Probability of car starting first time The Question

One a given day, Claude's car has an 80% chance to starting first time
and Andre's car has a 70% chance of the same, Given that one of the
cars has started first time, what is the chance that Andre's car
started the first time?

My Understanding

Event A: Claude's car starting first time
Event B: Andre's car starting first time

We want at least one car started first time, also event B occurs, so there are only these $2$ cases: AB and A'B. I calculated the probability of these $2$ separately and multiply them together, which is $0.8\times 0.7\ {+}\ 0.2 \times 0.7=0.7$, which makes sense since event A and B are independent events, the occurrence of A should not affect the probability of B. I have to try something else because this is not the answer.
Then I tried to calculate the probability of B', which is when Andre's car is not starting first time.
there's only one case for this to be happening while at least one car is starting first time. A B'.
the probability for this is $0.8 \times 0.3 =0.24$. then I subtracted this from $1$ to get $76\%$.
the answer for this is $\frac{35}{47}$
It's close but not really. Can someone help me with this? I can't think of more ways of solving it.
Let me rename the events, because I keep thinking event A refers to Andre.
Let A be the event that Andre's car starts the first time and let C be the event that Claude's car starts the first time.  We are asked for $$\Pr(A|A\cup C)=\frac{\Pr(A\cap(A\cup C))}{\Pr(A\cup C)}=\frac{\Pr(A)}{\Pr(A\cup C)}$$
Now, $\Pr(A\cup C)=\Pr(A)+\Pr(C)-\Pr(A\cap C) = .7+.8-.7\cdot.8=.94$, so the required probability is $$\frac{.7}{.94}=\frac{35}{47}$$
I think simplest way is using a tree diagram
enter image description here
P(At least one of them starts)=0.8X0.7+0.2X0.7+0.3X0.8=0.94
P(Andre starts first)=0.8X0.7+0.2X0.7=0.7
P(Andre given started)=0.7/0.94=35/47
Hope this helps