To clarify, Find the probability that the first coin was a dime given the 2nd was also a dime.
I'm very sorry for the extended title, but it says to be very specific.
I think the tree diagram method is very tedious when I have so many items to work with and I've been using that to help me with these types of problems, albeit with lower numbers and easier trees to draw.
Here is the approach I'm doing:
P(D1) = Probability of dime on first draw P(D2) = Probability of dime on second draw P(N1) = Probability of nickel on first draw P(N2) = Probability of nickel on second draw
P(D1|D2) = P(D2|D1) * P(D1) / P(D2) -> Bayes Theorem
We know: P(D1) = 10 / 22
Now this is where I'm messing up, I think I know the equation:
P(D2) = P(D1)*P(D2|D1) + P(N1)*P(D2|N1)
P(D2|D1) = P(D2 n D1)/P(D1) = ? P(D2|N1) = P(D2 n N1)/P(N1) = ?
I'm stuck as to where to go next? I could multiple D2 n D1, but I have no such values for D2 yet!
Instead of getting tangled up in equations, you might think about it this way. Since the second coin drawn was a dime, one of two things happened: we drew a dime and then another dime, or we drew a nickel and then a dime.
What’s the probability of the first of these? As you said, the probability that the first coin is a dime is $\frac{10}{22}=\frac5{11}$. For the second draw there are now $11$ dimes and $12$ nickels, so the probability of drawing a dime is $\frac{11}{23}$. Thus, the probability that the first two draws are both dimes is $\frac5{11}\cdot\frac{11}{23}=\frac5{23}$.
For the second alternative we have a probability of $\frac{12}{22}=\frac6{11}$ of drawing a nickel initially. The jar then has $10$ dimes and $13$ nickels for the second draw, so the probability of getting a dime is $\frac{10}{23}$, and the overall probability of this sequence of draws is $\frac6{11}\cdot\frac{10}{23}=\frac{60}{253}$.
The total probability of getting a dime on the second draw is therefore
$$\frac5{23}+\frac{60}{253}=\frac{55+60}{253}=\frac{115}{253}\;.$$
What fraction of this probability comes from the alternative of interest, the one in which the first coin was also a dime? That’s the probability that we’re in that alternative, given that we’re in one of the two. That case contributed $\frac5{23}$ to the total of $\frac{115}{253}$, so it contributed
$$\frac{5/23}{115/253}=\frac5{23}\cdot\frac{253}{115}=\frac{11}{23}$$
of the total. In other words, given that the second coin was a dime, the probability that the first was a dime (i.e., that we were in the first case) is $\frac{11}{23}$.
After working through that, you should probably try to match up the various parts of the calculation with the pieces of Bayes’ formula.