this is $240$ times.
Thus Abigail wins about $520$ times. In $240$ of these, Jared had a bid. So our required probability should be $\frac{240}{520}$. 
Another way: (Well, it really is the same way.) Draw a tree diagram. The first branching is Jared does not submit (write a $0.6$), Jared submits (write a $0.4$). From each of the two nodes, draw two branches, one for Abigail wins, the other for she loses. Write the appropriate probabilities along these $4$ branches. 
Look at the two paths that lead to an Abigail win. Multiplying probabilities along the branches, we find that the probability Abigail wins is $(0.4)(0.7)+(0.6)(0.4)$. (There are two paths along which Abigail can win.). This probability turns out to be $0.52$.
Now calculate the probability of the path that leads to an Abigail win and a Jared bid. This path has probability $(0.6)(0.4)=0.24$.
It follows that our conditional probability is $\dfrac{0.24}{0.52}$. 
A: $A$ will represent Abigail submitting a vote, and $J$ will represent Jared submitting a vote. 
We are given the following:
$P(J) = 0.6$
$P(A|J^c) = 0.7$
$P(A|J) = 0.4$
We want to find $P(J|A)$.
This becomes a relatively straightforward problem when you remember that $P(A|B) = \frac{P(A\cap B)}{P(B)}$.
Using this formula, we can expand the second equation:
$$P(A|J^c) = \frac{P(J^c\cap A)}{P(J^c)} = \frac{P(J^c\cap A)}{0.4} = 0.7$$
$$P(J^c\cap A) = 0.28$$
Now the third equation:
$$P(A|J) = \frac{P(J\cap A)}{P(J)} = \frac{P(J\cap A)}{0.6} = 0.4$$
$$P(J\cap A) = 0.24$$
I double checked, and you can prove that $P(J\cap A) + P(J^c\cap A) = P(A)$, and so we can find $P(A)$ to be equal to $0.24 + 0.28 = 0.52$. We then can expand $P(J|A)$ to find your answer:
$$P(J|A)$$
$$= \frac{P(J \cap A)}{P(A)}$$
$$= \frac{0.24}{0.52}$$
Which is approximately $0.46153846153846153846153846153846$.
A: Let us consider how Abigail might get the job
Abigail can win if either of E1 or E2 occur
E1 = Jared submits and Abigail wins when Jared submits
E2 = Jared doesnot submit and Abigail wins when Jared Does not submit
P(JS) = Probability of Jared submitting = .6
P(~JS) = Probability of Jared not submitting = .4  
P(A|JS) = Probability of Abigail winning when Jared submits = .4
P(A|~JS) = Probability of Abigail winning when Jared does not submit = .7
P(E1) = P(A|JS) * P(JS) = .6 * .4 = .24
P(E2) = P(A|~JS) * P(~JS) = .7 * .4 = .28
P(A) = P(E1) + P(E2)
P(A) = .52
P(JS|A) = P(E1) / P(A)
P(JS|A) = .24 / .52
P(JS|A) = .4615
