Suppose $n$ is repeated. There are 9 other numbers that can occur. And the other digit can occur in 4 possible positions giving $36$ possibilities.
There are $10$ possibilities for $n$ so the total number of combinations with exactly three digits the same is $360$.
The total number of passwords should be the number of 4-6 digit passwords using the characters 2, 3, 5, and 7 minus the number of passwords using just 3, 5, and 7. For this you compute
$$(4^4 + 4^5 + 4^6) - (3^4 + 3^5 + 3^6) = 5376 - 1053 = 4323.$$
The probability that your friend guesses on the first attempt is $\frac{1}{4323}$, as you said.
The probability that your friend guesses your password on the second attempt is the probability that she guesses wrong on the first attempt and right on the second attempt. That's $\frac{4322}{4323}\cdot\frac{1}{4322} = \frac{1}{4323}$.
Similarly, let's say you look at the third attempt: you want the probability that your friend guesses wrong on the first and second tries, and then guesses your password on the third try. That happens with probability $\frac{4322}{4323}\cdot\frac{4321}{4322}\cdot\frac{1}{4321} = \frac{1}{4323}$.
The same pattern continues: assuming your friend guesses a different password every time, the probability that she guesses on the $n^\text{th}$ attempt is $\frac{1}{4323}$.
Now, as you suggested, you can add the probabilities corresponding to the first try + second try + ... + tenth try, since you want to know the odds that she guesses right on the first try or the second try or the third try, and so on.
This gives you a probability of
$$10 \cdot \frac{1}{4323} = \frac{10}{4323}.$$
Best Answer
The simple approach is that there are $10000$ possible PINs and you have tried $3$ of them, so your chance of finding the right one is $\frac 3{10000}$. In your calculation, the denominators should decrease $10000,9999,9998$, so you will get the same result.$$1-\frac {9999}{10000}\cdot \frac {9998}{9999}\cdot\frac{9997}{9998}=1-\frac {9997}{10000}=\frac 3{10000}$$